6.3 Using the Central Limit Theorem
SLO 3
Describe and apply probability concepts and distributions.
The central limit theorem is the same idea applied to proportions now: a sample proportion is approximately normal with mean p and standard error root(pq/n), so you can find the probability that a survey result lands within a given range.
SLO 4
Demonstrate an understanding of, and ability to use, basic ideas of statistical processes, including hypothesis tests and confidence interval estimation.
Proportions are what most real polls and surveys report. This section gives you the sampling distribution for a sample proportion, which is the machinery behind every confidence interval and hypothesis test for a proportion later in the course.
Learning Objectives
By the end of this section, you will be able to:
- decide from the wording of a problem whether it is asking about a mean, a sum, or a single individual value, and pick the matching distribution;
- calculate probabilities and percentiles for a sample mean \(\overline{X}\) and for a sample sum \(\Sigma X\);
- explain what the law of large numbers says about a sample mean as the sample size grows;
- use the normal approximation to the binomial, including the continuity correction, and say when it is allowed.
The last two sections built two tools. Section 6.1 gave the central limit theorem for sample means, which says that \(\overline{X} \sim N\!\left(\mu, \frac{\sigma}{\sqrt{n}}\right)\) once the sample is large enough. Section 6.2 gave the version for sums, which says that \(\Sigma X \sim N\!\left(n\mu, \sqrt{n}\,\sigma\right)\). This section is about knowing which one a problem wants, and there are only three possibilities to sort between.
If the problem asks for the probability of a mean, use the central limit theorem for the mean. If it asks for the probability of a sum or a total, use the central limit theorem for sums. The same split applies to percentile questions. And if it asks about a single individual value, do not use the central limit theorem at all — go back to the distribution the individual values actually follow, whatever it happens to be.
Scan the sentence for "average", "mean", "total", "sum", or "one randomly selected". Those five words decide the whole problem. Everything after that is typing.
That third case is where most of the lost points in this section live, so it is worth stating on its own. A sample of 80 phone customers has an average excess time; that average is a mean, and the central limit theorem applies to it. One customer picked out of that sample has an excess time of their own; that is a single value from the original population, and the central limit theorem has nothing to say about it. The two numbers are not close to each other and they are not supposed to be. Averaging pulls values toward the centre, which is exactly why a mean is far less likely to land out in a tail than any individual value is. You will see both computed side by side in this section, on the same distribution, and they come out very differently — that difference is the point of the exercise, not a mistake in the arithmetic.
6.3.1 Law of Large Numbers
If you take samples of larger and larger size \(n\) from any population, then the sample mean \(\overline{x}\) tends to get closer and closer to the population mean \(\mu\). We can say that \(\mu\) is the value the sample means approach as \(n\) grows.
The \(\sqrt{n}\) is doing the work, and square roots grow lazily. To cut the spread of \(\overline{X}\) in half you need four times the data; to cut it to a tenth you need a hundred times the data.
The central limit theorem is what makes this work. It tells us that the sample means follow a normal distribution centred at \(\mu\) with standard deviation \(\frac{\sigma}{\sqrt{n}}\). That standard deviation is the whole story. The \(n\) sits under a square root in the denominator, so as \(n\) gets larger the standard deviation gets smaller, and the normal curve of sample means squeezes in tighter around \(\mu\). A tighter curve means a randomly drawn sample mean has less room to land far from the centre. So the sample mean \(\overline{x}\) must be close to the population mean \(\mu\), and it gets closer as you collect more data.
That square root is worth taking seriously before you plan a study. Going from 25 observations to 100 observations is four times the cost and only halves the spread of the sample mean. Going from 100 to 400 halves it again, at four times the cost again. This is why survey sample sizes tend to plateau in the low thousands rather than climbing forever — past a point, each additional respondent buys a vanishingly small improvement in precision, and the money is better spent on getting a less biased sample than a larger one. The law of large numbers promises that the sample mean converges to the population mean, but it makes no promise at all about speed, and it makes no promise whatsoever about a sample that was collected badly. A biased sampling method converges just as reliably — to the wrong number.
Before the worked examples, one consequence of the central limit theorem deserves its own name, because it explains why bigger samples are worth the trouble of collecting.
Definition 6.3.1 - The law of large numbers: bigger samples give sample means that cluster more tightly around the population mean.
A population has mean \(\mu = 50\) and standard deviation \(\sigma = 12\). Find the standard deviation of the sample mean \(\overline{X}\) for samples of size \(n = 9\), \(n = 36\), and \(n = 144\). What do the three answers show?
Solution
Step 1 — Use the central limit theorem formula. The standard deviation of \(\overline{X}\) is \(\frac{\sigma}{\sqrt{n}}\).
$$ n = 9: \quad \frac{12}{\sqrt{9}} = \frac{12}{3} = 4 $$ $$ n = 36: \quad \frac{12}{\sqrt{36}} = \frac{12}{6} = 2 $$ $$ n = 144: \quad \frac{12}{\sqrt{144}} = \frac{12}{12} = 1 $$Step 2 — Read the pattern. Each time the sample size is multiplied by 4, the standard deviation of the sample mean is cut in half. The curve of sample means keeps tightening around \(\mu = 50\), which is the law of large numbers in action.
Answer: 4, 2, and 1. Larger samples give sample means that cluster more tightly around the population mean, but four times the data is needed for each halving.
6.3.2 Examples of the Central Limit Theorem for the Mean and the Sum
The examples below all follow the same three moves. Read the question and decide whether it wants a mean, a sum, or an individual. Write down the correct distribution with its mean and standard deviation. Then run normalcdf for a probability or invNorm for a percentile. The setup is where the thinking happens.
Stress scores among students on a college campus follow a uniform distribution with the lowest stress score equal to one and the highest equal to five. This time, use a sample of 55 students.
a. Find \(P(\overline{x} < 7)\).
b. Find \(P(\Sigma x > 170)\).
c. Find the \(80^{th}\) percentile for the mean of 55 scores.
d. Find the \(85^{th}\) percentile for the sum of 55 scores.
Solution
Step 1 — Rebuild both distributions with \(n = 55\). We still have \(\mu = 3\) and \(\sigma = \sqrt{\frac{4}{3}} \approx 1.1547\).
$$ \overline{X} \sim N\!\left(3, \frac{1.1547}{\sqrt{55}}\right) = N(3, 0.1557) $$ $$ \Sigma X \sim N\!\left(55 \times 3, \sqrt{55} \times 1.1547\right) = N(165, 8.5635) $$Part a. The largest possible stress score is 5, so a mean of 7 is impossible to exceed. Every sample mean is below 7.
$$ P(\overline{x} < 7) = 1 $$Part b.
$$ P(\Sigma x > 170) = \text{normalcdf}(170, 10^{99}, 165, 8.5635) = 0.2796 $$Part c.
$$ k = \text{invNorm}(0.80, 3, 0.1557) = 3.13 $$Part d.
$$ k = \text{invNorm}(0.85, 165, 8.5635) = 173.88 $$Answer: a. 1 b. 0.2796 c. 3.13 d. 173.88
A study involving stress is conducted among the students on a college campus. The stress scores follow a uniform distribution with the lowest stress score equal to one and the highest equal to five. Using a sample of 75 students, find:
a. The probability that the mean stress score for the 75 students is less than two.
b. The \(90^{th}\) percentile for the mean stress score for the 75 students.
c. The probability that the total of the 75 stress scores is less than 200.
d. The \(90^{th}\) percentile for the total stress score for the 75 students.
Solution
Step 1 — Name the variable and sort the parts. Let \(X\) = one stress score. Parts a and b ask for a probability or a percentile for a mean. Parts c and d ask for a probability or a percentile for a total, or sum. The sample size \(n\) is 75.
Step 2 — Get the population mean and standard deviation. Since the individual stress scores follow a uniform distribution, \(X \sim U(1, 5)\) where \(a = 1\) and \(b = 5\).
$$ \mu_X = \frac{a + b}{2} = \frac{1 + 5}{2} = 3 $$ $$ \sigma_X = \sqrt{\frac{(b - a)^2}{12}} = \sqrt{\frac{(5 - 1)^2}{12}} = \sqrt{\frac{16}{12}} = \sqrt{\frac{4}{3}}, \text{ or approximately } 1.15 $$Step 3 — Build the distribution of the mean. For parts a and b, let \(\overline{X}\) = the mean stress score for the 75 students. Then
$$ \overline{X} \sim N\!\left(3, \frac{1.15}{\sqrt{75}}\right) $$Part a — Find \(P(\overline{x} < 2)\).
$$ P(\overline{x} < 2) = \text{normalcdf}\!\left(1, 2, 3, \frac{1.15}{\sqrt{75}}\right) = 0 $$The probability that the mean stress score is less than two is about zero. The lower bound is 1 because the smallest possible stress score is one.
Figure 6.3.1 - The distribution of the mean stress score for 75 students, centred at 3, with essentially no area to the left of 2.
Part b — Find the \(90^{th}\) percentile for the mean of 75 stress scores. Let \(k\) = the \(90^{th}\) percentile, so \(P(\overline{x} < k) = 0.90\).
$$ k = \text{invNorm}\!\left(0.90, 3, \frac{1.15}{\sqrt{75}}\right) = 3.17 $$
Figure 6.3.2 - The shaded area to the left of k is the probability 0.90, so k is the 90th percentile of the mean.
The \(90^{th}\) percentile for the mean of 75 scores is about 3.17. This tells us that 90% of all the means of 75 stress scores are at most 3.17, and that 10% are at least 3.17.
Step 4 — Build the distribution of the sum. For parts c and d, let \(\Sigma X\) = the sum of the 75 stress scores. Then \(\Sigma X \sim N\!\left[(75)(3), (\sqrt{75})(1.15)\right]\).
The mean of the sum of 75 stress scores is \((75)(3) = 225\). The standard deviation of the sum is
$$ \left(\sqrt{75}\right)\left(\sqrt{\tfrac{4}{3}}\right) = \sqrt{\frac{75 \times 4}{3}} = \sqrt{100} = 10 $$Part c — Find \(P(\Sigma x < 200)\).
$$ P(\Sigma x < 200) = \text{normalcdf}(75, 200, 225, 10) = 0.0062 $$The probability that the total of 75 scores is less than 200 is about 0.0062. The lower bound is 75, because the smallest total of 75 stress scores is 75 when every single score is one.
Figure 6.3.3 - The distribution of the sum of 75 stress scores, centred at 225, with a thin shaded tail left of 200.
Part d — Find the \(90^{th}\) percentile for the total of 75 stress scores. Let \(k\) = the \(90^{th}\) percentile, so \(P(\Sigma x < k) = 0.90\).
$$ k = \text{invNorm}\!\left(0.90, (75)(3), (\sqrt{75})(1.15)\right) = 237.8 $$
Figure 6.3.4 - The shaded area to the left of k is 0.90, so k is the 90th percentile of the sum.
The \(90^{th}\) percentile for the sum of 75 scores is about 237.8. This tells us that 90% of all the sums of 75 scores are no more than 237.8 and 10% are no less than 237.8.
Answer: a. 0 b. 3.17 c. 0.0062 d. 237.8
For cell phone customers who exceed the time allowance included in their basic contract, the excess time used follows an exponential distribution with a mean of 22 minutes. This time, use a sample of 144 such customers.
a. Find \(P(20 < \overline{x} < 30)\).
b. Find \(P(\Sigma x \text{ is at least } 3{,}000)\).
c. Find the \(75^{th}\) percentile for the sample mean excess time of 144 customers.
d. Find the \(85^{th}\) percentile for the sum of 144 excess times used by customers.
Solution
Step 1 — Rebuild both distributions with \(n = 144\). We have \(\mu = 22\) and \(\sigma = 22\).
$$ \overline{X} \sim N\!\left(22, \frac{22}{\sqrt{144}}\right) = N(22, 1.8333) $$ $$ \Sigma X \sim N\!\left(144 \times 22, \sqrt{144} \times 22\right) = N(3{,}168, 264) $$Part a — A between-question, so feed both edges to normalcdf.
Part b — "At least 3,000" is a right tail on the sum.
$$ P(\Sigma x \geq 3{,}000) = \text{normalcdf}(3{,}000, 10^{99}, 3{,}168, 264) = 0.7377 $$Part c.
$$ k = \text{invNorm}(0.75, 22, 1.8333) = 23.24 $$Part d.
$$ k = \text{invNorm}(0.85, 3{,}168, 264) = 3{,}441.6 $$Answer: a. 0.8623 b. 0.7377 c. 23.24 minutes d. 3,441.6 minutes
Mai Vang, a market research analyst for a cell phone company, conducts a study of the customers who exceed the time allowance included on their basic cell phone contract. She finds that for those people who exceed the time included in their basic contract, the excess time used follows an exponential distribution with a mean of 22 minutes.
Mai takes a random sample of 80 customers who exceed the time allowance included in their basic cell phone contract. Let \(X\) = the excess time used by one individual cell phone customer who exceeds their contracted time allowance, so \(X \sim \text{Exp}\!\left(\frac{1}{22}\right)\), and from previous chapters we know \(\mu = 22\) and \(\sigma = 22\).
a. Find the probability that the mean excess time used by the 80 customers in the sample is longer than 20 minutes.
b. Suppose Mai randomly selects one customer who exceeds the time limit. Find the probability that this individual customer's excess time is longer than 20 minutes.
c. Explain why the probabilities in parts a and b are different.
d. Find the \(95^{th}\) percentile for the sample mean excess time for samples of 80 customers.
Solution
Step 1 — Build the distribution of the mean. Let \(\overline{X}\) = the mean excess time used by a sample of \(n = 80\) customers. By the central limit theorem for sample means,
$$ \overline{X} \sim N\!\left(22, \frac{22}{\sqrt{80}}\right) $$Part a — This asks for a mean, so use the central limit theorem. We want \(P(\overline{x} > 20)\).
$$ P(\overline{x} > 20) = \text{normalcdf}\!\left(20, 10^{99}, 22, \frac{22}{\sqrt{80}}\right) = 0.7919 $$The probability is 0.7919 that the mean excess time used is more than 20 minutes, for a sample of 80 customers who exceed their contracted time allowance.
Figure 6.3.5 - The shaded area to the right of 20 is the probability that the mean excess time exceeds 20 minutes.
Part b — This asks about one individual, so do NOT use the central limit theorem. Go back to the exponential distribution the individual values follow, \(X \sim \text{Exp}\!\left(\frac{1}{22}\right)\).
$$ P(x > 20) = e^{-\left(\frac{1}{22}\right)(20)} = e^{-0.04545(20)} = 0.4029 $$Part c — Compare them. We have \(P(x > 20) = 0.4029\) but \(P(\overline{x} > 20) = 0.7919\). The probabilities are not equal because we use different distributions to calculate the probability for individuals and for means. When asked to find the probability of an individual value, use the stated distribution of its random variable; do not use the central limit theorem. Use the central limit theorem with the normal distribution when you are being asked to find the probability for a mean.
Part d — A percentile for the mean, so run invNorm. Let \(k\) = the \(95^{th}\) percentile, so \(P(\overline{x} < k) = 0.95\).
Figure 6.3.6 - The shaded area to the left of k is 0.95, so k is the 95th percentile of the sample mean.
The \(95^{th}\) percentile for the sample mean excess time used is about 26.0 minutes for random samples of 80 customers who exceed their contractual allowed time. Ninety-five percent of such samples would have means under 26 minutes; only five percent of such samples would have means above 26 minutes.
Answer: a. 0.7919 b. 0.4029 c. Different distributions — the central limit theorem governs means, not individuals d. 26.0 minutes
Try it in rāSHio
Every probability in this section is one normalcdf, and rāSHio will do it without a calculator. Choose Distributions → Normal, then set mean 22 and standard deviation 2.4597 — that second number is \(22/\sqrt{80}\), the standard error, and typing the standard error instead of the population standard deviation is the whole of what the central limit theorem asks you to do. Enter the cutoff 20 and the panel shades the region and reports its area. Part a wants the right-hand side, so read the shaded area's complement, 0.7919. The only thing that ever changes between a mean question and a sum question is which standard deviation you type.
Figure 6.3.7 — The rāSHio Normal distribution panel: Distributions → Normal. The walkthrough sets a mean and a standard deviation, enters one cutoff, and reads the shaded area under the curve. It uses its own demonstration numbers, not this example's.
Based on data from the National Health Survey, females between the ages of 18 and 24 have an average systolic blood pressure (in mm Hg) of 114.8 with a standard deviation of 13.1. Systolic blood pressure for females between the ages of 18 to 24 follows a normal distribution.
a. If one female from this population is randomly selected, find the probability that their systolic blood pressure is greater than 120.
b. If 40 females from this population are randomly selected, find the probability that their mean systolic blood pressure is greater than 120.
c. If the sample were four females between the ages of 18 to 24 and we did not know the original distribution, could the central limit theorem be used?
Solution
Part a — One individual, so use the population distribution directly. The population is normal with \(\mu = 114.8\) and \(\sigma = 13.1\).
$$ P(x > 120) = \text{normalcdf}(120, 10^{99}, 114.8, 13.1) = 0.3457 $$Part b — A mean, so use the central limit theorem. With \(n = 40\), the standard deviation of \(\overline{X}\) is \(\frac{13.1}{\sqrt{40}} = 2.0713\).
$$ P(\overline{x} > 120) = \text{normalcdf}(120, 10^{99}, 114.8, 2.0713) = 0.0060 $$Averaging 40 women pulls the result toward 114.8, so a mean above 120 is far rarer than an individual above 120.
Part c. No. With \(n = 4\) and an unknown population shape, the sample size is too small for the central limit theorem to guarantee that \(\overline{X}\) is approximately normal. The theorem needs a large sample (a common rule of thumb is \(n \geq 30\)) unless the original population is already known to be normal.
Answer: a. 0.3457 b. 0.0060 c. No — \(n = 4\) is too small when the population distribution is unknown
In the United States, a robbery occurs every two minutes, on average, according to a number of studies. Suppose the standard deviation is 0.5 minutes and the sample size is 100.
a. Find the median, the first quartile, and the third quartile for the sample mean time of robberies in the United States.
b. Find the median, the first quartile, and the third quartile for the sum of sample times of robberies in the United States.
c. Find the probability that a robbery occurs on the average between 1.75 and 1.85 minutes.
d. Find the value that is two standard deviations above the sample mean.
e. Find the IQR for the sum of the sample times.
Solution
Part a — Quartiles of the mean. We have \(\mu_{\overline{X}} = \mu = 2\) and \(\sigma_{\overline{X}} = \frac{\sigma}{\sqrt{n}} = \frac{0.5}{10} = 0.05\). A median is the \(50^{th}\) percentile, a first quartile is the \(25^{th}\), and a third quartile is the \(75^{th}\), so all three are invNorm questions.
- \(50^{th}\) percentile \(= \mu_{\overline{X}} = \mu = 2\)
- \(25^{th}\) percentile \(= \text{invNorm}(0.25, 2, 0.05) = 1.97\)
- \(75^{th}\) percentile \(= \text{invNorm}(0.75, 2, 0.05) = 2.03\)
Part b — Quartiles of the sum. We have \(\mu_{\Sigma X} = n(\mu_X) = 100(2) = 200\) and \(\sigma_{\Sigma X} = \sqrt{n}\,(\sigma_X) = 10(0.5) = 5\).
- \(50^{th}\) percentile \(= \mu_{\Sigma X} = 100(2) = 200\)
- \(25^{th}\) percentile \(= \text{invNorm}(0.25, 200, 5) = 196.63\)
- \(75^{th}\) percentile \(= \text{invNorm}(0.75, 200, 5) = 203.37\)
Part c — A between-question on the mean.
$$ P(1.75 < \overline{x} < 1.85) = \text{normalcdf}(1.75, 1.85, 2, 0.05) = 0.0013 $$Part d — Two standard deviations above the mean. Using the \(z\)-score equation \(z = \frac{\overline{x} - \mu_{\overline{x}}}{\sigma_{\overline{x}}}\) and solving for the value,
$$ \overline{x} = 2(0.05) + 2 = 2.1 $$Part e — IQR of the sum. The IQR is the \(75^{th}\) percentile minus the \(25^{th}\) percentile, and both came out of part b.
$$ \text{IQR} = 203.37 - 196.63 = 6.74 $$Answer: a. 2, 1.97, 2.03 b. 200, 196.63, 203.37 c. 0.0013 d. 2.1 e. 6.74
Amara Boateng is a design engineer on a 757 airliner programme. According to Boeing data, the 757 carries 200 passengers and has doors with a height of 72 inches. Assume that for a certain population of men we have a mean height of 69.0 inches and a standard deviation of 2.8 inches.
a. What doorway height would allow 95% of men to enter the aircraft without bending?
b. Assume that half of the 200 passengers are men. What mean doorway height satisfies the condition that there is a 0.95 probability that this height is greater than the mean height of 100 men?
c. For Amara and the other engineers designing the 757, which result is more relevant: the height from part a or part b? Why?
Solution
Part a — This is about individual men, so use the population distribution.
$$ k = \text{invNorm}(0.95, 69.0, 2.8) = 73.61 \text{ inches} $$Part b — This is about a mean of 100 men, so use the central limit theorem. The standard deviation of \(\overline{X}\) is \(\frac{2.8}{\sqrt{100}} = 0.28\).
$$ k = \text{invNorm}(0.95, 69.0, 0.28) = 69.46 \text{ inches} $$Part c. Part a. Doorways are walked through by individual passengers, not by averages, so Amara has to clear the tall end of the individual height distribution in her design. The 69.46-inch figure from part b describes how much the average of 100 men varies, which no passenger ever experiences. Building a 69.46-inch door would leave a large share of men ducking.
Answer: a. 73.61 inches b. 69.46 inches c. Part a, because doors are used by individuals rather than by sample means
A study was done regarding attendance at Broadway shows in New York City. The age range of the attendees was 14 to 61. The mean age was 30.9 years with a standard deviation of nine years.
a. In a sample of 25 attendees, what is the probability that the mean age is less than 35?
b. Is it likely that the mean age of the sample group could be more than 50 years? Interpret the results.
c. In a sample of 49 attendees, what is the probability that the sum of the ages is no less than 1,600?
d. Is it likely that the sum of the ages of the 49 attendees is at most 1,595? Interpret the results.
e. Find the \(95^{th}\) percentile for the sample mean age of 65 attendees. Interpret the results.
f. Find the \(90^{th}\) percentile for the sum of the ages of 65 attendees. Interpret the results.
Solution
Step 1 — Set up each sample size as it comes. With \(\mu = 30.9\) and \(\sigma = 9\), the standard deviation of \(\overline{X}\) is \(\frac{9}{\sqrt{n}}\) and the sum has mean \(30.9n\) with standard deviation \(9\sqrt{n}\).
Part a — \(n = 25\), so \(\sigma_{\overline{X}} = \frac{9}{5} = 1.8\).
$$ P(\overline{x} < 35) = \text{normalcdf}(-10^{99}, 35, 30.9, 1.8) = 0.9886 $$Part b.
$$ P(\overline{x} > 50) = \text{normalcdf}(50, 10^{99}, 30.9, 1.8) \approx 0 $$For this sample group, it is almost impossible for the group's average age to be more than 50. However, it is still possible for an individual in this group to have an age greater than 50 — the age range runs to 61.
Part c — \(n = 49\), so the sum has mean \(49(30.9) = 1{,}514.10\) and standard deviation \(9\sqrt{49} = 63\).
$$ P(\Sigma x \geq 1{,}600) = \text{normalcdf}(1{,}600, 10^{99}, 1{,}514.10, 63) = 0.0864 $$Part d.
$$ P(\Sigma x \leq 1{,}595) = \text{normalcdf}(-10^{99}, 1{,}595, 1{,}514.10, 63) = 0.9005 $$This means there is about a 90% chance that the sum of the ages for the sample group of \(n = 49\) is at most 1,595. So yes, it is likely.
Part e — \(n = 65\), so \(\sigma_{\overline{X}} = \frac{9}{\sqrt{65}} = 1.1\).
$$ k = \text{invNorm}(0.95, 30.9, 1.1) = 32.7 $$This indicates that in 95% of samples of 65 attendees, the average age is younger than 32.7 years.
Part f — The sum for \(n = 65\) has mean \(65(30.9) = 2{,}008.5\) and standard deviation \(9\sqrt{65} = 72.56\).
$$ k = \text{invNorm}(0.90, 2{,}008.5, 72.56) = 2{,}101.5 $$This indicates that in 90% of samples of 65 attendees, the sum of the ages is less than 2,101.5 years.
Answer: a. 0.9886 b. Almost 0 — very unlikely c. 0.0864 d. 0.9005 — likely e. 32.7 years f. 2,101.5 years
6.3.3 Normal Approximation to the Binomial
To get the best normal approximation to a binomial probability, add 0.5 to \(x\) or subtract 0.5 from \(x\) — that is, use \(x + 0.5\) or \(x - 0.5\). The number 0.5 is called the continuity correction factor.
Which direction you move depends on whether the endpoint is included in the event. Ask yourself whether the whole number named in the problem belongs to the outcomes you are counting. If it does — "at least 150", "at most 160", "exactly 175" — the bar over that number has to be inside the shaded region, so the boundary moves outward by 0.5 to swallow it. If it does not — "more than 155", "fewer than 147" — the bar over that number sits outside the region, so the boundary moves inward by 0.5 to exclude it. Written out as a rule: \(P(X \geq a)\) becomes \(P(Y \geq a - 0.5)\), \(P(X \leq a)\) becomes \(P(Y \leq a + 0.5)\), \(P(X > a)\) becomes \(P(Y > a + 0.5)\), and \(P(X < a)\) becomes \(P(Y < a - 0.5)\). An exact value \(P(X = a)\) becomes the strip \(P(a - 0.5 < Y < a + 0.5)\), which is the only reason a continuous distribution can answer an "exactly" question at all. Getting the direction backwards shifts the answer by roughly one bar's worth of probability, which is small but consistently wrong, so it is worth reading the inequality out loud before typing.
Historically, being able to compute binomial probabilities was one of the most important applications of the central limit theorem. Binomial probabilities with a small value for \(n\) (say, 20) were displayed in a table in a book. To calculate the probabilities with large values of \(n\), you had to use the binomial formula, which could be very complicated. Using the normal approximation to the binomial distribution simplified the process.
To compute the normal approximation to the binomial distribution, take a simple random sample from a population. You must meet the conditions for a binomial distribution:
- there are a certain number \(n\) of independent trials;
- the outcomes of any trial are success or failure;
- each trial has the same probability of a success \(p\).
Recall that if \(X\) is the binomial random variable, then \(X \sim B(n, p)\). The shape of the binomial distribution needs to be similar to the shape of the normal distribution. To ensure this, the quantities \(np\) and \(nq\) must both be greater than five (\(np > 5\) and \(nq > 5\); the approximation is better if they are both greater than or equal to 10). Then the binomial can be approximated by the normal distribution with mean \(\mu = np\) and standard deviation \(\sigma = \sqrt{npq}\), where \(q = 1 - p\).
A binomial distribution is a row of separate bars, one per whole number. A normal curve is continuous. The correction widens each bar by half a unit on both sides so the curve covers the same ground the bars did.
It is worth being clear about which part of this the correction actually fixes. The approximation is not doing anything clever about which bars belong in your event — you decide that by reading the inequality in the problem. What the half-unit repairs is the mismatch between a bar that occupies real width on the number line and a curve that only has area. The bar sitting over the whole number 250 stretches from 249.5 to 250.5, so if that outcome belongs in your event, the shaded region under the curve has to begin at 249.5 rather than at 250. Begin it at 250 and you have quietly discarded half of that bar. The error this introduces is never dramatic, which is exactly what makes it worth guarding against: an answer computed without the correction still looks reasonable, still lands in the right decimal place, and still survives a rough sanity check. It is simply wrong by about half a bar's worth of probability every time, and always in the same predictable direction. So the working order is fixed — confirm that \(np\) and \(nq\) both clear five, compute \(\mu = np\) and \(\sigma = \sqrt{npq}\), and only then move the boundary half a unit in whichever direction the inequality demands.
Definition 6.3.2 - The continuity correction: a binomial bar has real width, so the boundary slides half a unit to cover it.
In a city, 46 percent of the population favor the incumbent, Rosa Delgado, for mayor. A simple random sample of 500 is taken. Using the continuity correction factor, find the probability that at least 250 favor Rosa Delgado for mayor.
Solution
Step 1 — Set up the binomial and check the conditions. Let \(X\) = the number who favor the incumbent, so \(X \sim B(500, 0.46)\). Here \(np = 230 > 5\) and \(nq = 270 > 5\), so the normal approximation is allowed.
$$ \mu = np = 500(0.46) = 230 $$ $$ \sigma = \sqrt{npq} = \sqrt{500(0.46)(0.54)} = \sqrt{124.2} = 11.1445 $$Step 2 — Apply the continuity correction. "At least 250" includes 250, so the boundary moves outward: \(P(X \geq 250)\) becomes \(P(Y \geq 249.5)\).
$$ \text{normalcdf}(249.5, 10^{99}, 230, 11.1445) = 0.0401 $$Answer: 0.0401. Only about a 4% chance that at least half the sample of 500 favors the incumbent.
Suppose in a local Kindergarten through \(12^{th}\) grade (K–12) school district, 53 percent of the population favor a charter school for grades K through 5. A simple random sample of 300 is surveyed.
a. Find the probability that at least 150 favor a charter school.
b. Find the probability that at most 160 favor a charter school.
c. Find the probability that more than 155 favor a charter school.
d. Find the probability that fewer than 147 favor a charter school.
e. Find the probability that exactly 175 favor a charter school.
Solution
Step 1 — Set up the binomial and check the conditions. Let \(X\) = the number that favor a charter school for grades K through 5. Then \(X \sim B(n, p)\) where \(n = 300\) and \(p = 0.53\). Since \(np = 159 > 5\) and \(nq = 141 > 5\), we may use the normal approximation to the binomial.
$$ \mu = np = 300(0.53) = 159 $$ $$ \sigma = \sqrt{npq} = \sqrt{300(0.53)(0.47)} = 8.6447 $$The random variable for the normal distribution is \(Y\), so \(Y \sim N(159, 8.6447)\).
Part a — 150 is included, so widen outward. \(P(X \geq 150)\) has normal approximation \(P(Y \geq 149.5)\).
$$ \text{normalcdf}(149.5, 10^{99}, 159, 8.6447) = 0.8641 $$Part b — 160 is included, so widen outward. \(P(X \leq 160)\) has normal approximation \(P(Y \leq 160.5)\).
$$ \text{normalcdf}(0, 160.5, 159, 8.6447) = 0.5689 $$Part c — 155 is excluded, so pull inward. \(P(X > 155)\) has normal approximation \(P(Y > 155.5)\).
$$ \text{normalcdf}(155.5, 10^{99}, 159, 8.6447) = 0.6572 $$Part d — 147 is excluded, so pull inward. \(P(X < 147)\) has normal approximation \(P(Y < 146.5)\).
$$ \text{normalcdf}(0, 146.5, 159, 8.6447) = 0.0741 $$Part e — An exact value becomes a strip half a unit wide on each side. \(P(X = 175)\) has normal approximation \(P(174.5 < Y < 175.5)\).
$$ \text{normalcdf}(174.5, 175.5, 159, 8.6447) = 0.0083 $$Step 2 — Compare against the exact binomial. Because of calculators and computer software that let you calculate binomial probabilities for large values of \(n\) easily, it is not necessary to use the normal approximation when you have access to these tools. Using \(n = 300\) and \(p = 0.53\) directly:
$$ P(X \geq 150): 1 - \text{binomialcdf}(300, 0.53, 149) = 0.8641 $$ $$ P(X \leq 160): \text{binomialcdf}(300, 0.53, 160) = 0.5684 $$ $$ P(X > 155): 1 - \text{binomialcdf}(300, 0.53, 155) = 0.6576 $$ $$ P(X < 147): \text{binomialcdf}(300, 0.53, 146) = 0.0742 $$ $$ P(X = 175): \text{binomialpdf}(300, 0.53, 175) = 0.0083 $$The two columns agree to three decimal places, which is what a well-conditioned approximation looks like.
Answer: a. 0.8641 b. 0.5689 c. 0.6572 d. 0.0741 e. 0.0083
Try it in rāSHio
This example is worth running both ways, because the whole point is that the two columns agree. rāSHio has the exact binomial built in: choose Calculators → Binomial, set n to 300 and p to 0.53, and ask for \(P(X \geq 150)\) to get 0.8641 without any continuity correction at all. Then go back to Distributions → Normal with mean 159 and standard deviation 8.6447 and shade from 149.5 — the approximation the correction buys you. Seeing them land on the same number is more convincing than being told they do.
Figure 6.3.8 — The rāSHio Binomial calculator: Calculators → Binomial. It computes the exact binomial probability, so you can check any normal approximation against the number it is approximating. The walkthrough uses its own demonstration numbers, not this example's.
Problem Set 6.3
Use the following information to answer the next ten exercises: A manufacturer produces 25-pound lifting weights. The lowest actual weight is 24 pounds, and the highest is 26 pounds. Each weight is equally likely so the distribution of weights is uniform. A sample of 100 weights is taken.
Problem 1. a) What is the distribution for the weights of one 25-pound lifting weight? What is the mean and standard deviation?
b) What is the distribution for the mean weight of 100 25-pound lifting weights?
c) Find the probability that the mean actual weight for the 100 weights is less than 24.9.
Problem 1 Solution
Step 1 — Identify the individual distribution: the weights are equally likely between 24 and 26 pounds, which is the definition of a uniform distribution.
$$ X \sim U(24, 26), \quad \mu = \frac{24 + 26}{2} = 25, \quad \sigma = \sqrt{\frac{(26 - 24)^2}{12}} = \sqrt{\frac{4}{12}} = 0.5774 $$Step 2 — Apply the central limit theorem to get the distribution of the mean: with \(n = 100\), the sample mean is approximately normal with the same centre and a standard deviation shrunk by \(\sqrt{n}\).
$$ \overline{X} \sim N\!\left(25, \frac{0.5774}{\sqrt{100}}\right) = N(25, 0.0577) $$Step 3 — Find the probability: this asks for an area to the left of 24.9 on the distribution of the mean.
$$ P(\overline{x} < 24.9) = \text{normalcdf}(-10^{99}, 24.9, 25, 0.0577) = 0.0416 $$Answer: a) \(X \sim U(24, 26)\) with \(\mu = 25\) and \(\sigma = 0.5774\) b) \(\overline{X} \sim N(25, 0.0577)\) c) 0.0416
Problem 2. Draw the graph from Problem 6.3.1.
Problem 2 Solution
Step 1 — Draw the axis and centre the curve: sketch a normal curve for \(\overline{X}\) centred at \(\mu_{\overline{X}} = 25\) pounds, and label the horizontal axis \(\overline{x}\).
Step 2 — Scale the axis by the standard deviation: the standard deviation of the sample mean is 0.0577, so mark tick marks roughly every 0.06 pounds. Almost the whole curve lives between 24.83 and 25.17.
Step 3 — Shade the region the probability describes: draw a vertical line at \(\overline{x} = 24.9\) and shade everything to its LEFT. That value is about 1.73 standard deviations below the centre, so the shaded sliver is thin — its area is the 0.0416 from Problem 6.3.1.
Answer: A normal curve centred at 25 with the region left of a vertical line at 24.9 shaded; the shaded area equals 0.0416.
Problem 3. Find the probability that the mean actual weight for the 100 weights is greater than 25.2.
Problem 3 Solution
Step 1 — Use the same distribution of the mean: from Problem 6.3.1, \(\overline{X} \sim N(25, 0.0577)\).
Step 2 — Recognize this as a right tail: "greater than 25.2" is the area to the right of 25.2.
$$ P(\overline{x} > 25.2) = \text{normalcdf}(25.2, 10^{99}, 25, 0.0577) = 0.0003 $$A quick sanity check: 25.2 sits \(\frac{25.2 - 25}{0.0577} = 3.46\) standard deviations above the centre, and areas that far out in a tail are always tiny.
Answer: 0.0003
Problem 4. Draw the graph from Problem 6.3.3.
Problem 4 Solution
Step 1 — Draw the same curve: a normal curve for \(\overline{X}\) centred at 25 with standard deviation 0.0577.
Step 2 — Shade the opposite side this time: draw a vertical line at \(\overline{x} = 25.2\) and shade everything to its RIGHT.
Step 3 — Check the picture against the number: 25.2 is about 3.46 standard deviations above the centre, so the shaded right tail is barely visible — consistent with the area 0.0003 from Problem 6.3.3.
Answer: A normal curve centred at 25 with the thin region right of 25.2 shaded; the shaded area equals 0.0003.
Problem 5. Find the \(90^{th}\) percentile for the mean weight for the 100 weights.
Problem 5 Solution
Step 1 — Recognize the direction of the question: a percentile hands you an area and asks for a value on the axis, so this is an invNorm question, not a normalcdf question.
Step 2 — Run it on the distribution of the mean: we want \(k\) with \(P(\overline{x} < k) = 0.90\).
$$ k = \text{invNorm}(0.90, 25, 0.0577) = 25.07 $$Step 3 — Say what it means: 90% of all samples of 100 weights have a mean weight at or below 25.07 pounds.
Answer: 25.07 pounds
Problem 6. Draw the graph from Problem 6.3.5.
Problem 6 Solution
Step 1 — Draw the curve: a normal curve for \(\overline{X}\) centred at 25 with standard deviation 0.0577.
Step 2 — Mark the percentile and shade left: draw a vertical line at \(k = 25.07\) and shade everything to its LEFT, because a percentile is defined by the area below it.
Step 3 — Label the shaded area: write 0.90 inside the shaded region and label the line \(k = 25.07\). The unshaded right tail carries the remaining 0.10.
Answer: A normal curve centred at 25 with the region left of \(k = 25.07\) shaded and labelled 0.90.
Problem 7. a) What is the distribution for the sum of the weights of 100 25-pound lifting weights?
b) Find \(P(\Sigma x < 2{,}450)\).
Problem 7 Solution
Step 1 — Build the distribution of the sum: the central limit theorem for sums multiplies the mean by \(n\) and the standard deviation by \(\sqrt{n}\).
$$ \mu_{\Sigma X} = n\mu = 100(25) = 2{,}500 $$ $$ \sigma_{\Sigma X} = \sqrt{n}\,\sigma = \sqrt{100}\,(0.5774) = 5.7735 $$ $$ \Sigma X \sim N(2{,}500, 5.7735) $$Step 2 — Find the left-tail probability:
$$ P(\Sigma x < 2{,}450) = \text{normalcdf}(-10^{99}, 2{,}450, 2{,}500, 5.7735) = 0 $$A total of 2,450 is \(\frac{2{,}450 - 2{,}500}{5.7735} = -8.66\) standard deviations below the centre, which is far past anything a normal curve gives measurable area to.
Answer: a) \(\Sigma X \sim N(2{,}500, 5.7735)\) b) 0
Problem 8. Draw the graph from Problem 6.3.7.
Problem 8 Solution
Step 1 — Draw the curve for the sum: a normal curve centred at \(\Sigma x = 2{,}500\) with standard deviation 5.7735, so the visible range runs roughly from 2,483 to 2,517.
Step 2 — Shade to the left of 2,450: draw a vertical line at 2,450. It falls so far left that it lies off the drawn portion of the curve entirely.
Step 3 — Read the picture: there is no visible shaded area, which is the graphical version of the probability 0 from Problem 6.3.7.
Answer: A normal curve centred at 2,500 with 2,450 lying far outside the left edge; the shaded area is 0.
Problem 9. Find the \(90^{th}\) percentile for the total weight of the 100 weights.
Problem 9 Solution
Step 1 — Percentile means invNorm: we want \(k\) with \(P(\Sigma x < k) = 0.90\), using the distribution of the sum from Problem 6.3.7.
Step 2 — Interpret it: 90% of samples of 100 weights have a total weight at or below 2,507.40 pounds.
Answer: 2,507.40 pounds
Problem 10. Draw the graph from Problem 6.3.9.
Problem 10 Solution
Step 1 — Draw the curve for the sum: centred at 2,500 with standard deviation 5.7735.
Step 2 — Mark the percentile and shade left: draw a vertical line at \(k = 2{,}507.40\), a bit more than one standard deviation to the right of centre, and shade everything to its left.
Step 3 — Label: write 0.90 in the shaded region; the unshaded right tail carries 0.10.
Answer: A normal curve centred at 2,500 with the region left of \(k = 2{,}507.40\) shaded and labelled 0.90.
Use the following information to answer the next five exercises: The length of time a particular smartphone's battery lasts follows an exponential distribution with a mean of ten months. A sample of 64 of these smartphones is taken.
Problem 11. a) What is the standard deviation?
b) What is the parameter \(m\)?
Problem 11 Solution
Step 1 — Use the defining property of the exponential distribution: for an exponential random variable the standard deviation equals the mean.
$$ \sigma = \mu = 10 \text{ months} $$Step 2 — Get the decay parameter: the parameter \(m\) is the reciprocal of the mean.
$$ m = \frac{1}{\mu} = \frac{1}{10} = 0.1 $$Answer: a) \(\sigma = 10\) months b) \(m = \frac{1}{10} = 0.1\)
Problem 12. What is the distribution for the length of time one battery lasts?
Problem 12 Solution
Step 1 — Read what is being asked: "one battery" means a single individual value, so the central limit theorem does not apply. Use the distribution the problem states for individuals.
Step 2 — Write it with its parameter: the lifetime is exponential with mean ten months, so \(m = \frac{1}{10}\).
Answer: \(X \sim \text{Exp}\!\left(\frac{1}{10}\right)\), with mean 10 months and standard deviation 10 months.
Problem 13. What is the distribution for the mean length of time 64 batteries last?
Problem 13 Solution
Step 1 — This one asks about a mean, so the central limit theorem applies: with \(n = 64\) the sample mean is approximately normal even though the individual values are exponential.
Step 2 — Compute the two parameters:
$$ \mu_{\overline{X}} = 10, \qquad \sigma_{\overline{X}} = \frac{\sigma}{\sqrt{n}} = \frac{10}{\sqrt{64}} = \frac{10}{8} = 1.25 $$Answer: \(\overline{X} \sim N\!\left(10, \frac{10}{8}\right) = N(10, 1.25)\)
Problem 14. What is the distribution for the total length of time 64 batteries last?
Problem 14 Solution
Step 1 — Use the central limit theorem for sums: multiply the mean by \(n\) and the standard deviation by \(\sqrt{n}\).
$$ \mu_{\Sigma X} = 64(10) = 640, \qquad \sigma_{\Sigma X} = \sqrt{64}\,(10) = 80 $$Answer: \(\Sigma X \sim N(640, 80)\)
Problem 15. Find the probability that the sample mean is between seven and 11.
Problem 15 Solution
Step 1 — Recognize a between-question: normalcdf already accepts a lower and an upper bound, so supply both, using \(\overline{X} \sim N(10, 1.25)\) from Problem 6.3.13.
Step 2 — Sanity-check the bounds: 7 is 2.4 standard deviations below the centre and 11 is 0.8 above, so most but not all of the curve is captured — a value near 0.78 is what that picture should give.
Answer: 0.7799
Problem 16. Find the \(80^{th}\) percentile for the total length of time 64 batteries last.
Problem 16 Solution
Step 1 — Percentile on the sum, so invNorm with the sum's parameters: from Problem 6.3.14, \(\Sigma X \sim N(640, 80)\).
Step 2 — Interpret: in 80% of samples of 64 batteries, the 64 lifetimes total 707.33 months or less.
Answer: 707.33 months
Problem 17. Find the IQR for the mean amount of time 64 batteries last.
Problem 17 Solution
Step 1 — Translate IQR into two percentiles: the interquartile range is the \(75^{th}\) percentile minus the \(25^{th}\) percentile, so run invNorm twice on \(\overline{X} \sim N(10, 1.25)\).
Step 2 — Subtract:
$$ \text{IQR} = 10.843 - 9.157 = 1.69 $$Answer: 1.69 months
Problem 18. Find the middle 80% for the total amount of time 64 batteries last.
Problem 18 Solution
Step 1 — Translate "middle 80%" into two percentiles: leaving 80% in the middle leaves 10% in each tail, so the boundaries are the \(10^{th}\) and \(90^{th}\) percentiles of the sum, \(\Sigma X \sim N(640, 80)\).
$$ \text{lower} = \text{invNorm}(0.10, 640, 80) = 537.48 $$ $$ \text{upper} = \text{invNorm}(0.90, 640, 80) = 742.52 $$Step 2 — Check the symmetry: both boundaries sit 102.52 months from the centre, which they should, because a normal curve is symmetric.
Answer: The middle 80% of total lifetimes runs from 537.48 to 742.52 months.
Use the following information to answer the next eight exercises: A uniform distribution has a minimum of six and a maximum of ten. A sample of 50 is taken.
Problem 19. Find \(P(\Sigma x > 420)\).
Problem 19 Solution
Step 1 — Get the individual mean and standard deviation: for \(U(6, 10)\),
$$ \mu = \frac{6 + 10}{2} = 8, \qquad \sigma = \sqrt{\frac{(10 - 6)^2}{12}} = \sqrt{\frac{16}{12}} = 1.1547 $$Step 2 — Build the distribution of the sum for \(n = 50\):
$$ \mu_{\Sigma X} = 50(8) = 400, \qquad \sigma_{\Sigma X} = \sqrt{50}\,(1.1547) = 8.1650 $$Step 3 — Find the right-tail area:
$$ P(\Sigma x > 420) = \text{normalcdf}(420, 10^{99}, 400, 8.1650) = 0.0072 $$Answer: 0.0072
Problem 20. Find the \(90^{th}\) percentile for the sums.
Problem 20 Solution
Step 1 — Percentile on the sum, so invNorm: using \(\Sigma X \sim N(400, 8.1650)\) from Problem 6.3.19.
Answer: 410.46
Problem 21. Find the \(15^{th}\) percentile for the sums.
Problem 21 Solution
Step 1 — Same distribution, lower area: a \(15^{th}\) percentile leaves 15% of the area to its left, so it lands below the centre.
$$ k = \text{invNorm}(0.15, 400, 8.1650) = 391.54 $$Answer: 391.54
Problem 22. Find the first quartile for the sums.
Problem 22 Solution
Step 1 — Recognize the first quartile as the \(25^{th}\) percentile:
$$ Q_1 = \text{invNorm}(0.25, 400, 8.1650) = 394.49 $$Answer: 394.49
Problem 23. Find the third quartile for the sums.
Problem 23 Solution
Step 1 — Recognize the third quartile as the \(75^{th}\) percentile:
$$ Q_3 = \text{invNorm}(0.75, 400, 8.1650) = 405.51 $$Step 2 — Check against Problem 6.3.22: \(Q_1\) and \(Q_3\) sit the same distance (5.51) either side of 400, as symmetry requires.
Answer: 405.51
Problem 24. Find the \(80^{th}\) percentile for the sums.
Problem 24 Solution
Step 1 — Run invNorm one more time on the sum:
Answer: 406.87
Problem 25. The attention span of a two-year-old is exponentially distributed with a mean of about eight minutes. Suppose we randomly survey 60 two-year-olds.
a) In words, \(X =\) ____
b) \(X \sim\) ____(____,____)
c) In words, \(\overline{X} =\) ____
d) \(\overline{X} \sim\) ____(____,____)
e) Before doing any calculations, which do you think will be higher? Explain why.
i. The probability that an individual attention span is less than ten minutes.
ii. The probability that the average attention span for the 60 children is less than ten minutes.
f) Calculate the probabilities in part e.
g) Explain why the distribution for \(\overline{X}\) is not exponential.
Problem 25 Solution
Step 1 — Name the variables in words. \(X\) is one child's attention span; \(\overline{X}\) is the average attention span of the 60 children in the sample.
Step 2 — Write the distributions. An exponential with mean 8 has \(m = \frac{1}{8}\). For the sample mean, the exponential standard deviation equals its mean, so \(\sigma = 8\) and
$$ \sigma_{\overline{X}} = \frac{8}{\sqrt{60}} = 1.0328 $$Step 3 — Predict before computing (part e). The average should be more likely to fall under ten minutes. A single exponential value has a long right tail, so a fair number of individual children run well past ten minutes. Averaging 60 children pulls the result in tight around 8, and ten minutes sits about two standard deviations above that centre — so almost every sample mean lands below it.
Step 4 — Compute both (part f). For the individual, use the exponential formula \(P(X < x) = 1 - e^{-mx}\).
$$ P(X < 10) = 1 - e^{-\left(\frac{1}{8}\right)(10)} = 1 - e^{-1.25} = 0.7135 $$ $$ P(\overline{X} < 10) = \text{normalcdf}(-10^{99}, 10, 8, 1.0328) = 0.9736 $$The prediction holds.
Step 5 — Explain the shape (part g). \(\overline{X}\) is not exponential because it is an average of 60 values, and the central limit theorem says an average of that many independent values is approximately normal no matter what shape the individual values have. The skew of the exponential washes out in the averaging.
Answer: a) \(X\) = the attention span of one two-year-old, in minutes b) \(X \sim \text{Exp}\!\left(\frac{1}{8}\right)\) c) \(\overline{X}\) = the mean attention span of the 60 two-year-olds d) \(\overline{X} \sim N(8, 1.0328)\) e) the average, because averaging shrinks the spread f) 0.7135 and 0.9736 g) the central limit theorem makes an average of 60 values approximately normal
Problem 26. The closing stock prices of 35 U.S. semiconductor manufacturers are given as follows.
8.625; 30.25; 27.625; 46.75; 32.875; 18.25; 5; 0.125; 2.9375; 6.875; 28.25; 24.25; 21; 1.5; 30.25; 71; 43.5; 49.25; 2.5625; 31; 16.5; 9.5; 18.5; 18; 9; 10.5; 16.625; 1.25; 18; 12.87; 7; 12.875; 2.875; 60.25; 29.25
a) In words, \(X =\) ____
b) i. \(\overline{x} =\) ____ ii. \(s_X =\) ____ iii. \(n =\) ____
c) Construct a histogram of the distribution of the averages. Start at \(x = -0.0005\). Use bar widths of ten.
d) In words, describe the distribution of stock prices.
e) Randomly average five stock prices together. (Use a random number generator.) Continue averaging five pieces together until you have ten averages. List those ten averages.
f) Use the ten averages from part e to calculate the following.
i. \(\overline{x} =\) ____ ii. \(s_X =\) ____
g) Construct a histogram of the distribution of the averages. Start at \(x = -0.0005\). Use bar widths of ten.
h) Does this histogram look like the graph in part c?
i) In one or two complete sentences, explain why the graphs either look the same or look different.
j) Based upon the theory of the central limit theorem, \(\overline{X} \sim\) ____(____,____)
Problem 26 Solution
Step 1 — Name the variable (part a). \(X\) is the closing stock price of one U.S. semiconductor manufacturer.
Step 2 — Summarize the data (part b). Entering all 35 prices into a calculator's one-variable statistics gives
$$ \overline{x} = \$20.71, \qquad s_X = \$17.31, \qquad n = 35 $$Step 3 — Build the histogram and describe it (parts c and d). Starting at \(x = -0.0005\) with bar widths of ten puts the prices into bins 0–10, 10–20, 20–30, and so on. The tallest bar is the first one and the bars fall off steadily to the right, with a long thin tail out past $70. That is a right-skewed shape, well modelled by an exponential distribution, \(X \sim \text{Exp}\!\left(\frac{1}{20.71}\right)\).
Step 4 — Build ten averages of five (parts e and f). Using a random number generator to pick five prices at a time and averaging, then repeating until you have ten averages, answers will vary. The mean of the ten averages should land near the population mean of $20.71, while their standard deviation should be much smaller than 17.31 — around \(\frac{17.31}{\sqrt{5}} = \$7.74\) in theory, and roughly $11.14 in a typical run of only ten averages.
Step 5 — Compare the two histograms (parts g, h, i). The second histogram is noticeably narrower and more symmetric than the first, and its bars are clustered near $20. The graphs look different because the first one shows individual prices, which are strongly right-skewed, while the second shows averages of five prices, and averaging pulls values toward the centre and starts to build the bell shape. With only five values per average the shape is not fully normal yet, but the movement is unmistakable.
Step 6 — State the theoretical distribution (part j). By the central limit theorem, averages of five prices have the population mean and a standard deviation divided by \(\sqrt{5}\).
Answer: a) \(X\) = the closing stock price of one U.S. semiconductor manufacturer b) \(\overline{x} = \$20.71\), \(s_X = \$17.31\), \(n = 35\) c), e), f), g), h), i) answers will vary d) right-skewed, \(X \sim \text{Exp}\!\left(\frac{1}{20.71}\right)\) j) \(\overline{X} \sim N\!\left(20.71, \frac{17.31}{\sqrt{5}}\right)\)
Use the following information to answer the next three exercises: Ramírez Furniture Company delivers furniture from 10 A.M. to 2 P.M. continuously and uniformly. We are interested in how long (in hours) past the 10 A.M. start time that individuals wait for their delivery.
Problem 27. \(X \sim\) ____(____,____)
a) \(U(0,4)\)
b) \(U(10,2)\)
c) \(\text{Exp}(2)\)
d) \(N(2,1)\)
Problem 27 Solution
Step 1 — Read the wait window off the problem: deliveries run from 10 A.M. to 2 P.M., which is a four-hour window, and we measure hours past 10 A.M. So the wait runs from 0 to 4 hours.
Step 2 — Read the shape: "continuously and uniformly" says every moment in that window is equally likely, which is a uniform distribution on \([0, 4]\).
Answer: a) \(U(0,4)\)
Problem 28. The average wait time is:
a) one hour.
b) two hours.
c) two and a half hours.
d) four hours.
Problem 28 Solution
Step 1 — Use the uniform mean formula: for \(U(a, b)\) the mean is the midpoint of the interval.
$$ \mu = \frac{a + b}{2} = \frac{0 + 4}{2} = 2 $$Answer: b) two hours
Problem 29. Suppose that it is now past noon on a delivery day. The probability that a person must wait at least one and a half more hours is:
a) \(\frac{1}{4}\)
b) \(\frac{1}{2}\)
c) \(\frac{2}{3}\)
d) \(\frac{3}{8}\)
Problem 29 Solution
Step 1 — Notice this is a conditional probability: "it is now past noon" tells us two hours have already passed, so we know \(X > 2\). We want the probability of waiting at least 1.5 more hours, which means \(X \geq 3.5\).
Step 2 — Restrict the uniform to what is left: given \(X > 2\), the remaining possibilities are spread uniformly over \([2, 4]\), an interval of length 2. The part of that interval satisfying \(X \geq 3.5\) has length \(4 - 3.5 = 0.5\).
$$ P(X \geq 3.5 \mid X > 2) = \frac{4 - 3.5}{4 - 2} = \frac{0.5}{2} = \frac{1}{4} $$Answer: a) \(\frac{1}{4}\)
Use the following information to answer the next two exercises: The time to wait for a particular rural bus is distributed uniformly from zero to 75 minutes. One hundred riders are randomly sampled to learn how long they waited.
Problem 30. The \(90^{th}\) percentile sample average wait time (in minutes) for a sample of 100 riders is:
a) 315.0
b) 40.3
c) 38.5
d) 65.2
Problem 30 Solution
Step 1 — Get the individual parameters: the wait is \(U(0, 75)\), so
$$ \mu = \frac{0 + 75}{2} = 37.5, \qquad \sigma = \sqrt{\frac{75^2}{12}} = 21.65 $$Step 2 — Build the distribution of the mean for \(n = 100\):
$$ \sigma_{\overline{X}} = \frac{21.65}{\sqrt{100}} = 2.1651, \qquad \overline{X} \sim N(37.5, 2.1651) $$Step 3 — Run invNorm for the \(90^{th}\) percentile:
Answer: b) 40.3 minutes
Problem 31. Would you be surprised, based upon numerical calculations, if the sample average wait time (in minutes) for 100 riders was less than 30 minutes?
a) yes
b) no
c) There is not enough information.
Problem 31 Solution
Step 1 — Compute how likely that outcome is: using \(\overline{X} \sim N(37.5, 2.1651)\) from Problem 6.3.30,
$$ P(\overline{x} < 30) = \text{normalcdf}(-10^{99}, 30, 37.5, 2.1651) = 0.00027 $$Step 2 — Judge it: 30 minutes is about 3.46 standard deviations below the centre. An outcome that happens roughly 3 times in 10,000 samples is genuinely surprising.
Answer: a) yes
Use the following to answer the next two exercises: The cost of unleaded gasoline in the Bay Area once followed an unknown distribution with a mean of $4.59 and a standard deviation of $0.10. Sixteen gas stations from the Bay Area are randomly chosen. We are interested in the average cost of gasoline for the 16 gas stations.
Problem 32. What is the approximate probability that the average price for 16 gas stations is over $4.69?
a) almost zero
b) 0.1587
c) 0.0943
d) unknown
Problem 32 Solution
Step 1 — Build the distribution of the mean: with \(\mu = \$4.59\), \(\sigma = \$0.10\), and \(n = 16\),
$$ \sigma_{\overline{X}} = \frac{0.10}{\sqrt{16}} = 0.025 $$Step 2 — Find the right-tail area:
$$ P(\overline{x} > 4.69) = \text{normalcdf}(4.69, 10^{99}, 4.59, 0.025) = 0.00003 $$The value $4.69 sits exactly four standard deviations above the centre, which is why the area is so small.
Answer: a) almost zero
Problem 33. Find the probability that the average price for 30 gas stations is less than $4.55.
a) 0.6554
b) 0.3446
c) 0.0142
d) 0.9858
e) 0
Problem 33 Solution
Step 1 — Recompute the standard deviation for the new sample size: with \(n = 30\),
$$ \sigma_{\overline{X}} = \frac{0.10}{\sqrt{30}} = 0.018257 $$Step 2 — Find the left-tail area:
$$ P(\overline{x} < 4.55) = \text{normalcdf}(-10^{99}, 4.55, 4.59, 0.018257) = 0.0142 $$Answer: c) 0.0142
Problem 34. Suppose in a local Kindergarten through \(12^{th}\) grade (K–12) school district, 53 percent of the population favor a charter school for grades K through five. A simple random sample of 300 is surveyed. Calculate the following using the normal approximation to the binomial distribution.
a) Find the probability that less than 100 favor a charter school for grades K through 5.
b) Find the probability that 170 or more favor a charter school for grades K through 5.
c) Find the probability that no more than 140 favor a charter school for grades K through 5.
d) Find the probability that there are fewer than 130 that favor a charter school for grades K through 5.
e) Find the probability that exactly 150 favor a charter school for grades K through 5.
If you have access to an appropriate calculator or computer software, try calculating these probabilities using the technology.
Problem 34 Solution
Step 1 — Set up the normal approximation: with \(n = 300\) and \(p = 0.53\), we have \(np = 159 > 5\) and \(nq = 141 > 5\), so the approximation is allowed.
$$ \mu = np = 159, \qquad \sigma = \sqrt{npq} = \sqrt{300(0.53)(0.47)} = 8.6447 $$Step 2 — Apply the continuity correction to each part. Move the boundary outward when the named value is included in the event, inward when it is excluded.
$$ \text{a) } P(X < 100) \to P(Y < 99.5) = \text{normalcdf}(0, 99.5, 159, 8.6447) = 0 $$ $$ \text{b) } P(X \geq 170) \to P(Y \geq 169.5) = \text{normalcdf}(169.5, 10^{99}, 159, 8.6447) = 0.1123 $$ $$ \text{c) } P(X \leq 140) \to P(Y \leq 140.5) = \text{normalcdf}(0, 140.5, 159, 8.6447) = 0.0162 $$ $$ \text{d) } P(X < 130) \to P(Y < 129.5) = \text{normalcdf}(0, 129.5, 159, 8.6447) = 0.0003 $$ $$ \text{e) } P(X = 150) \to P(149.5 < Y < 150.5) = \text{normalcdf}(149.5, 150.5, 159, 8.6447) = 0.0268 $$Note on the technology check: running the exact binomial with binomialcdf(300, 0.53, …) and binomialpdf(300, 0.53, 150) returns values that agree with these to about three decimal places, which is what a well-conditioned approximation should do.
Answer: a) 0 b) 0.1123 c) 0.0162 d) 0.0003 e) 0.0268
Problem 35. Four friends — Lucía, Alex, Kathy and Roberta — decided to carpool together to get to school. Each day the driver would be chosen by randomly selecting one of the four names, so on any given morning each of them is equally likely to drive. They carpool to school for 96 days. Use the normal approximation to the binomial to calculate the following probabilities. Round the standard deviation to four decimal places.
a) Find the probability that Lucía is the driver at most 20 days.
b) Find the probability that Roberta is the driver more than 16 days.
c) Find the probability that Alex, who uses they/them pronouns, drives exactly 24 of those 96 days.
Problem 35 Solution
Step 1 — Recognize the binomial setup: each of the 96 days is an independent trial, and for any one named friend the trial is a success if that friend is picked. With four names drawn at random, \(p = \frac{1}{4}\).
$$ \mu = np = 96\left(\tfrac{1}{4}\right) = 24, \qquad \sigma = \sqrt{npq} = \sqrt{96\left(\tfrac{1}{4}\right)\left(\tfrac{3}{4}\right)} = \sqrt{18} = 4.2426 $$Since \(np = 24 > 5\) and \(nq = 72 > 5\), the normal approximation applies. The same \(\mu\) and \(\sigma\) work for all three parts because every friend has the same chance of driving.
Step 2 — Part a, "at most 20 days": 20 is included, so widen outward to 20.5.
$$ P(X \leq 20) \to P(Y \leq 20.5) = \text{normalcdf}(0, 20.5, 24, 4.2426) = 0.2047 $$Step 3 — Part b, "more than 16 days": 16 is excluded, so pull inward to 16.5.
$$ P(X > 16) \to P(Y > 16.5) = \text{normalcdf}(16.5, 10^{99}, 24, 4.2426) = 0.9615 $$Step 4 — Part c, "exactly 24 days": an exact value becomes a strip half a unit wide on each side.
$$ P(X = 24) \to P(23.5 < Y < 24.5) = \text{normalcdf}(23.5, 24.5, 24, 4.2426) = 0.0938 $$Answer: a) 0.2047 b) 0.9615 c) 0.0938
Problem 36. \(X \sim N(60, 9)\). Suppose that you form random samples of 25 from this distribution. Let \(\overline{X}\) be the random variable of averages. Let \(\Sigma X\) be the random variable of sums. For parts c through f, sketch the graph, shade the region, label and scale the horizontal axis for \(\overline{X}\), and find the probability.
a) Sketch the distributions of \(X\) and \(\overline{X}\) on the same graph.
b) \(\overline{X} \sim\) ____(____,____)
c) \(P(\overline{x} < 60) =\) ____
d) Find the \(30^{th}\) percentile for the mean.
e) \(P(56 < \overline{x} < 62) =\) ____
f) \(P(18 < \overline{x} < 58) =\) ____
g) \(\Sigma x \sim\) ____(____,____)
h) Find the minimum value for the upper quartile for the sum.
i) \(P(1{,}400 < \Sigma x < 1{,}550) =\) ____
Problem 36 Solution
Step 1 — Sketch both distributions (part a). Draw two normal curves on the same axis, both centred at 60. The curve for \(X\) has standard deviation 9 and is wide; the curve for \(\overline{X}\) has standard deviation \(\frac{9}{\sqrt{25}} = 1.8\) and is tall and narrow. The averaging is what compresses the second curve.
Step 2 — Write the distribution of the mean (part b).
$$ \overline{X} \sim N\!\left(60, \frac{9}{\sqrt{25}}\right) = N(60, 1.8) $$Step 3 — Part c. The value 60 is the centre of a symmetric curve, so exactly half the area lies below it.
$$ P(\overline{x} < 60) = 0.5000 $$Step 4 — Part d, a percentile, so invNorm.
Step 5 — Parts e and f, between-questions.
$$ P(56 < \overline{x} < 62) = \text{normalcdf}(56, 62, 60, 1.8) = 0.8536 $$ $$ P(18 < \overline{x} < 58) = \text{normalcdf}(18, 58, 60, 1.8) = 0.1333 $$In part f the lower bound 18 is more than 23 standard deviations below the centre, so it contributes nothing; the answer is really just the area below 58.
Step 6 — Part g, the distribution of the sum.
$$ \Sigma X \sim N\!\left(25(60), \sqrt{25}\,(9)\right) = N(1{,}500, 45) $$Step 7 — Part h, the upper quartile of the sum. The upper quartile begins at the \(75^{th}\) percentile, so its minimum value is that percentile.
$$ Q_3 = \text{invNorm}(0.75, 1{,}500, 45) = 1{,}530.35 $$Step 8 — Part i, a between-question on the sum.
$$ P(1{,}400 < \Sigma x < 1{,}550) = \text{normalcdf}(1{,}400, 1{,}550, 1{,}500, 45) = 0.8536 $$This has to match part e, and it does. Dividing the sum bounds by \(n = 25\) gives \(\frac{1{,}400}{25} = 56\) and \(\frac{1{,}550}{25} = 62\) — the exact bounds of part e. Asking about the sum of 25 values landing between 1,400 and 1,550 is the same question as asking about their average landing between 56 and 62.
Answer: b) \(N(60, 1.8)\) c) 0.5000 d) 59.06 e) 0.8536 f) 0.1333 g) \(N(1{,}500, 45)\) h) 1,530.35 i) 0.8536
Problem 37. Suppose that the length of research papers is uniformly distributed from ten to 25 pages. We survey a class in which 55 research papers were turned in to a professor. The 55 research papers are considered a random collection of all papers. We are interested in the average length of the research papers.
a) In words, \(X =\) ____
b) \(X \sim\) ____(____,____)
c) \(\mu_X =\) ____
d) \(\sigma_X =\) ____
e) In words, \(\overline{X} =\) ____
f) \(\overline{X} \sim\) ____(____,____)
g) In words, \(\Sigma X =\) ____
h) \(\Sigma X \sim\) ____(____,____)
i) Without doing any calculations, do you think that it is likely that the professor will need to read a total of more than 1,050 pages? Why?
j) Calculate the probability that the professor will need to read a total of more than 1,050 pages.
k) Why is it so unlikely that the average length of the papers will be less than 12 pages?
Problem 37 Solution
Step 1 — Name the variables and get the individual parameters (parts a–d). \(X\) is the length of one research paper, uniform from 10 to 25 pages.
$$ \mu_X = \frac{10 + 25}{2} = 17.5, \qquad \sigma_X = \sqrt{\frac{(25 - 10)^2}{12}} = \frac{15}{\sqrt{12}} = 4.3301 $$Step 2 — Build the distribution of the mean (parts e and f). \(\overline{X}\) is the mean length of the 55 papers.
$$ \overline{X} \sim N\!\left(17.5, \frac{4.3301}{\sqrt{55}}\right) = N(17.5, 0.5839) $$Step 3 — Build the distribution of the sum (parts g and h). \(\Sigma X\) is the total number of pages across all 55 papers.
$$ \Sigma X \sim N\!\left(55(17.5), \sqrt{55}\,(4.3301)\right) = N(962.5, 32.11) $$Step 4 — Predict before computing (part i). The expected total is 962.5 pages, and 1,050 is about \(\frac{1{,}050 - 962.5}{32.11} = 2.7\) standard deviations above that. So no, it is not likely.
Step 5 — Compute it (part j).
$$ P(\Sigma x > 1{,}050) = \text{normalcdf}(1{,}050, 10^{99}, 962.5, 32.11) = 0.0032 $$Step 6 — Explain part k. The standard deviation of the mean is only 0.5839 pages, so 12 pages sits \(\frac{12 - 17.5}{0.5839} = -9.4\) standard deviations below the centre. Individual papers of 12 pages are perfectly ordinary, but an average of 55 papers coming out that low is essentially impossible — that is the shrinking effect of \(\sqrt{n}\).
Answer: a) \(X\) = the length of one research paper, in pages b) \(X \sim U(10, 25)\) c) 17.5 d) 4.3301 e) \(\overline{X}\) = the mean length of the 55 papers f) \(N(17.5, 0.5839)\) g) \(\Sigma X\) = the total pages of all 55 papers h) \(N(962.5, 32.11)\) i) no, 1,050 is about 2.7 standard deviations above the expected total j) 0.0032 k) an average that low is over nine standard deviations from centre
Problem 38. Salaries for teachers in a particular elementary school district are normally distributed with a mean of $44,000 and a standard deviation of $6,500. We randomly survey ten teachers from that district.
a) Find the \(90^{th}\) percentile for an individual teacher's salary.
b) Find the \(90^{th}\) percentile for the average teacher's salary.
Problem 38 Solution
Step 1 — Part a is about one teacher, so use the population distribution directly. The salaries are normal with \(\mu = \$44{,}000\) and \(\sigma = \$6{,}500\).
$$ k = \text{invNorm}(0.90, 44{,}000, 6{,}500) = \$52{,}330 $$Step 2 — Part b is about an average, so apply the central limit theorem. With \(n = 10\),
$$ \sigma_{\overline{X}} = \frac{6{,}500}{\sqrt{10}} = \$2{,}055.60 $$ $$ k = \text{invNorm}(0.90, 44{,}000, 2{,}055.60) = \$46{,}634 $$Step 3 — Notice the gap. The two answers differ by nearly $5,700 even though both are \(90^{th}\) percentiles. Individual salaries spread widely; averages of ten do not.
Answer: a) $52,330 b) $46,634
Problem 39. The average length of a maternity stay in a U.S. hospital is said to be 2.4 days with a standard deviation of 0.9 days. We randomly survey 80 women who recently bore children in a U.S. hospital.
a) In words, \(X =\) ____
b) In words, \(\overline{X} =\) ____
c) \(\overline{X} \sim\) ____(____,____)
d) In words, \(\Sigma X =\) ____
e) \(\Sigma X \sim\) ____(____,____)
f) Is it likely that an individual stayed more than five days in the hospital? Why or why not?
g) Is it likely that the average stay for the 80 women was more than five days? Why or why not?
h) Which is more likely?
i. An individual stayed more than five days.
ii. The average stay of 80 women was more than five days.
i) If we were to sum up the women's stays, is it likely that collectively they spent more than a year in the hospital? Why or why not?
Problem 39 Solution
Step 1 — Name the variables (parts a, b, d). \(X\) is the length of one woman's maternity stay in days; \(\overline{X}\) is the mean stay of the 80 women sampled; \(\Sigma X\) is the total number of days across all 80 stays.
Step 2 — Write the two distributions (parts c and e).
$$ \overline{X} \sim N\!\left(2.4, \frac{0.9}{\sqrt{80}}\right) = N(2.4, 0.1006) $$ $$ \Sigma X \sim N\!\left(80(2.4), \sqrt{80}\,(0.9)\right) = N(192, 8.05) $$Step 3 — Judge the individual (part f). For one woman, five days is \(\frac{5 - 2.4}{0.9} = 2.89\) standard deviations above the mean. That is uncommon but entirely possible — complicated deliveries do run that long — so yes, an individual stay over five days is plausible.
Step 4 — Judge the average (part g). For the mean of 80 women, five days is \(\frac{5 - 2.4}{0.1006} = 25.8\) standard deviations above the centre.
$$ P(\overline{x} > 5) = \text{normalcdf}(5, 10^{99}, 2.4, 0.1006) \approx 0 $$No, that is essentially impossible.
Step 5 — Compare (part h). The individual is far more likely, by an enormous margin. Averaging 80 stays collapses the spread from 0.9 days down to about 0.1 days.
Step 6 — Judge the total (part i). More than a year collectively means \(\Sigma x > 365\) days, which is \(\frac{365 - 192}{8.05} = 21.5\) standard deviations above the expected total. No, that is not likely.
Answer: a) \(X\) = one woman's maternity stay, in days b) \(\overline{X}\) = the mean stay of the 80 women c) \(N(2.4, 0.1006)\) d) \(\Sigma X\) = the total days across all 80 stays e) \(N(192, 8.05)\) f) yes, about 2.89 standard deviations out for an individual g) no, about 25.8 standard deviations out for the mean h) i, the individual i) no, about 21.5 standard deviations out
For each of the remaining problems, wherever possible, provide graphs and use the calculator.
Problem 40. NeverReady batteries has engineered a newer, longer lasting AAA battery. The company claims this battery has an average life span of 17 hours with a standard deviation of 0.8 hours. Your statistics class questions this claim. As a class, you randomly select 30 batteries and find that the sample mean life span is 16.7 hours. If the process is working properly, what is the probability of getting a random sample of 30 batteries in which the sample mean lifetime is 16.7 hours or less? Is the company's claim reasonable?
Problem 40 Solution
Step 1 — Write down what the company claims and what you observed. The claim gives \(\mu = 17\) and \(\sigma = 0.8\); your class collected \(n = 30\) batteries with \(\overline{x} = 16.7\).
Step 2 — Build the distribution the claim predicts for the sample mean.
$$ \sigma_{\overline{X}} = \frac{0.8}{\sqrt{30}} = 0.1461 $$Step 3 — Find how likely a mean that low would be if the claim were true.
$$ P(\overline{x} \leq 16.7) = \text{normalcdf}(-10^{99}, 16.7, 17, 0.1461) = 0.0200 $$Step 4 — Judge the claim. If the process really is working as advertised, only about 2% of samples of 30 batteries would come out with a mean lifetime of 16.7 hours or less. That is unusual enough to make the class's scepticism reasonable. It is not proof the company is wrong — 2% events do happen — but it is grounds to ask for more evidence.
Answer: 0.0200. The class is justified in questioning the claim, since a result this low happens only about 2% of the time when the claim holds.
Problem 41. Men have an average weight of 172 pounds with a standard deviation of 29 pounds.
a) Find the probability that 20 randomly selected men will have a sum weight greater than 3,600 pounds.
b) If 20 men have a sum weight greater than 3,500 pounds, then their total weight exceeds the safety limits for water taxis. Based on part a, is this a safety concern? Explain.
Problem 41 Solution
Step 1 — Build the distribution of the sum for 20 men.
$$ \mu_{\Sigma X} = 20(172) = 3{,}440, \qquad \sigma_{\Sigma X} = \sqrt{20}\,(29) = 129.69 $$Step 2 — Part a, the probability of exceeding 3,600 pounds.
$$ P(\Sigma x > 3{,}600) = \text{normalcdf}(3{,}600, 10^{99}, 3{,}440, 129.69) = 0.1087 $$Step 3 — Part b, check the actual safety limit. The limit that matters is 3,500 pounds, not 3,600, so compute that one too.
$$ P(\Sigma x > 3{,}500) = \text{normalcdf}(3{,}500, 10^{99}, 3{,}440, 129.69) = 0.3218 $$Step 4 — Judge it. Yes, this is a real safety concern. Roughly one in three loads of 20 randomly chosen men would push the water taxi past its stated safety limit, and part a shows that even the higher 3,600-pound threshold is crossed about 11% of the time. A limit that gets exceeded on a third of trips is not a limit that is being respected in practice.
Answer: a) 0.1087 b) Yes. About 32% of 20-man loads exceed the 3,500-pound safety limit, which is far too often to ignore.
Problem 42. M&M candies large candy bags have a claimed net weight of 396.9 g. The standard deviation for the weight of the individual candies is 0.017 g. The following table is from a stats experiment conducted by a statistics class.
| Red | Orange | Yellow | Brown | Blue | Green |
|---|---|---|---|---|---|
| 0.751 | 0.735 | 0.883 | 0.696 | 0.881 | 0.925 |
| 0.841 | 0.895 | 0.769 | 0.876 | 0.863 | 0.914 |
| 0.856 | 0.865 | 0.859 | 0.855 | 0.775 | 0.881 |
| 0.799 | 0.864 | 0.784 | 0.806 | 0.854 | 0.865 |
| 0.966 | 0.852 | 0.824 | 0.840 | 0.810 | 0.865 |
| 0.859 | 0.866 | 0.858 | 0.868 | 0.858 | 1.015 |
| 0.857 | 0.859 | 0.848 | 0.859 | 0.818 | 0.876 |
| 0.942 | 0.838 | 0.851 | 0.982 | 0.868 | 0.809 |
| 0.873 | 0.863 | 0.803 | 0.865 | ||
| 0.809 | 0.888 | 0.932 | 0.848 | ||
| 0.890 | 0.925 | 0.842 | 0.940 | ||
| 0.878 | 0.793 | 0.832 | 0.833 | ||
| 0.905 | 0.977 | 0.807 | 0.845 | ||
| 0.850 | 0.841 | 0.852 | |||
| 0.830 | 0.932 | 0.778 | |||
| 0.856 | 0.833 | 0.814 | |||
| 0.842 | 0.881 | 0.791 | |||
| 0.778 | 0.818 | 0.810 | |||
| 0.786 | 0.864 | 0.881 | |||
| 0.853 | 0.825 | ||||
| 0.864 | 0.855 | ||||
| 0.873 | 0.942 | ||||
| 0.880 | 0.825 | ||||
| 0.882 | 0.869 | ||||
| 0.931 | 0.912 | ||||
| 0.887 |
The bag contained 465 candies and the listed weights in the table came from randomly selected candies. Count the weights.
a) Find the mean sample weight and the standard deviation of the sample weights of candies in the table.
b) Find the sum of the sample weights in the table and the standard deviation of the sum of the weights.
c) If 465 M&Ms are randomly selected, find the probability that their weights sum to at least 396.9.
d) Is the Mars Company's M&M labeling accurate?
Problem 42 Solution
Step 1 — Compute the sample statistics from the table (part a). Counting every entry gives \(n = 99\) candies. Entering them into one-variable statistics gives
$$ \overline{x} = 0.8562 \text{ g}, \qquad s = 0.0520 \text{ g} $$Step 2 — Find the sum of the sample and its standard deviation (part b).
$$ \Sigma x = 84.76 \text{ g}, \qquad \sigma_{\Sigma X} = \sqrt{n}\,\sigma = \sqrt{99}\,(0.052) = 0.517 \text{ g} $$Step 3 — Scale up to the full bag of 465 candies (part c). Here the problem hands us the population standard deviation for a single candy, \(\sigma = 0.017\) g, so use that rather than the sample \(s\).
$$ \mu_{\Sigma X} = 465(0.8562) = 398.13 \text{ g}, \qquad \sigma_{\Sigma X} = \sqrt{465}\,(0.017) = 0.3666 \text{ g} $$ $$ P(\Sigma x \geq 396.9) = \text{normalcdf}(396.9, 10^{99}, 398.13, 0.3666) \approx 1 $$Step 4 — Judge the label (part d). The claimed net weight of 396.9 g sits about 3.4 standard deviations below the expected bag weight, so essentially every bag of 465 candies comes in at or above the printed weight. The Mars Company's labeling is accurate — and slightly conservative, which is what a manufacturer wants.
Answer: a) \(\overline{x} = 0.8562\) g, \(s = 0.0520\) g b) \(\Sigma x = 84.76\) g with standard deviation 0.517 g c) approximately 1 d) yes, the labeling is accurate
Problem 43. The Screw Right Company claims their \(\frac{3}{4}\) inch screws are within \(\pm 0.23\) of the claimed mean diameter of 0.750 inches with a standard deviation of 0.115 inches. The following data were recorded.
| 0.757 | 0.723 | 0.754 | 0.737 | 0.757 | 0.741 | 0.722 | 0.741 | 0.743 | 0.742 |
| 0.740 | 0.758 | 0.724 | 0.739 | 0.736 | 0.735 | 0.760 | 0.750 | 0.759 | 0.754 |
| 0.744 | 0.758 | 0.765 | 0.756 | 0.738 | 0.742 | 0.758 | 0.757 | 0.724 | 0.757 |
| 0.744 | 0.738 | 0.763 | 0.756 | 0.760 | 0.768 | 0.761 | 0.742 | 0.734 | 0.754 |
| 0.758 | 0.735 | 0.740 | 0.743 | 0.737 | 0.737 | 0.725 | 0.761 | 0.758 | 0.756 |
The screws were randomly selected from the local home repair store.
a) Find the mean diameter and standard deviation for the sample.
b) Find the probability that 50 randomly selected screws will be within the stated tolerance levels. Is the company's diameter claim plausible?
Problem 43 Solution
Step 1 — Compute the sample statistics (part a). The table holds \(n = 50\) measured diameters. One-variable statistics gives
$$ \overline{x} = 0.7468 \text{ inches}, \qquad s = 0.0123 \text{ inches} $$Step 2 — Translate the tolerance into an interval (part b). "Within \(\pm 0.23\) of 0.750" means the acceptable range runs from \(0.750 - 0.23 = 0.520\) to \(0.750 + 0.23 = 0.980\) inches.
Step 3 — Build the distribution of the sample mean under the company's claim. Using the company's stated \(\sigma = 0.115\),
$$ \sigma_{\overline{X}} = \frac{0.115}{\sqrt{50}} = 0.01626 $$ $$ P(0.520 < \overline{x} < 0.980) = \text{normalcdf}(0.520, 0.980, 0.750, 0.01626) \approx 1 $$Step 4 — Judge the claim. Yes, the diameter claim is plausible, and comfortably so. The observed mean of 0.7468 inches is only about 0.2 standard deviations below the claimed 0.750. In fact the screws are far more consistent than the company promises: the observed sample standard deviation of 0.0123 inches is under a ninth of the claimed 0.115, so the real tolerance is much tighter than \(\pm 0.23\).
Answer: a) \(\overline{x} = 0.7468\) inches, \(s = 0.0123\) inches b) approximately 1; the claim is plausible, and the screws are considerably more consistent than claimed
Problem 44. Your company has a contract to perform preventive maintenance on thousands of air-conditioners in a large city. Based on service records from previous years, the time that a technician spends servicing a unit averages one hour with a standard deviation of one hour. In the coming week, your company will service a simple random sample of 70 units in the city. You plan to budget an average of 1.1 hours per technician to complete the work. Will this be enough time?
Problem 44 Solution
Step 1 — Build the distribution of the mean service time. With \(\mu = 1\) hour, \(\sigma = 1\) hour, and \(n = 70\),
$$ \sigma_{\overline{X}} = \frac{1}{\sqrt{70}} = 0.1195 \text{ hours} $$Step 2 — Find the probability the budget holds. Budgeting 1.1 hours per unit on average means the job fits if the sample mean service time comes in at or below 1.1 hours.
$$ P(\overline{x} < 1.1) = \text{normalcdf}(-10^{99}, 1.1, 1, 0.1195) = 0.7986 $$Step 3 — Judge it. There is about an 80% chance the budgeted time is enough — which also means a 20% chance it is not. One week in five, the crew would run over. Whether that is acceptable is a business call, but it would be wise to schedule some slack rather than treat 1.1 hours as safe.
Answer: 0.7986. The budget works about 80% of the time, so there is a real 20% chance of running over; scheduling extra time would be prudent.
Problem 45. A typical adult has an average IQ score of 105 with a standard deviation of 20. If 20 randomly selected adults are given an IQ test, what is the probability that the sample mean score will be between 85 and 125 points?
Problem 45 Solution
Step 1 — Build the distribution of the mean IQ score. With \(\mu = 105\), \(\sigma = 20\), and \(n = 20\),
$$ \sigma_{\overline{X}} = \frac{20}{\sqrt{20}} = 4.4721 $$Step 2 — Recognize a between-question and run normalcdf with both bounds.
Step 3 — Read the result. Both bounds sit about 4.47 standard deviations from the centre, so the interval captures essentially the whole curve. Note how different this is from the individual case: a single adult's score of 85 or 125 is only one standard deviation from the mean and is completely ordinary.
Answer: 0.9999, or essentially 1
Problem 46. Certain coins have an average weight of 5.201 grams with a standard deviation of 0.065 g. If a vending machine is designed to accept coins whose weights range from 5.111 g to 5.291 g, what is the expected number of rejected coins when 280 randomly selected coins are inserted into the machine?
Problem 46 Solution
Step 1 — Recognize this as an individual-value question. The machine weighs each coin one at a time, so use the population distribution of a single coin, not a sampling distribution. Assume the weights are normally distributed with \(\mu = 5.201\) g and \(\sigma = 0.065\) g.
Step 2 — Find the probability one coin is accepted.
$$ P(5.111 < x < 5.291) = \text{normalcdf}(5.111, 5.291, 5.201, 0.065) = 0.8338 $$Step 3 — Convert to an expected count of rejections. A coin is rejected with probability \(1 - 0.8338 = 0.1662\), and 280 coins are inserted.
$$ (1 - 0.8338)(280) = 0.1662(280) = 46.5 \approx 47 $$Answer: About 47 of the 280 coins are expected to be rejected.
Key Terms
law of large numbers — as the sample size \(n\) grows, the sample mean \(\overline{x}\) gets closer and closer to the population mean \(\mu\).
normal approximation to the binomial — using a normal distribution with \(\mu = np\) and \(\sigma = \sqrt{npq}\) to estimate binomial probabilities, valid when \(np > 5\) and \(nq > 5\).
continuity correction factor — the 0.5 added to or subtracted from \(x\) when a continuous normal curve is used to approximate a discrete binomial probability.