6.2 The Central Limit Theorem for Sums

Aligned outcomes:

Learning Objectives

By the end of this section, you will be able to:

In this section, you will learn to:

state the central limit theorem for sums and give the mean and standard deviation of the sampling distribution of $\Sigma X$;
compute the $z$-score of a sum and convert a $z$-score back into a sum;
use technology to find the probability that a sum falls above, below, or between given values;
find the value of a sum at a stated percentile and say what it means in the original units.

The last section took a sample, averaged it, and watched those averages pile up into a normal curve. This section changes one word. Instead of averaging the sample, we add it up. Everything else survives the change, because a sum is just an average that never got divided — and that means the central limit theorem works on totals too.

6.2.1 Sums Add Up to a Normal Curve

Definition 6.2.1: The Central Limit Theorem for Sums

If $X$ has mean $\mu_X$ and standard deviation $\sigma_X$, and samples of size $n$ are drawn at random, then the sum $\Sigma X$ is approximately normal for large $n$:

$$ \Sigma X \sim N\!\left((n)(\mu_X),\ (\sqrt{n})(\sigma_X)\right) $$

Ten shopping trips, not ten times the surprise

Ten trips cost about ten times one trip, so the mean scales by 10. But the ups and downs do not all line up — a cheap week offsets an expensive one — so the spread only grows by $\sqrt{10} \approx 3.16$.

Definition 6.2.1 - Sums from any population turn normal as the sample size grows.

Unpack that one piece at a time. Start with a random variable $X$ whose distribution may be known or unknown. It can be any distribution at all: skewed, lumpy, uniform, whatever the population happens to look like. Write $\mu_X$ for its mean and $\sigma_X$ for its standard deviation. Now draw random samples of size $n$ and, for each sample, add the $n$ values together. That total is a new random variable, written $\Sigma X$, and as $n$ grows it tends to be normally distributed no matter what $X$ looked like to begin with.

Read as a sentence, the theorem says that if you repeatedly draw samples of a given size — repeatedly rolling ten dice, say, and writing down the total each time — those totals tend to follow a normal distribution. As the sample size increases, the fit gets better. The new normal curve has a mean equal to the original mean multiplied by the sample size, and a standard deviation equal to the original standard deviation multiplied by the square root of the sample size.

The two pieces are worth naming separately, because students mix them up constantly. The mean multiplies by $n$. The standard deviation multiplies by $\sqrt{n}$, not by $n$. Adding more values pushes the total up in a straight line, but it spreads the total out much more slowly than that, because high draws and low draws partly cancel each other on the way to the total.

Once you know the mean and standard deviation of $\Sigma X$, the sum is just another normal variable, so it gets a $z$-score exactly like every other normal variable in this book. Nothing new has to be learned; only the two parameters change.

Definition 6.2.2: The $z$-Score of a Sum

Let $\Sigma x$ be one observed sum. Its $z$-score is

$$ z = \frac{\Sigma x - (n)(\mu_X)}{(\sqrt{n})(\sigma_X)} $$

where $(n)(\mu_X)$ is the mean of $\Sigma X$ and $(\sqrt{n})(\sigma_X)$ is the standard deviation of $\Sigma X$.

Totals are what the world actually bills you for

A store cares about the day's total sales, not the average sale. A payroll office cares about total salary, not the mean salary. Those totals are exactly what the next definition puts a score on.

Definition 6.2.2 - The z-score of a sum is its gap from the mean measured in spreads of the sums.

The same equation, solved for $\Sigma x$, runs the other direction. If a problem tells you how many standard deviations above or below the mean a total sits and asks for the total itself, use

$$ \Sigma x = (n)(\mu_X) + (z)(\sqrt{n})(\sigma_X). $$

Both forms come up in the same problem often enough that it is worth writing them side by side once. Reading left to right, the first turns a total into a location on the standard normal curve; the second turns a location back into a total.

Try It Now 6.2.1

An unknown distribution has a mean of 20 and a standard deviation of four. A sample of size 36 is drawn. What are the mean and standard deviation of $\Sigma X$?

Solution

The mean multiplies by $n$, and the standard deviation multiplies by $\sqrt{n}$.

$$ \mu_{\Sigma X} = (n)(\mu_X) = (36)(20) = 720 $$ $$ \sigma_{\Sigma X} = (\sqrt{n})(\sigma_X) = (\sqrt{36})(4) = (6)(4) = 24 $$

Answer: $\Sigma X \sim N(720, 24)$.

6.2.2 Finding Probabilities for Sums

A probability question about a sum shades a region under the $\Sigma X$ curve, and every technology that handles normal distributions can hand you that area. The only thing you have to get right is which two parameters you feed it: not the original $\mu_X$ and $\sigma_X$, but the sum's own $(n)(\mu_X)$ and $(\sqrt{n})(\sigma_X)$. Feeding in the original pair is the single most common mistake in this section, and nothing on the screen will warn you.

On a TI-83/83+/84/84+: press 2nd DISTR, then 2:normalcdf(. The syntax is

normalcdf(lower value of the area, upper value of the area, (n)(mean), (sqrt(n))(standard deviation))

where the mean and standard deviation you type are the ones from the original distribution, and the $n$ and $\sqrt{n}$ factors do the converting for you. For a right tail with no upper bound, use 1E99 as the upper value — that is $10^{99}$, close enough to infinity for any real distribution. Press the EE key to get the E. For a left tail with no lower bound, use -1E99.

Before you type anything, sketch the curve and shade what the sentence is asking for. "More than 7,500" shades a right tail, "less than 7,000" shades a left tail, and "between 1,500 and 1,800" shades a strip in the middle. The sketch is what stops you from swapping the bounds, and a swapped pair returns a negative area with no error message. Once the picture is shaded, reading off the two bounds is mechanical: the lower edge of the shaded region goes in first, the upper edge goes in second, and an edge that runs off the page gets the appropriate 1E99.

Try It Now 6.2.2

An unknown distribution has a mean of 45 and a standard deviation of eight. A sample size of 50 is drawn randomly from the population. Find the probability that the sum of the 50 values is more than 2,400.

Solution

Step 1 — Find the parameters of the sum. With $\mu_X = 45$, $\sigma_X = 8$, and $n = 50$:

$$ \mu_{\Sigma X} = (50)(45) = 2{,}250 \qquad \sigma_{\Sigma X} = (\sqrt{50})(8) \approx 56.57 $$

Step 2 — Shade a right tail. "More than 2,400" is the area above 2,400.

$$ P(\Sigma x > 2{,}400) = \text{normalcdf}(2400, 1E99, (50)(45), (\sqrt{50})(8)) = 0.0040 $$

Answer: $P(\Sigma x > 2{,}400) = 0.0040$, so a total that high happens about 4 times in 1,000 samples.

Example 6.2.1: A Sum Above 7,500

An unknown distribution has a mean of 90 and a standard deviation of 15. A sample of size 80 is drawn randomly from the population.

a. Find the probability that the sum of the 80 values (or the total of the 80 values) is more than 7,500.

b. Find the sum that is 1.5 standard deviations above the mean of the sums.

Solution

Step 1 — Name the variables. Let $X$ be one value from the original unknown population, and let $\Sigma X$ be the sum or total of 80 values. The question asks about the total, not about a single value, so $\Sigma X$ is the variable we work with.

Step 2 — Find the mean and standard deviation of the sums. With $\mu_X = 90$, $\sigma_X = 15$, and $n = 80$:

$$ \mu_{\Sigma X} = (n)(\mu_X) = (80)(90) = 7{,}200 $$ $$ \sigma_{\Sigma X} = (\sqrt{n})(\sigma_X) = (\sqrt{80})(15) \approx 134.16 $$

so $\Sigma X \sim N\!\left((80)(90),\ (\sqrt{80})(15)\right)$.

Step 3 — Part a: shade the right tail above 7,500.

Figure 6.2.1 - The shaded right tail beyond a total of 7,500 under the distribution of sums.

$$ P(\Sigma x > 7{,}500) = \text{normalcdf}(7500, 1E99, (80)(90), (\sqrt{80})(15)) = 0.0127 $$

Step 4 — Part b: run the $z$-score backwards. We want the sum sitting 1.5 standard deviations above the mean of the sums, so $z = 1.5$.

$$ \Sigma x = (n)(\mu_X) + (z)(\sqrt{n})(\sigma_X) = (80)(90) + (1.5)(\sqrt{80})(15) = 7{,}401.2 $$

Answer: a. $P(\Sigma x > 7{,}500) = 0.0127$; b. the sum 1.5 standard deviations above the mean is 7,401.2.

Try it in rāSHio

Open rāSHio and choose Distributions → Normal. Set the mean to 7200 and the standard deviation to 134.16 — those are the sum’s own $(80)(90)$ and $(\sqrt{80})(15)$, not the original 90 and 15 — then pick Right with cutoff 7500. The 0.0127 this example computed comes back with the tail already shaded.

Figure 6.2.2 — The rāSHio Normal panel: set the mean and standard deviation by hand, then read the area to one side of a cutoff. The clip walks the Left tab; the note above uses Right on the distribution of sums.

6.2.3 Finding Percentiles for Sums

The other direction gives you an area and asks for the total that cuts it off. That is a percentile question, and on a calculator it is a different command with a different first argument: an area instead of a boundary. The $k^{\text{th}}$ percentile of the sums is the total with $k$ percent of all possible totals at or below it.

On a TI-83/83+/84/84+: press 2nd DISTR, then 3:invNorm(. The syntax is

invNorm(area to the left of k, (n)(mean), (sqrt(n))(standard deviation))

where $k$ is the percentile you want, and the mean and standard deviation you type are again the ones from the original distribution. The one rule to hold onto is that invNorm always wants the area to the left. If a problem describes the top 10%, convert it to a left area of 0.90 before typing anything, and if it describes the bottom quartile, use 0.25.

Telling the two commands apart is nearly the whole skill here, and the tell is in the question, not the numbers. If the sentence hands you a total and asks how likely, you want normalcdf and the answer will be a decimal between 0 and 1. If the sentence hands you a percentage and asks for a total, you want invNorm and the answer will come back in the units of the original problem — years, minutes, dollars. Getting an answer of 0.83 when the question asked for a number of years is the clearest possible sign you reached for the wrong command, so read the units of your answer before you write it down.

Try It Now 6.2.3

In a recent study, Mei Chen reported that the mean age of iPad users is 35 years. Suppose the standard deviation is ten years. Her sample size is 39.

a. What are the mean and standard deviation for the sum of the ages of iPad users? What is the distribution?

b. Find the probability that the sum of the ages is between 1,400 and 1,500 years.

c. Find the $90^{\text{th}}$ percentile for the sum of the 39 ages.

Solution

Part a — Parameters of the sum.

$$ \mu_{\Sigma X} = n\mu_X = 39(35) = 1{,}365 \qquad \sigma_{\Sigma X} = \sqrt{n}\,\sigma_X = (\sqrt{39})(10) \approx 62.45 $$

By the central limit theorem the distribution of sums is normal, so $\Sigma X \sim N(1{,}365,\ 62.45)$.

Part b — A strip in the middle.

$$ P(1{,}400 < \Sigma x < 1{,}500) = \text{normalcdf}(1400, 1500, (39)(35), (\sqrt{39})(10)) = 0.2723 $$

Part c — A percentile, so invNorm. The $90^{\text{th}}$ percentile has an area of 0.90 to its left.

$$ k = \text{invNorm}(0.90, (39)(35), (\sqrt{39})(10)) = 1{,}445.0 $$

Answer: a. mean 1,365 and standard deviation 62.45, normal by the central limit theorem; b. 0.2723; c. 1,445.0 years.

Example 6.2.2: Ages of iPad Users

In a recent study, Robin Ashford reported that the mean age of iPad users is 34 years. Suppose the standard deviation is 15 years. Their sample size is 50.

a. What are the mean and standard deviation for the sum of the ages of iPad users? What is the distribution?

b. Find the probability that the sum of the ages is between 1,500 and 1,800 years.

c. Find the $80^{\text{th}}$ percentile for the sum of the 50 ages.

Solution

Step 1 — Part a: scale the two parameters.

$$ \mu_{\Sigma X} = n\mu_X = 50(34) = 1{,}700 $$ $$ \sigma_{\Sigma X} = \sqrt{n}\,\sigma_X = (\sqrt{50})(15) = 106.07 $$

The distribution is normal for sums by the central limit theorem, so $\Sigma X \sim N(1{,}700,\ 106.07)$.

Step 2 — Part b: a between-question, so normalcdf with two real bounds.

$$ P(1{,}500 < \Sigma x < 1{,}800) = \text{normalcdf}(1500, 1800, (50)(34), (\sqrt{50})(15)) = 0.7974 $$

Step 3 — Part c: a percentile, so invNorm. Let $k$ be the $80^{\text{th}}$ percentile; the area to its left is 0.80.

$$ k = \text{invNorm}(0.80, (50)(34), (\sqrt{50})(15)) = 1{,}789.3 $$

Answer: a. mean 1,700, standard deviation 106.07, normal; b. 0.7974; c. the $80^{\text{th}}$ percentile is 1,789.3 years, so 80% of samples of 50 users have a combined age at or below 1,789.3 years.

Try it in rāSHio

This example runs both directions on one distribution, and rāSHio runs them the same two ways. Choose Distributions → Normal, set the mean to 1700 and the standard deviation to 106.07, then pick Between with bounds 1500 and 1800 for part b’s 0.7974 — and switch to Inverse (Left) with probability 0.80 for part c’s 1,789.3.

6.2.4 Sums in Context: Screen Time and Salaries

Both commands earn their keep when a single problem asks for a percentile and a probability on the same distribution, which is how most exam questions are built. The other habit worth practicing here is unit conversion. A problem will hand you a mean in minutes and then ask about "ten hours", and the arithmetic is trivial but forgetting it is not: convert everything to one unit before you touch the calculator, and state which unit your answer is in when you write it down.

The interpretation sentence matters as much as the number. A percentile of 587.76 minutes means nothing on its own; what it means is that ninety-five percent of the sums of app engagement times are at most 587.76 minutes. Whenever a question says "interpret this value in a complete sentence", it is asking you to name the percentage, name the quantity, and name the unit, in that order.

Try It Now 6.2.4

Carmen Ibarra, a product analyst, finds that the mean number of minutes for app engagement by a tablet user is 8.2 minutes. Suppose the standard deviation is one minute. She takes a sample size of 70.

a. What is the probability that the sum of the sample is between seven hours and ten hours? What does this mean in context of the problem?

b. Find the $84^{\text{th}}$ and $16^{\text{th}}$ percentiles for the sum of the sample. Interpret these values in context.

Solution

Step 1 — Convert to one unit and find the parameters. Seven hours is 420 minutes and ten hours is 600 minutes.

$$ \mu_{\Sigma X} = 70(8.2) = 574 \text{ minutes} \qquad \sigma_{\Sigma X} = (\sqrt{70})(1) \approx 8.37 \text{ minutes} $$

Part a — A between-question.

$$ P(420 < \Sigma x < 600) = \text{normalcdf}(420, 600, (70)(8.2), (\sqrt{70})(1)) = 0.9991 $$

In context: for a sample of 70 tablet users, there is about a 99.91% chance that their combined app engagement time falls between seven and ten hours.

Part b — Two percentiles.

$$ k_{84} = \text{invNorm}(0.84, 574, 8.37) = 582.32 \text{ minutes} $$ $$ k_{16} = \text{invNorm}(0.16, 574, 8.37) = 565.68 \text{ minutes} $$

In context: 84% of the sums of app engagement times are at most 582.32 minutes, and 16% of them are at most 565.68 minutes. Between them sits the middle 68% of all such totals.

Answer: a. 0.9991; b. $84^{\text{th}}$ percentile 582.32 minutes, $16^{\text{th}}$ percentile 565.68 minutes.

Example 6.2.3: App Engagement Time

Luis Ontiveros tracks app engagement on tablets and finds a mean of 8.2 minutes per user. Suppose the standard deviation is one minute. He takes a sample of size 70.

a. What are the mean and standard deviation for the sums?

b. Find the $95^{\text{th}}$ percentile for the sum of the sample. Interpret this value in a complete sentence.

c. Find the probability that the sum of the sample is at least ten hours.

Solution

Step 1 — Part a: scale the parameters.

$$ \mu_{\Sigma X} = n\mu_X = 70(8.2) = 574 \text{ minutes} $$ $$ \sigma_{\Sigma X} = (\sqrt{n})(\sigma_X) = (\sqrt{70})(1) = 8.37 \text{ minutes} $$

Step 2 — Part b: a percentile, so invNorm. Let $k$ be the $95^{\text{th}}$ percentile.

$$ k = \text{invNorm}(0.95, (70)(8.2), (\sqrt{70})(1)) = 587.76 \text{ minutes} $$

Ninety five percent of the sums of app engagement times are at most 587.76 minutes.

Step 3 — Part c: convert the unit first. Ten hours is 600 minutes, and "at least" shades the right tail.

$$ P(\Sigma x \geq 600) = \text{normalcdf}(600, 1E99, (70)(8.2), (\sqrt{70})(1)) = 0.0009 $$

Answer: a. mean 574 minutes, standard deviation 8.37 minutes; b. the $95^{\text{th}}$ percentile is 587.76 minutes, meaning ninety five percent of the sums of app engagement times are at most 587.76 minutes; c. 0.0009.

Problem Set 6.2

Use the following information to answer the next four exercises: An unknown distribution has a mean of 80 and a standard deviation of 12. A sample size of 95 is drawn randomly from the population.

Problem 1. Find the probability that the sum of the 95 values is greater than 7,650.

Solution

Step 1 — Scale the two parameters: the mean of the sums multiplies by $n$ and the standard deviation multiplies by $\sqrt{n}$.

$$ \mu_{\Sigma X} = (95)(80) = 7{,}600 \qquad \sigma_{\Sigma X} = (\sqrt{95})(12) \approx 116.96 $$

Step 2 — Shade the right tail above 7,650: "greater than" means the area to the right.

$$ P(\Sigma x > 7{,}650) = \text{normalcdf}(7650, 1E99, (95)(80), (\sqrt{95})(12)) = 0.3345 $$

Answer: $P(\Sigma x > 7{,}650) = 0.3345$

Problem 2. Find the probability that the sum of the 95 values is less than 7,400.

Solution

Step 1 — Use the same parameters: $\mu_{\Sigma X} = 7{,}600$ and $\sigma_{\Sigma X} \approx 116.96$.

Step 2 — Shade the left tail below 7,400: "less than" means the area to the left, so the lower bound runs off to negative infinity.

$$ P(\Sigma x < 7{,}400) = \text{normalcdf}(-1E99, 7400, (95)(80), (\sqrt{95})(12)) = 0.0436 $$

Answer: $P(\Sigma x < 7{,}400) = 0.0436$

Problem 3. Find the sum that is two standard deviations above the mean of the sums.

Solution

Step 1 — Recognize this as a $z$-score run backwards: "two standard deviations above the mean" is $z = 2$, so use $\Sigma x = (n)(\mu_X) + (z)(\sqrt{n})(\sigma_X)$.

Step 2 — Substitute:

$$ \Sigma x = 7{,}600 + (2)(116.96) = 7{,}833.92 $$

Answer: The sum two standard deviations above the mean of the sums is 7,833.92.

Problem 4. Find the sum that is 1.5 standard deviations below the mean of the sums.

Solution

Step 1 — Below the mean makes $z$ negative: $z = -1.5$.

Step 2 — Substitute:

$$ \Sigma x = 7{,}600 + (-1.5)(116.96) = 7{,}600 - 175.44 = 7{,}424.56 $$

Answer: The sum 1.5 standard deviations below the mean of the sums is 7,424.56.

Use the following information to answer the next five exercises: The distribution of results from a cholesterol test has a mean of 180 and a standard deviation of 20. A sample size of 40 is drawn randomly.

Problem 5. Find the probability that the sum of the 40 values is greater than 7,500.

Solution

Step 1 — Scale the parameters for a sample of 40:

$$ \mu_{\Sigma X} = (40)(180) = 7{,}200 \qquad \sigma_{\Sigma X} = (\sqrt{40})(20) \approx 126.49 $$

Step 2 — Right tail above 7,500:

$$ P(\Sigma x > 7{,}500) = \text{normalcdf}(7500, 1E99, (40)(180), (\sqrt{40})(20)) = 0.0089 $$

Answer: $P(\Sigma x > 7{,}500) = 0.0089$

Problem 6. Find the probability that the sum of the 40 values is less than 7,000.

Solution

Step 1 — Same parameters: $\mu_{\Sigma X} = 7{,}200$, $\sigma_{\Sigma X} \approx 126.49$.

Step 2 — Left tail below 7,000:

$$ P(\Sigma x < 7{,}000) = \text{normalcdf}(-1E99, 7000, (40)(180), (\sqrt{40})(20)) = 0.0569 $$

Answer: $P(\Sigma x < 7{,}000) = 0.0569$

Problem 7. Find the sum that is one standard deviation above the mean of the sums.

Solution

Step 1 — Set $z = 1$: one standard deviation above the mean.

Step 2 — Substitute:

$$ \Sigma x = 7{,}200 + (1)(126.49) = 7{,}326.49 $$

Answer: The sum one standard deviation above the mean of the sums is 7,326.49.

Problem 8. Find the sum that is 1.5 standard deviations below the mean of the sums.

Solution

Step 1 — Set $z = -1.5$: below the mean, so the $z$-score is negative.

Step 2 — Substitute:

$$ \Sigma x = 7{,}200 + (-1.5)(126.49) = 7{,}200 - 189.74 = 7{,}010.26 $$

Answer: The sum 1.5 standard deviations below the mean of the sums is 7,010.26.

Problem 9. Find the percentage of sums between 1.5 standard deviations below the mean of the sums and one standard deviation above the mean of the sums.

Solution

Step 1 — Translate the sentence into $z$-scores: the region runs from $z = -1.5$ to $z = 1$. Because the bounds are given in standard deviations, you can work on the standard normal curve directly.

Step 2 — Subtract the two areas to the left:

$$ P(-1.5 < z < 1) = 0.8413 - 0.0668 = 0.7745 $$

Answer: About 77.45% of the sums fall in that region.

Use the following information to answer the next six exercises: Priya Raman measures the amount of sugar in several cans of the same soda. The mean is 39.01 with a standard deviation of 0.5. She randomly selects a sample of 100.

Problem 10. Find the probability that the sum of the 100 values is greater than 3,910.

Solution

Step 1 — Scale the parameters for a sample of 100:

$$ \mu_{\Sigma X} = (100)(39.01) = 3{,}901 \qquad \sigma_{\Sigma X} = (\sqrt{100})(0.5) = 5 $$

Step 2 — Right tail above 3,910:

$$ P(\Sigma x > 3{,}910) = \text{normalcdf}(3910, 1E99, 3901, 5) = 0.0359 $$

Answer: $P(\Sigma x > 3{,}910) = 0.0359$

Problem 11. Find the probability that the sum of the 100 values is less than 3,900.

Solution

Step 1 — Same parameters: $\mu_{\Sigma X} = 3{,}901$, $\sigma_{\Sigma X} = 5$.

Step 2 — Left tail below 3,900:

$$ P(\Sigma x < 3{,}900) = \text{normalcdf}(-1E99, 3900, 3901, 5) = 0.4207 $$

Answer: $P(\Sigma x < 3{,}900) = 0.4207$

Problem 12. Find the probability that the sum of the 100 values falls between the numbers you found in Problem 6.2.10 and Problem 6.2.11.

Solution

Step 1 — Identify the two boundaries: the previous two problems used 3,900 and 3,910, so this is the strip between them.

Step 2 — Run one normalcdf with both bounds:

$$ P(3{,}900 < \Sigma x < 3{,}910) = \text{normalcdf}(3900, 3910, 3901, 5) = 0.5433 $$

Step 3 — Sanity check: the two tails were 0.4207 and 0.0359, and $1 - 0.4207 - 0.0359 = 0.5434$, which agrees to rounding.

Answer: $P(3{,}900 < \Sigma x < 3{,}910) = 0.5433$

Problem 13. Find the sum with a $z$-score of $-2.5$.

Solution

Step 1 — Run the $z$-score backwards: $\Sigma x = (n)(\mu_X) + (z)(\sqrt{n})(\sigma_X)$.

Step 2 — Substitute $z = -2.5$:

$$ \Sigma x = 3{,}901 + (-2.5)(5) = 3{,}901 - 12.5 = 3{,}888.5 $$

Answer: The sum with a $z$-score of $-2.5$ is 3,888.5.

Problem 14. Find the sum with a $z$-score of 0.5.

Solution

Step 1 — Same formula, positive $z$:

$$ \Sigma x = 3{,}901 + (0.5)(5) = 3{,}901 + 2.5 = 3{,}903.5 $$

Answer: The sum with a $z$-score of 0.5 is 3,903.5.

Problem 15. Find the probability that the sums will fall between the $z$-scores $-2$ and 1.

Solution

Step 1 — The bounds are already $z$-scores: no conversion is needed, so work on the standard normal curve.

Step 2 — Subtract the two areas to the left:

$$ P(-2 < z < 1) = 0.8413 - 0.0228 = 0.8186 $$

Answer: $P(-2 < z < 1) = 0.8186$

Use the following information to answer the next four exercises: An unknown distribution has a mean 12 and a standard deviation of one. A sample size of 25 is taken. Let $X$ = the object of interest.

Problem 16. What is the mean of $\Sigma X$?

Solution

Step 1 — The mean of the sums multiplies by $n$:

$$ \mu_{\Sigma X} = (n)(\mu_X) = (25)(12) = 300 $$

Answer: The mean of $\Sigma X$ is 300.

Problem 17. What is the standard deviation of $\Sigma X$?

Solution

Step 1 — The standard deviation multiplies by $\sqrt{n}$, not $n$:

$$ \sigma_{\Sigma X} = (\sqrt{n})(\sigma_X) = (\sqrt{25})(1) = 5 $$

Answer: The standard deviation of $\Sigma X$ is 5.

Problem 18. What is $P(\Sigma x = 290)$?

Solution

Step 1 — Ask whether $\Sigma X$ is discrete or continuous: by the central limit theorem $\Sigma X$ is normal, and a normal distribution is continuous.

Step 2 — A single point has no width: on a continuous distribution the area over one exact value is zero, so the probability of landing on it exactly is zero.

Answer: $P(\Sigma x = 290) = 0$

Problem 19. What is $P(\Sigma x > 290)$?

Solution

Step 1 — Use $\mu_{\Sigma X} = 300$ and $\sigma_{\Sigma X} = 5$:

$$ z = \frac{290 - 300}{5} = -2 $$

Step 2 — Take the area to the right of $z = -2$:

$$ P(\Sigma x > 290) = 1 - 0.0228 = 0.9772 $$

Answer: $P(\Sigma x > 290) = 0.9772$

Problem 20. True or False: only the sums of normal distributions are also normal distributions.

Solution

Step 1 — Recall what the central limit theorem actually claims: it says that sums from any distribution — skewed, uniform, bimodal, anything — approach a normal distribution as the sample size grows. The starting distribution does not have to be normal.

Step 2 — Test the statement against that: the claim restricts the result to normal starting distributions, which is narrower than the theorem.

Answer: False. Sums from any distribution tend toward normal as $n$ increases.

Problem 21. In order for the sums of a distribution to approach a normal distribution, what must be true?

Solution

Step 1 — Identify the condition in the theorem: the approximation improves as more values go into each sum.

Answer: The sample size $n$ must get larger.

Problem 22. What three things must you know about a distribution to find the probability of sums?

Solution

Step 1 — List what normalcdf needs for a sum: it needs the mean of the sums $(n)(\mu_X)$ and the standard deviation of the sums $(\sqrt{n})(\sigma_X)$, and both are built out of three ingredients.

Answer: The mean of the original distribution, the standard deviation of the original distribution, and the sample size.

Problem 23. An unknown distribution has a mean of 25 and a standard deviation of six. Let $X$ = one object from this distribution. What is the sample size if the standard deviation of $\Sigma X$ is 42?

Solution

Step 1 — Write the formula and put the known value in:

$$ \sigma_{\Sigma X} = (\sqrt{n})(\sigma_X) \quad \Longrightarrow \quad 42 = (\sqrt{n})(6) $$

Step 2 — Solve for $n$:

$$ \sqrt{n} = \frac{42}{6} = 7 \quad \Longrightarrow \quad n = 7^2 = 49 $$

Answer: The sample size is 49.

Problem 24. An unknown distribution has a mean of 19 and a standard deviation of 20. Let $X$ = the object of interest. What is the sample size if the mean of $\Sigma X$ is 15,200?

Solution

Step 1 — Use the mean formula this time:

$$ \mu_{\Sigma X} = (n)(\mu_X) \quad \Longrightarrow \quad 15{,}200 = (n)(19) $$

Step 2 — Solve for $n$:

$$ n = \frac{15{,}200}{19} = 800 $$

Answer: The sample size is 800. (The standard deviation of 20 is not needed here.)

Use the following information to answer the next three exercises. Jared Whitlow, a market researcher, analyzes how many electronics devices customers buy in a single purchase. The distribution has a mean of three with a standard deviation of 0.7. He samples 400 customers.

Problem 25. What is the $z$-score for $\Sigma x = 840$?

Solution

Step 1 — Scale the parameters for 400 customers:

$$ \mu_{\Sigma X} = (400)(3) = 1{,}200 \qquad \sigma_{\Sigma X} = (\sqrt{400})(0.7) = (20)(0.7) = 14 $$

Step 2 — Apply the $z$-score formula:

$$ z = \frac{840 - 1{,}200}{14} = \frac{-360}{14} \approx -25.71 $$

Answer: $z \approx -25.71$. A total that far below the mean is effectively impossible for this sample size.

Problem 26. What is the $z$-score for $\Sigma x = 1{,}186$?

Solution

Step 1 — Same parameters: $\mu_{\Sigma X} = 1{,}200$, $\sigma_{\Sigma X} = 14$.

Step 2 — Apply the $z$-score formula:

$$ z = \frac{1{,}186 - 1{,}200}{14} = \frac{-14}{14} = -1 $$

Answer: $z = -1.00$

Problem 27. What is $P(\Sigma x < 1{,}186)$?

Solution

Step 1 — Reuse the $z$-score from the previous problem: $\Sigma x = 1{,}186$ sits at $z = -1$.

Step 2 — Read the area to the left of $z = -1$:

$$ P(\Sigma x < 1{,}186) = 0.1587 $$

Answer: $P(\Sigma x < 1{,}186) = 0.1587$

Use the following information to answer the next three exercises: An unknown distribution has a mean of 100, a standard deviation of 100, and a sample size of 100. Let $X$ = one object of interest.

Problem 28. What is the mean of $\Sigma X$?

Solution

Step 1 — Multiply the mean by the sample size:

$$ \mu_{\Sigma X} = (100)(100) = 10{,}000 $$

Answer: The mean of $\Sigma X$ is 10,000.

Problem 29. What is the standard deviation of $\Sigma X$?

Solution

Step 1 — Multiply the standard deviation by $\sqrt{n}$:

$$ \sigma_{\Sigma X} = (\sqrt{100})(100) = (10)(100) = 1{,}000 $$

Answer: The standard deviation of $\Sigma X$ is 1,000.

Problem 30. What is $P(\Sigma x > 9{,}000)$?

Solution

Step 1 — Find the $z$-score of 9,000:

$$ z = \frac{9{,}000 - 10{,}000}{1{,}000} = -1 $$

Step 2 — Take the area to the right:

$$ P(\Sigma x > 9{,}000) = 1 - 0.1587 = 0.8413 $$

Answer: $P(\Sigma x > 9{,}000) = 0.8413$

Problem 31. Which of the following is NOT TRUE about the theoretical distribution of sums?

a) The mean, median and mode are equal.

b) The area under the curve is one.

c) The curve never touches the $x$-axis.

d) The curve is skewed to the right.

Solution

Step 1 — Check each statement against the shape of a normal curve. A normal curve is symmetric, its total area is 1, and its tails approach the horizontal axis without ever touching it.

Step 2 — Test them one at a time: a is true because symmetry forces the mean, median, and mode to coincide; b is true because every probability distribution has total area 1; c is true because the tails are asymptotic to the $x$-axis.

Step 3 — That leaves d: a normal curve is symmetric, so it is not skewed in either direction.

Answer: d. The curve is skewed to the right — this is the statement that is NOT true.

Problem 32. Suppose that the duration of a particular type of criminal trial is known to have a mean of 21 days and a standard deviation of seven days. We randomly sample nine trials.

a) In words, $\Sigma X$ = ____

b) $\Sigma X \sim$ ____

c) Find the probability that the total length of the nine trials is at least 225 days.

d) Ninety percent of the total of nine of these types of trials will last at least how long?

Solution

Step 1 — Part a: say what the sum means in words. $\Sigma X$ is the total length of time, in days, for nine criminal trials of this type.

Step 2 — Part b: scale the parameters.

$$ \mu_{\Sigma X} = (9)(21) = 189 \qquad \sigma_{\Sigma X} = (\sqrt{9})(7) = (3)(7) = 21 $$

so $\Sigma X \sim N(189, 21)$.

Step 3 — Part c: "at least 225 days" is a right tail.

$$ P(\Sigma x \geq 225) = \text{normalcdf}(225, 1E99, 189, 21) = 0.0432 $$

Step 4 — Part d: "ninety percent will last at least how long" is a percentile. If 90% last at least that long, then only 10% fall below it, so the area to the left is 0.10.

$$ k = \text{invNorm}(0.10, 189, 21) = 162.09 $$

Answer: a. the total length of time for nine criminal trials; b. $\Sigma X \sim N(189, 21)$; c. 0.0432; d. 162.09 days — ninety percent of the totals of nine such trials will last 162 days or more.

Problem 33. Suppose that the weight of open boxes of cereal in a home with children is uniformly distributed from two to six pounds with a mean of four pounds and standard deviation of 1.1547. We randomly survey 64 homes with children.

a) In words, $X$ = ____

b) The distribution is ____.

c) In words, $\Sigma X$ = ____

d) $\Sigma X \sim$ ____

e) Find the probability that the total weight of open boxes is less than 250 pounds.

f) Find the $35^{\text{th}}$ percentile for the total weight of open boxes of cereal.

Solution

Step 1 — Part a: name the single value. $X$ is the weight, in pounds, of one open box of cereal in a home with children.

Step 2 — Part b: name its distribution. The weights are spread evenly between two and six pounds, so $X \sim U(2, 6)$ — a uniform distribution.

Step 3 — Part c: name the sum. $\Sigma X$ is the total weight, in pounds, of the open boxes of cereal in the 64 surveyed homes.

Step 4 — Part d: scale the parameters.

$$ \mu_{\Sigma X} = (64)(4) = 256 \qquad \sigma_{\Sigma X} = (\sqrt{64})(1.1547) = (8)(1.1547) \approx 9.24 $$

so $\Sigma X \sim N(256, 9.24)$ by the central limit theorem, even though $X$ itself is uniform.

Step 5 — Part e: a left tail below 250 pounds.

$$ P(\Sigma x < 250) = \text{normalcdf}(-1E99, 250, 256, 9.24) = 0.2578 $$

Step 6 — Part f: a percentile, so invNorm.

$$ k = \text{invNorm}(0.35, 256, 9.24) = 252.44 $$

Answer: a. the weight of one open box of cereal; b. uniform, $X \sim U(2, 6)$; c. the total weight of the 64 open boxes; d. $\Sigma X \sim N(256, 9.24)$; e. 0.2578; f. 252.44 pounds.

Problem 34. 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) In words, $X$ = ____

b) $X \sim$ ____

c) In words, $\Sigma X$ = ____

d) $\Sigma X \sim$ ____

e) Find the probability that the teachers earn a total of over $400,000.

f) Find the $90^{\text{th}}$ percentile for an individual teacher's salary.

g) Find the $90^{\text{th}}$ percentile for the sum of ten teachers' salary.

h) If we surveyed 70 teachers instead of ten, graphically, how would that change the distribution in part d?

i) If each of the 70 teachers received a $3,000 raise, graphically, how would that change the distribution in part b?

Solution

Step 1 — Part a: name the single value. $X$ is the salary of one elementary school teacher in the district.

Step 2 — Part b: name its distribution. $X \sim N(44{,}000,\ 6{,}500)$.

Step 3 — Part c: name the sum. $\Sigma X$ is the sum of the salaries of the ten sampled teachers.

Step 4 — Part d: scale the parameters.

$$ \mu_{\Sigma X} = (10)(44{,}000) = 440{,}000 \qquad \sigma_{\Sigma X} = (\sqrt{10})(6{,}500) \approx 20{,}554.80 $$

so $\Sigma X \sim N(440{,}000,\ 20{,}554.80)$.

Step 5 — Part e: "a total of over $400,000" is a right tail.

$$ P(\Sigma x > 400{,}000) = \text{normalcdf}(400000, 1E99, 440000, 20554.80) = 0.9742 $$

Step 6 — Part f: the 90th percentile of ONE salary uses the original parameters.

$$ k = \text{invNorm}(0.90,\ 44{,}000,\ 6{,}500) = 52{,}330.09 $$

Step 7 — Part g: the 90th percentile of the SUM uses the scaled parameters.

$$ k = \text{invNorm}(0.90,\ 440{,}000,\ 20{,}554.80) = 466{,}342.04 $$

Step 8 — Part h: raising $n$ from 10 to 70 changes the spread. The mean of the sums rises to $(70)(44{,}000)$ and the standard deviation rises to $(\sqrt{70})(6{,}500)$, so the curve sits farther right and is wider — and with more values in each sum, the shape is an even better symmetrical normal curve.

Step 9 — Part i: a flat raise shifts, it does not spread. Adding $3,000 to every salary adds $3,000 to the mean of $X$ and leaves the standard deviation alone, so the distribution of $X$ slides right by $3,000 to a mean of $47,000 with the same shape.

Answer: a. the salary of one elementary school teacher; b. $X \sim N(44{,}000, 6{,}500)$; c. the sum of ten teachers' salaries; d. $\Sigma X \sim N(440{,}000, 20{,}554.80)$; e. 0.9742; f. $52,330.09; g. $466,342.04; h. more spread out and more symmetrical; i. shifted right by $3,000, to a mean of $47,000.

Key Terms

sum of a sample — written $\Sigma X$, the total obtained by adding all $n$ values in a random sample.

central limit theorem for sums — the result that $\Sigma X$ is approximately normal for large $n$, with mean $(n)(\mu_X)$ and standard deviation $(\sqrt{n})(\sigma_X)$.

mean of the sums — $(n)(\mu_X)$, the original population mean multiplied by the sample size.

standard deviation of the sums — $(\sqrt{n})(\sigma_X)$, the original population standard deviation multiplied by the square root of the sample size.