A.4 The Uniform Distribution
Learning Objectives
By the end of this section, you will be able to:
- recognize when a situation is uniformly distributed, and write it in the notation \(X \sim U(a, b)\);
- write the probability density function for a uniform distribution and explain why its graph is a rectangle;
- compute the theoretical mean and standard deviation of a uniform distribution and say what each one measures;
- find probabilities and percentiles by treating area under the density as (base)(height);
- solve conditional uniform problems two ways — by shrinking the sample space and by using the conditional probability formula.
Section A.2 set up the rule that governs every continuous distribution: probability is area under a curve, and the total area is 1. This section takes the simplest curve there is — a flat line — and shows you how far that one idea will carry you.
A uniform distribution describes outcomes that are all equally likely. Nothing in the range is favored over anything else. When you round a measurement to the nearest whole number, the leftover error is uniform. When a bus runs every 15 minutes and you arrive without checking the schedule, your wait is uniform. When a computer picks a "random number between 0 and 1", it is drawing from a uniform distribution.
Because every outcome is equally likely, the density curve is a flat horizontal line, and the region under it is a rectangle. That means every probability you will ever compute in this section is the area of a rectangle: base times height. There is no calculus and no table to look up. The whole section is one formula applied carefully.
Casinos, lotteries, and random number generators all advertise the same promise — no outcome is favored. A uniform distribution is the mathematical statement of that promise, and its flat graph is what "fair" looks like on paper.
The one thing you do have to watch is the endpoints. Problems will tell you whether the range is inclusive or exclusive of its endpoints, and the wording matters for how you set the problem up — though, as you will see, it never changes the answer, because a single point in a continuous distribution has zero probability.
A.4.1 Reading a Uniform Distribution
A continuous random variable \(X\) has a uniform distribution on the interval from \(a\) to \(b\) when every value in that interval is equally likely to occur. We write this
$$X \sim U(a, b)$$where \(a\) is the lowest value \(X\) can take and \(b\) is the highest.
Definition A.4.1
The symbol \(\sim\) is read "is distributed as." So \(X \sim U(0, 23)\) is read "\(X\) is uniformly distributed between 0 and 23."
Now we need the height of that flat line. The total area under any density curve has to be 1, and the region is a rectangle with base \(b - a\). If the base is \(b - a\) and the area must be 1, the height has no choice.
If \(X \sim U(a, b)\), the probability density function of \(X\) is
$$f(x) = \frac{1}{b - a} \quad \text{for } a \leq x \leq b$$and \(f(x) = 0\) everywhere else.
Squeeze the interval and the rectangle has to grow taller to keep its area at 1, like the same amount of water poured into a narrower glass. That is the entire content of \(f(x) = \frac{1}{b-a}\).
Definition A.4.2
Notice what that height is not: it is not a probability. A density of \(\frac{1}{15}\) does not mean "there is a 1 in 15 chance." It is the height of a rectangle whose area is the probability. Densities can even exceed 1 — if \(X \sim U(0, 0.5)\), then \(f(x) = 2\) — and nothing is broken, because \((0.5)(2) = 1\) is still a perfectly good total area.
The center and the spread of a uniform distribution follow from the same picture. A rectangle balances at its midpoint, so the mean sits exactly halfway between \(a\) and \(b\). The standard deviation is less obvious, but it depends only on the width of the interval.
If \(X \sim U(a, b)\), the theoretical mean and standard deviation of \(X\) are
$$\mu = \frac{a + b}{2} \qquad \text{and} \qquad \sigma = \sqrt{\frac{(b - a)^{2}}{12}}$$Definition A.4.3
These are the population values. When you collect real data from a uniform process, the sample mean \(\bar{x}\) and the sample standard deviation \(s\) will land near them without matching exactly.
Rosa Delgado runs the weigh station at a harbor, and she logged the total weight, to the nearest pound, of fish caught by passengers on 35 different charter fishing boats on one summer day. Her data follow a uniform distribution where all values between and including zero and 14 are equally likely. The sample mean is 7.9 pounds and the sample standard deviation is 4.33 pounds.
State the values of \(a\) and \(b\). Write the distribution in proper notation, and calculate the theoretical mean and standard deviation.
| 1 | 12 | 4 | 10 | 4 | 14 | 11 |
| 7 | 11 | 4 | 13 | 2 | 4 | 6 |
| 3 | 10 | 0 | 12 | 6 | 9 | 10 |
| 5 | 13 | 4 | 10 | 14 | 12 | 11 |
| 6 | 10 | 11 | 0 | 11 | 13 | 2 |
Solution
Step 1 — Read off \(a\) and \(b\). The problem says all values between and including zero and 14 are equally likely, so \(a = 0\) and \(b = 14\).
Step 2 — Write the distribution.
$$X \sim U(0, 14)$$with density \(f(x) = \frac{1}{14 - 0} = \frac{1}{14}\) for \(0 \leq x \leq 14\).
Step 3 — Theoretical mean.
$$\mu = \frac{a + b}{2} = \frac{0 + 14}{2} = 7 \text{ pounds}$$Step 4 — Theoretical standard deviation.
$$\sigma = \sqrt{\frac{(b - a)^{2}}{12}} = \sqrt{\frac{(14 - 0)^{2}}{12}} = \sqrt{\frac{196}{12}} \approx 4.04 \text{ pounds}$$Answer: \(a = 0\), \(b = 14\), \(X \sim U(0, 14)\), \(\mu = 7\) pounds, \(\sigma \approx 4.04\) pounds. Again the theoretical values sit close to the sample mean of 7.9 and the sample standard deviation of 4.33.
Owen is eight weeks old, and his dads have been timing how long each of his smiles lasts. The data in Table A.4.2 are 55 of those smiling times, in seconds.
| 10.4 | 19.6 | 18.8 | 13.9 | 17.8 | 16.8 | 21.6 | 17.9 | 12.5 | 11.1 | 4.9 |
| 12.8 | 14.8 | 22.8 | 20.0 | 15.9 | 16.3 | 13.4 | 17.1 | 14.5 | 19.0 | 22.8 |
| 1.3 | 0.7 | 8.9 | 11.9 | 10.9 | 7.3 | 5.9 | 3.7 | 17.9 | 19.2 | 9.8 |
| 5.8 | 6.9 | 2.6 | 5.8 | 21.7 | 11.8 | 3.4 | 2.1 | 4.5 | 6.3 | 10.7 |
| 8.9 | 9.4 | 9.4 | 7.6 | 10.0 | 3.3 | 6.7 | 7.8 | 11.6 | 13.8 | 18.6 |
The sample mean is 11.65 seconds and the sample standard deviation is 6.08 seconds. We will assume his smiling times follow a uniform distribution between zero and 23 seconds, inclusive — any smiling time from zero up to and including 23 seconds is equally likely. A histogram built from this sample is an empirical distribution that closely matches the theoretical uniform distribution.
Write the distribution in proper notation, state the probability density function, and find the theoretical mean and standard deviation. Compare them to the sample values.
Solution
Step 1 — Define the random variable. Let \(X\) = the length, in seconds, of one of Owen's smiles.
Step 2 — Identify \(a\) and \(b\). The lowest possible value is \(a = 0\) and the highest is \(b = 23\), so
$$X \sim U(0, 23)$$Step 3 — Write the density. Using \(f(x) = \frac{1}{b-a}\),
$$f(x) = \frac{1}{23 - 0} = \frac{1}{23} \quad \text{for } 0 \leq x \leq 23$$
Figure A.4.1 — The probability density function for \(X \sim U(0, 23)\). The graph is a rectangle of height \(\frac{1}{23}\) sitting over the interval from 0 to 23 seconds; its total area is \((23)\left(\frac{1}{23}\right) = 1\).
Step 4 — Compute the theoretical mean and standard deviation.
$$\mu = \frac{a + b}{2} = \frac{0 + 23}{2} = 11.50 \text{ seconds}$$ $$\sigma = \sqrt{\frac{(b - a)^{2}}{12}} = \sqrt{\frac{(23 - 0)^{2}}{12}} = 6.64 \text{ seconds}$$Answer: \(X \sim U(0, 23)\), \(f(x) = \frac{1}{23}\) for \(0 \leq x \leq 23\), \(\mu = 11.50\) seconds, and \(\sigma = 6.64\) seconds. The theoretical values sit close to the sample mean of 11.65 seconds and the sample standard deviation of 6.08 seconds, which is what we would expect if the uniform model is a reasonable description of his smiling.
A.4.2 Areas, Percentiles, and Conditional Probability
The \(p^{th}\) percentile of a distribution is the value \(k\) such that \(p\) percent of the data fall at or below \(k\). For a continuous distribution this means
$$P(x < k) = \frac{p}{100}$$A probability question hands you the base and asks for the area. A percentile question hands you the area and asks for the base. Same rectangle, same equation — you are just solving for the other letter.
Definition A.4.4
The value \(k\) is sometimes called a critical value.
Once you have the rectangle, every question in this section is a question about part of its area. Three question types come up over and over, and all three reduce to (base)(height).
"What is the probability that \(X\) falls between two values?" Shade the strip between them. Its base is the distance between the two values and its height is \(f(x)\), so \(P(c < x < d) = (d - c)\left(\frac{1}{b-a}\right)\).
"What value has 90% of the data below it?" Now you know the area and you are solving for the base. Set (base)(height) equal to 0.90 and solve for the unknown endpoint. That endpoint is a percentile.
"What is the probability of \(X\) being large, given that we already know \(X\) is at least this big?" Knowing something has already happened throws away part of the rectangle, so you rebuild the density over what is left.
Because the answers are areas, and a single point has no width, \(P(x = c) = 0\) for every value \(c\). That is why the inclusive-versus-exclusive wording never changes a numerical answer: \(P(2 < x < 18)\) and \(P(2 \leq x \leq 18)\) are the same number.
A distribution is given as \(X \sim U(0, 20)\). What is \(P(2 < x < 18)\)? Find the \(90^{th}\) percentile.
Solution
Step 1 — Write the density. With \(a = 0\) and \(b = 20\),
$$f(x) = \frac{1}{20 - 0} = \frac{1}{20} \quad \text{for } 0 \leq x \leq 20$$Step 2 — Find the probability. The base runs from 2 to 18, a width of 16:
$$P(2 < x < 18) = (18 - 2)\left(\frac{1}{20}\right) = \frac{16}{20} = 0.8$$Step 3 — Find the \(90^{th}\) percentile. Set the area from 0 to \(k\) equal to 0.90 and solve:
$$(k - 0)\left(\frac{1}{20}\right) = 0.90$$ $$k = (20)(0.90) = 18$$Answer: \(P(2 < x < 18) = 0.8\), and the \(90^{th}\) percentile is 18.
Refer to Example A.4.1, where \(X \sim U(0, 23)\) is the smiling time of an eight-week-old baby.
a. What is the probability that a randomly chosen eight-week-old baby smiles between two and 18 seconds?
b. Find the \(90^{th}\) percentile for an eight-week-old baby's smiling time.
c. Find the probability that a random eight-week-old baby smiles more than 12 seconds, knowing that the baby smiles more than eight seconds.
Solution
Part a — a probability, so find the area. The base runs from 2 to 18 and the height is \(\frac{1}{23}\):
$$P(2 < x < 18) = (\text{base})(\text{height}) = (18 - 2)\left(\frac{1}{23}\right) = \frac{16}{23} \approx 0.6957$$
Figure A.4.2 — \(P(2 < x < 18)\) is the shaded strip between \(x = 2\) and \(x = 18\) under the density \(f(x) = \frac{1}{23}\).
Part b — a percentile, so solve for the base. Ninety percent of the smiling times fall below the \(90^{th}\) percentile, \(k\), so \(P(x < k) = 0.90\). The shaded region runs from 0 to \(k\):
$$(\text{base})(\text{height}) = 0.90$$ $$(k - 0)\left(\frac{1}{23}\right) = 0.90$$ $$k = (23)(0.90) = 20.7$$
Figure A.4.3 — The \(90^{th}\) percentile of smiling time. The shaded region from 0 to \(k = 20.7\) seconds holds 90% of the area.
Ninety percent of eight-week-old smiles last 20.7 seconds or less.
Part c — a conditional, so shrink the sample space. We are asked for \(P(x > 12 \mid x > 8)\). There are two ways to do this.
First way: rebuild the density on what is left. You already know the baby smiled more than eight seconds, so the interval below 8 is gone. The new interval runs from 8 to 23, a width of 15, so the new height is
$$f(x) = \frac{1}{23 - 8} = \frac{1}{15} \quad \text{for } 8 < x < 23$$Now the question is an ordinary area question on the new rectangle:
$$P(x > 12 \mid x > 8) = (23 - 12)\left(\frac{1}{15}\right) = \frac{11}{15} \approx 0.7333$$
Figure A.4.4 — The reduced sample space. Knowing \(x > 8\) throws away everything to the left of 8, so the density is rebuilt at height \(\frac{1}{15}\) over the interval from 8 to 23, and \(P(x > 12 \mid x > 8)\) is the shaded strip from 12 to 23.
Second way: use the conditional probability formula from Probability Topics, with the original distribution \(X \sim U(0, 23)\):
$$P(A \mid B) = \frac{P(A \text{ AND } B)}{P(B)}$$Here \(A\) is \((x > 12)\) and \(B\) is \((x > 8)\). Since any \(x\) greater than 12 is automatically greater than 8, the event \(A \text{ AND } B\) is just \(A\):
$$P(x > 12 \mid x > 8) = \frac{P(x > 12 \text{ AND } x > 8)}{P(x > 8)} = \frac{P(x > 12)}{P(x > 8)} = \frac{\frac{11}{23}}{\frac{15}{23}} = \frac{11}{15}$$Figure A.4.5 — The original density from 0 to 23 with the region right of 8 shaded and the region right of 12 shaded darker.
Answer: a. \(\frac{16}{23} \approx 0.6957\); b. \(k = 20.7\) seconds; c. \(\frac{11}{15} \approx 0.7333\).
A.4.3 Working a Complete Uniform Problem
Most exam problems bundle all three question types into one setup: define the variable, write the distribution, find a probability, find the mean and standard deviation, and find a percentile. The next example walks that whole sequence once so you have a template.
The order matters more than it looks. Define the random variable in words before you touch a formula — half of the mistakes people make in this section come from being fuzzy about what \(X\) actually measures. Then read \(a\) and \(b\) straight out of the problem statement and write \(f(x)\); once that height is on paper, every remaining part is arithmetic on a rectangle. Sketch the graph for each part and shade the region you are after, even when you can see the answer without it. The sketch is what tells you whether you are solving for an area or for a base, and it is the only thing that catches a reversed inequality before it costs you the problem.
Hannah Brennan and her wife keep a season log for their favorite club. The total duration of baseball games in the major league in a typical season is uniformly distributed between 447 hours and 521 hours inclusive.
a. Find \(a\) and \(b\) and describe what they represent.
b. Write the distribution.
c. Find the mean and the standard deviation.
d. What is the probability that the duration of games for a team for a single season is between 480 and 500 hours?
e. What is the \(65^{th}\) percentile for the duration of games for a team in a single season?
Solution
Part a. \(a = 447\) and \(b = 521\). Here \(a\) is the shortest total season duration, in hours, and \(b\) is the longest.
Part b.
$$X \sim U(447, 521) \qquad f(x) = \frac{1}{521 - 447} = \frac{1}{74} \quad \text{for } 447 \leq x \leq 521$$Part c.
$$\mu = \frac{447 + 521}{2} = 484 \text{ hours}$$ $$\sigma = \sqrt{\frac{(521 - 447)^{2}}{12}} = \sqrt{\frac{5476}{12}} \approx 21.36 \text{ hours}$$Part d. The base runs from 480 to 500, a width of 20:
$$P(480 < x < 500) = (500 - 480)\left(\frac{1}{74}\right) = \frac{20}{74} \approx 0.2703$$Part e. Let \(k\) be the \(65^{th}\) percentile. The shaded region starts at \(a = 447\), not at 0:
$$(k - 447)\left(\frac{1}{74}\right) = 0.65$$ $$k - 447 = (0.65)(74) = 48.1$$ $$k = 495.1 \text{ hours}$$Answer: a. \(a = 447\) hours (shortest), \(b = 521\) hours (longest); b. \(X \sim U(447, 521)\); c. \(\mu = 484\) hours, \(\sigma \approx 21.36\) hours; d. about 0.2703; e. about 495.1 hours.
The amount of time, in minutes, that a person must wait for a bus is uniformly distributed between zero and 15 minutes, inclusive.
a. What is the probability that a person waits fewer than 12.5 minutes?
b. On average, how long must a person wait? Find the mean, \(\mu\), and the standard deviation, \(\sigma\).
c. Ninety percent of the time, the time a person must wait falls below what value? (This asks for the \(90^{th}\) percentile.)
Solution
Step 1 — Define the variable and the distribution. Let \(X\) = the number of minutes a person must wait for a bus. Then \(a = 0\) and \(b = 15\), so
$$X \sim U(0, 15) \qquad f(x) = \frac{1}{15 - 0} = \frac{1}{15} \quad \text{for } 0 \leq x \leq 15$$Part a — find \(P(x < 12.5)\). Draw the graph and shade from 0 to 12.5:
$$P(x < 12.5) = (\text{base})(\text{height}) = (12.5 - 0)\left(\frac{1}{15}\right) = 0.8333$$
Figure A.4.6 — \(P(x < 12.5)\) for a wait uniformly distributed on 0 to 15 minutes. The shaded strip covers \(\frac{12.5}{15}\) of the rectangle.
The probability a person waits less than 12.5 minutes is 0.8333.
Part b — mean and standard deviation.
$$\mu = \frac{a + b}{2} = \frac{0 + 15}{2} = 7.5$$On average, a person must wait 7.5 minutes.
$$\sigma = \sqrt{\frac{(b - a)^{2}}{12}} = \sqrt{\frac{(15 - 0)^{2}}{12}} = 4.3$$The standard deviation is 4.3 minutes.
Part c — the \(90^{th}\) percentile. Let \(k\) be the \(90^{th}\) percentile. Draw the graph and shade from 0 to \(k\):
$$P(x < k) = (\text{base})(\text{height}) = (k - 0)\left(\frac{1}{15}\right)$$ $$0.90 = (k)\left(\frac{1}{15}\right)$$ $$k = (0.90)(15) = 13.5$$
Figure A.4.7 — The \(90^{th}\) percentile of bus-wait time. The shaded region from 0 to \(k = 13.5\) minutes holds 90% of the area.
Answer: a. 0.8333; b. \(\mu = 7.5\) minutes and \(\sigma = 4.3\) minutes; c. the \(90^{th}\) percentile is 13.5 minutes, so ninety percent of the time a person waits at most 13.5 minutes.
A.4.4 Two Ways to Handle a Conditional
Example A.4.2 part c showed both methods for a conditional probability. They are worth naming, because on a test you want to be able to pick whichever one is faster and then check with the other.
To find \(P(x > c \mid x > d)\) when \(X \sim U(a, b)\), use either method:
Method 1 — Reduce the sample space. The condition tells you \(X\) is somewhere in the smaller interval from \(d\) to \(b\). Write a new density on that interval, \(f(x) = \frac{1}{b - d}\), and compute the ordinary area you want on the new rectangle.
Method 2 — Use the conditional formula. Keep the original density and compute \(P(x > c \mid x > d) = \frac{P(x > c \text{ AND } x > d)}{P(x > d)}\). When \(c > d\), the numerator simplifies to \(P(x > c)\).
Method 1 tends to be faster when the condition chops off one clean end of the interval, which is almost always. Method 2 is safer when the events overlap in a way that is easy to misread, because the formula forces you to write down exactly which event is in the numerator. Whichever you choose, the sanity check is the same: a conditional probability computed on a shrunken sample space should come out larger than the unconditional one, since you have thrown away outcomes that would have counted against you.
Kai Lor tutors in the campus math lab, and they have noticed that the time it takes a student to finish the weekly quiz is uniformly distributed between six and 15 minutes, inclusive. Let \(X\) = the time, in minutes, it takes a student to finish a quiz. Then \(X \sim U(6, 15)\).
Find the probability that a randomly selected student needs at least eight minutes to complete the quiz. Then find the probability that a different student needs at least eight minutes to finish the quiz given that they have already taken more than seven minutes.
Solution
Step 1 — Write the density. With \(a = 6\) and \(b = 15\),
$$f(x) = \frac{1}{15 - 6} = \frac{1}{9} \quad \text{for } 6 \leq x \leq 15$$Step 2 — The unconditional probability. "At least eight minutes" is the strip from 8 to 15:
$$P(x \geq 8) = (15 - 8)\left(\frac{1}{9}\right) = \frac{7}{9} \approx 0.7778$$Step 3 — The conditional probability, by reducing the sample space. Knowing the student has already taken more than seven minutes, the interval now runs from 7 to 15, a width of 8, so the new density is \(f(x) = \frac{1}{8}\):
$$P(x \geq 8 \mid x > 7) = (15 - 8)\left(\frac{1}{8}\right) = \frac{7}{8} = 0.875$$Step 4 — Check with the formula. Kai can confirm the same number without redrawing anything:
$$P(x \geq 8 \mid x > 7) = \frac{P(x \geq 8)}{P(x > 7)} = \frac{\frac{7}{9}}{\frac{8}{9}} = \frac{7}{8}$$Answer: \(P(x \geq 8) = \frac{7}{9} \approx 0.7778\) and \(P(x \geq 8 \mid x > 7) = \frac{7}{8} = 0.875\). The conditional probability is larger, exactly as it should be — one minute of waiting has already been ruled out.
Andi Ramírez is nine, and their moms have timed enough Sunday donuts to know the job takes between 0.5 and 4 minutes, inclusive. Let \(X\) = the time, in minutes, it takes a nine-year-old child to eat a donut. Then \(X \sim U(0.5, 4)\).
a. The probability that a randomly selected nine-year-old child eats a donut in at least two minutes is ____.
b. Find the probability that a different nine-year-old child eats a donut in more than two minutes given that the child has already been eating the donut for more than 1.5 minutes.
Solution
Step 1 — Write the density. With \(a = 0.5\) and \(b = 4\),
$$f(x) = \frac{1}{4 - 0.5} = \frac{1}{3.5} = \frac{2}{7} \quad \text{for } 0.5 \leq x \leq 4$$Part a — an ordinary area. "At least two minutes" means the strip from 2 to 4:
$$P(x \geq 2) = (4 - 2)\left(\frac{2}{7}\right) = \frac{4}{7} \approx 0.5714$$Part b — a conditional, done both ways.
First way: reduce the sample space. Since you know the child has already been eating for more than 1.5 minutes, you are no longer starting at \(a = 0.5\) minutes. Your starting point is 1.5 minutes, so the new interval has width \(4 - 1.5 = 2.5\) and the new density is
$$f(x) = \frac{1}{4 - 1.5} = \frac{2}{5} \quad \text{for } 1.5 \leq x \leq 4$$
Figure A.4.8 — The reduced sample space for the donut problem. Knowing the child has eaten for more than 1.5 minutes rebuilds the density at height \(\frac{2}{5}\) over the interval from 1.5 to 4 minutes.
$$P(x > 2 \mid x > 1.5) = (\text{base})(\text{new height}) = (4 - 2)\left(\frac{2}{5}\right) = \frac{4}{5}$$Second way: use the conditional formula on the original distribution \(X \sim U(0.5, 4)\):
$$P(x > 2 \mid x > 1.5) = \frac{P(x > 2 \text{ AND } x > 1.5)}{P(x > 1.5)} = \frac{P(x > 2)}{P(x > 1.5)} = \frac{\frac{2}{3.5}}{\frac{2.5}{3.5}} = 0.8 = \frac{4}{5}$$Both routes give the same number, which is the point of having two of them.
Answer: a. 0.5714; b. \(\frac{4}{5}\). The probability that a nine-year-old child eats a donut in more than two minutes, given that the child has already been eating the donut for more than 1.5 minutes, is \(\frac{4}{5}\).
A.4.5 Percentiles from Both Tails
So far every percentile we found was measured from the left: "90% of the data is below \(k\)." Problems also ask the question from the right — "the longest 25% of repair times take at least how long?" — and that is where people lose points.
The fix is mechanical. If a problem describes the upper portion, either shade the right tail and set its area equal to that proportion, or convert to a left-tail statement first. The longest 25% start at the same place the shortest 75% end, so the value that cuts off the top quarter is the \(75^{th}\) percentile. Say that conversion out loud before you compute, and the arithmetic takes care of itself.
The amount of time a service technician needs to change the oil in a car is uniformly distributed between 11 and 21 minutes. Let \(X\) = the time needed to change the oil on a car.
a. Write the random variable \(X\) in words. \(X\) = ____.
b. Write the distribution.
c. Graph the distribution.
d. Find \(P(x > 19)\).
e. Find the \(50^{th}\) percentile.
Solution
Part a. \(X\) = the number of minutes it takes a service technician to change the oil in a car.
Part b. With \(a = 11\) and \(b = 21\),
$$X \sim U(11, 21) \qquad f(x) = \frac{1}{21 - 11} = \frac{1}{10} \quad \text{for } 11 \leq x \leq 21$$Part c. The graph is a rectangle of height \(\frac{1}{10}\) sitting over the interval from 11 to 21 on the horizontal axis, with the horizontal axis labelled "time in minutes" and the vertical axis labelled \(f(x)\). Its area is \((21 - 11)\left(\frac{1}{10}\right) = 1\).
Part d. The base runs from 19 to 21, a width of 2:
$$P(x > 19) = (21 - 19)\left(\frac{1}{10}\right) = 0.2$$Part e. The \(50^{th}\) percentile of a uniform distribution is its midpoint, but solve it out to be sure:
$$(k - 11)\left(\frac{1}{10}\right) = 0.50$$ $$k - 11 = 5$$ $$k = 16 \text{ minutes}$$Answer: a. the time in minutes needed to change the oil on a car; b. \(X \sim U(11, 21)\) with \(f(x) = \frac{1}{10}\) on \(11 \leq x \leq 21\); c. a rectangle of height \(\frac{1}{10}\) from 11 to 21; d. 0.2; e. 16 minutes, which is also the mean.
Ace Heating and Air Conditioning Service finds that the amount of time a repair technician needs to fix a furnace is uniformly distributed between 1.5 and four hours. Let \(X\) = the time needed to fix a furnace. Then \(X \sim U(1.5, 4)\).
a. Find the probability that a randomly selected furnace repair requires more than two hours.
b. Find the probability that a randomly selected furnace repair requires less than three hours.
c. Find the \(30^{th}\) percentile of furnace repair times.
d. The longest 25% of furnace repair times take at least how long? (In other words, find the minimum time for the longest 25% of repair times.) What percentile does this represent?
e. Find the mean and standard deviation.
Solution
Step 1 — Write the density. With \(a = 1.5\) and \(b = 4\),
$$f(x) = \frac{1}{4 - 1.5} = \frac{1}{2.5} = 0.4 \quad \text{for } 1.5 \leq x \leq 4$$Part a — \(P(x > 2)\).
$$P(x > 2) = (\text{base})(\text{height}) = (4 - 2)(0.4) = 0.8$$
Figure A.4.9 — Uniform distribution between 1.5 and four hours, with the area between two and four shaded to represent the probability that the repair time \(x\) is greater than two hours.
Part b — \(P(x < 3)\).
$$P(x < 3) = (\text{base})(\text{height}) = (3 - 1.5)(0.4) = 0.6$$The rectangle is the same as before, but the shading now runs between \(x = 1.5\) and \(x = 3\). Note that the shaded area starts at \(x = 1.5\) rather than at \(x = 0\) — since \(X \sim U(1.5, 4)\), \(x\) cannot be less than 1.5.
Figure A.4.10 — Uniform distribution between 1.5 and four hours, with the area between 1.5 and three shaded to represent the probability that the repair time \(x\) is less than three hours.
Part c — the \(30^{th}\) percentile. Set the left-hand area equal to 0.30 and solve for \(k\):
$$P(x < k) = 0.30$$ $$P(x < k) = (\text{base})(\text{height}) = (k - 1.5)(0.4)$$ $$0.30 = (k - 1.5)(0.4)$$ $$0.75 = k - 1.5 \qquad \text{dividing both sides by } 0.4$$ $$k = 2.25 \qquad \text{adding 1.5 to both sides}$$
Figure A.4.11 — Uniform distribution between 1.5 and 4 hours with an area of 0.30 shaded to the left, representing the shortest 30% of repair times.
The \(30^{th}\) percentile of repair times is 2.25 hours: 30% of repair times are 2.25 hours or less.
Part d — the longest 25%. Now the area is on the right:
$$P(x > k) = 0.25$$ $$P(x > k) = (\text{base})(\text{height}) = (4 - k)(0.4)$$ $$0.25 = (4 - k)(0.4)$$ $$0.625 = 4 - k \qquad \text{dividing both sides by } 0.4$$ $$k = 3.375 \qquad \text{after subtracting four from both sides}$$
Figure A.4.12 — Uniform distribution between 1.5 and 4 hours with an area of 0.25 shaded to the right, representing the longest 25% of repair times.
The longest 25% of furnace repairs take at least 3.375 hours. Since 25% of repair times are 3.375 hours or longer, 75% of repair times are 3.375 hours or less — so 3.375 hours is the \(75^{th}\) percentile of furnace repair times.
Part e — mean and standard deviation.
$$\mu = \frac{a + b}{2} = \frac{1.5 + 4}{2} = 2.75 \text{ hours}$$ $$\sigma = \sqrt{\frac{(b - a)^{2}}{12}} = \sqrt{\frac{(4 - 1.5)^{2}}{12}} = 0.7217 \text{ hours}$$Answer: a. 0.8; b. 0.6; c. 2.25 hours; d. 3.375 hours, which is the \(75^{th}\) percentile; e. \(\mu = 2.75\) hours and \(\sigma = 0.7217\) hours.
Problem Set A.4
Use the following information to answer the next ten questions. Grant Whitfield appraises houses in the county where he and his husband live, and the data that follow are the square footage (in 1,000 feet squared) of 28 homes he measured. The sample mean is 2.50 and the sample standard deviation is 0.8302. The distribution can be written as \(X \sim U(1.5, 4.5)\).
| 1.5 | 2.4 | 3.6 | 2.6 | 1.6 | 2.4 | 2.0 |
| 3.5 | 2.5 | 1.8 | 2.4 | 2.5 | 3.5 | 4.0 |
| 2.6 | 1.6 | 2.2 | 1.8 | 3.8 | 2.5 | 1.5 |
| 2.8 | 1.8 | 4.5 | 1.9 | 1.9 | 3.1 | 1.6 |
Problem 1. What type of distribution is this?
Solution
Step 1 — Read the description: every value between 1.5 and 4.5 is equally likely, and square footage can take any value in that range, not just whole numbers.
Answer: It is a continuous uniform distribution, written \(X \sim U(1.5, 4.5)\).
Problem 2. In this distribution, outcomes are equally likely. What does this mean?
Solution
Step 1 — Translate "equally likely" into a statement about \(x\): in a uniform distribution no value in the interval is favored over any other, so the density has the same height everywhere on the interval.
Answer: It means the value of \(x\) is just as likely to be any number between 1.5 and 4.5. No range of a given width inside the interval is more likely than another range of the same width.
Problem 3. What is the height of \(f(x)\) for the continuous probability distribution?
Solution
Step 1 — Use the uniform density formula: for \(X \sim U(a, b)\), the height is \(f(x) = \frac{1}{b-a}\).
Step 2 — Substitute \(a = 1.5\) and \(b = 4.5\):
$$f(x) = \frac{1}{4.5 - 1.5} = \frac{1}{3} \approx 0.3333$$Answer: \(f(x) = \frac{1}{3} \approx 0.3333\).
Problem 4. What are the constraints for the values of \(x\)?
Solution
Step 1 — Read \(a\) and \(b\) off the distribution: \(X \sim U(1.5, 4.5)\) means \(x\) cannot fall below 1.5 or above 4.5.
Answer: \(1.5 \leq x \leq 4.5\).
Problem 5. Graph \(P(2 < x < 3)\).
Solution
Step 1 — Draw the density: put square footage (in thousands) on the horizontal axis and \(f(x)\) on the vertical axis. Draw a rectangle of height \(\frac{1}{3}\) stretching from \(x = 1.5\) to \(x = 4.5\).
Step 2 — Shade the region asked for: shade the strip between \(x = 2\) and \(x = 3\). Its base is 1 unit wide and its height is \(\frac{1}{3}\).
Answer: The graph is a rectangle of height \(\frac{1}{3}\) over the interval from 1.5 to 4.5, with the strip from \(x = 2\) to \(x = 3\) shaded. That shaded strip is \(P(2 < x < 3)\).
Problem 6. What is \(P(2 < x < 3)\)?
Solution
Step 1 — Set up the area: probability is (base)(height), with base \(3 - 2\) and height \(\frac{1}{3}\).
$$P(2 < x < 3) = (3 - 2)\left(\frac{1}{3}\right) = \frac{1}{3}$$Answer: \(P(2 < x < 3) = \frac{1}{3} \approx 0.3333\).
Problem 7. What is \(P(x < 3.5 \mid x < 4)\)?
Solution
Step 1 — Recognize the conditional: knowing \(x < 4\) throws away everything above 4, so the sample space shrinks to the interval from 1.5 to 4.
Step 2 — Rebuild the density on the reduced interval: the new width is \(4 - 1.5 = 2.5\), so
$$f(x) = \frac{1}{2.5} = 0.4 \quad \text{for } 1.5 \leq x < 4$$Step 3 — Find the area below 3.5 on the new rectangle:
$$P(x < 3.5 \mid x < 4) = (3.5 - 1.5)(0.4) = (2)(0.4) = 0.8$$Answer: \(P(x < 3.5 \mid x < 4) = 0.8\).
Problem 8. What is \(P(x = 1.5)\)?
Solution
Step 1 — Remember that probability is area: a single point has no width, so the rectangle above it has base 0.
$$P(x = 1.5) = (0)\left(\frac{1}{3}\right) = 0$$Answer: \(P(x = 1.5) = 0\). In any continuous distribution the probability of landing on one exact value is zero.
Problem 9. What is the \(90^{th}\) percentile of square footage for homes?
Solution
Step 1 — Write what the percentile means: let \(k\) be the \(90^{th}\) percentile, so \(P(x < k) = 0.90\). The shaded region runs from 1.5 to \(k\), not from 0.
Step 2 — Set (base)(height) equal to the area and solve:
$$(k - 1.5)\left(\frac{1}{3}\right) = 0.90$$ $$k - 1.5 = (0.90)(3) = 2.7$$ $$k = 4.2$$Answer: The \(90^{th}\) percentile is 4.2 thousand square feet, or 4,200 square feet. Ninety percent of these homes are 4,200 square feet or smaller.
Problem 10. Find the probability that a randomly selected home has more than 3,000 square feet given that you already know the house has more than 2,000 square feet.
Solution
Step 1 — Put the question in the units of the distribution: the data are in thousands of square feet, so 3,000 square feet is \(x = 3\) and 2,000 square feet is \(x = 2\). We want \(P(x > 3 \mid x > 2)\).
Step 2 — Reduce the sample space: knowing \(x > 2\) leaves the interval from 2 to 4.5, a width of 2.5, so the new height is \(\frac{1}{2.5} = 0.4\).
Step 3 — Find the area above 3 on the new rectangle:
$$P(x > 3 \mid x > 2) = (4.5 - 3)(0.4) = (1.5)(0.4) = 0.6$$Answer: 0.6.
Use the following information to answer the next eight exercises. A distribution is given as \(X \sim U(0, 12)\).
Problem 11. What is \(a\)? What does it represent?
Solution
Step 1 — Read the first parameter out of \(X \sim U(0, 12)\): the first number in the notation is always the left endpoint.
Answer: \(a = 0\), and it represents the lowest value \(x\) can take.
Problem 12. What is \(b\)? What does it represent?
Solution
Step 1 — Read the second parameter out of \(X \sim U(0, 12)\): the second number is the right endpoint.
Answer: \(b = 12\), and it represents the highest value \(x\) can take.
Problem 13. What is the probability density function?
Solution
Step 1 — Apply \(f(x) = \frac{1}{b-a}\):
$$f(x) = \frac{1}{12 - 0} = \frac{1}{12} \quad \text{for } 0 \leq x \leq 12$$Answer: \(f(x) = \frac{1}{12}\) for \(0 \leq x \leq 12\), and \(f(x) = 0\) everywhere else.
Problem 14. What is the theoretical mean?
Solution
Step 1 — Use the midpoint formula: a rectangle balances at its center.
$$\mu = \frac{a + b}{2} = \frac{0 + 12}{2} = 6$$Answer: \(\mu = 6\).
Problem 15. What is the theoretical standard deviation?
Solution
Step 1 — Apply the uniform standard deviation formula:
$$\sigma = \sqrt{\frac{(b - a)^{2}}{12}} = \sqrt{\frac{(12 - 0)^{2}}{12}} = \sqrt{\frac{144}{12}} = \sqrt{12}$$Answer: \(\sigma = \sqrt{12} \approx 3.4641\).
Problem 16. Draw the graph of the distribution for \(P(x > 9)\).
Solution
Step 1 — Draw the density: a rectangle of height \(\frac{1}{12}\) over the interval from \(x = 0\) to \(x = 12\), with \(x\) on the horizontal axis and \(f(x)\) on the vertical axis.
Step 2 — Shade the region for \(P(x > 9)\): shade everything to the right of \(x = 9\), a strip 3 units wide running from 9 to 12.
Answer: The graph is a rectangle of height \(\frac{1}{12}\) from 0 to 12 with the strip from \(x = 9\) to \(x = 12\) shaded.
Problem 17. Find \(P(x > 9)\).
Solution
Step 1 — Compute the shaded area: base \(12 - 9 = 3\), height \(\frac{1}{12}\).
$$P(x > 9) = (12 - 9)\left(\frac{1}{12}\right) = \frac{3}{12} = 0.25$$Answer: \(P(x > 9) = 0.25\).
Problem 18. Find the \(40^{th}\) percentile.
Solution
Step 1 — Write the percentile equation: let \(k\) be the \(40^{th}\) percentile, so \(P(x < k) = 0.40\).
$$(k - 0)\left(\frac{1}{12}\right) = 0.40$$Step 2 — Solve for \(k\):
$$k = (0.40)(12) = 4.8$$Answer: The \(40^{th}\) percentile is 4.8.
Use the following information to answer the next eleven exercises. The age of cars in the staff parking lot of a suburban college is uniformly distributed from six months (0.5 years) to 9.5 years.
Problem 19. What is being measured here?
Solution
Step 1 — Identify the quantity the data record: each observation is one car's age, recorded in years.
Answer: The age, in years, of the cars parked in the staff parking lot of a suburban college.
Problem 20. In words, define the random variable \(X\).
Solution
Step 1 — Write the random variable as a sentence: a random variable is always "the [measurement] of [one randomly chosen unit]".
Answer: \(X\) = the age (in years) of a randomly chosen car in the staff parking lot.
Problem 21. Are the data discrete or continuous?
Solution
Step 1 — Ask whether the values can be counted or must be measured: age can be any value in the interval, including fractions of a year, so it is measured rather than counted.
Answer: Continuous.
Problem 22. The interval of values for \(x\) is ____.
Solution
Step 1 — Convert six months to years: six months is 0.5 years, and the problem gives 9.5 years as the upper end.
Answer: \(0.5 \leq x \leq 9.5\), that is, 0.5 to 9.5 years.
Problem 23. The distribution for \(X\) is ____.
Solution
Step 1 — Write the distribution in proper notation using \(a = 0.5\) and \(b = 9.5\).
Answer: \(X \sim U(0.5, 9.5)\).
Problem 24. Write the probability density function.
Solution
Step 1 — Find the width of the interval: \(9.5 - 0.5 = 9\).
Step 2 — Take the reciprocal:
$$f(x) = \frac{1}{9} \quad \text{for } 0.5 \leq x \leq 9.5$$Answer: \(f(x) = \frac{1}{9}\) where \(x\) is between 0.5 and 9.5, inclusive.
Problem 25. Graph the probability distribution.
a) Sketch the graph of the probability distribution.
b) Identify the following values:
i. Lowest value for \(x\): ____
ii. Highest value for \(x\): ____
iii. Height of the rectangle: ____
iv. Label for x-axis (words): ____
v. Label for y-axis (words): ____
Solution
Part a — sketch. Draw a rectangle whose left edge sits at \(x = 0.5\), whose right edge sits at \(x = 9.5\), and whose height is \(\frac{1}{9}\). Check the area: \((9)\left(\frac{1}{9}\right) = 1\), as it must be.
Part b — the five values.
i. Lowest value for \(x\): 0.5
ii. Highest value for \(x\): 9.5
iii. Height of the rectangle: \(\frac{1}{9} \approx 0.1111\)
iv. Label for x-axis (words): age of the car, in years
v. Label for y-axis (words): probability density, \(f(x)\)
Answer: A rectangle of height \(\frac{1}{9}\) running from 0.5 to 9.5 on an axis labelled "age of the car, in years", with the vertical axis labelled \(f(x)\).
Problem 26. Find the average age of the cars in the lot.
Solution
Step 1 — The average of a uniform distribution is its midpoint:
$$\mu = \frac{a + b}{2} = \frac{0.5 + 9.5}{2} = 5$$Answer: \(\mu = 5\) years.
Problem 27. Find the probability that a randomly chosen car in the lot was less than four years old.
a) Sketch the graph, and shade the area of interest.
b) Find the probability. \(P(x < 4) =\) ____
Solution
Part a — sketch. Draw the rectangle of height \(\frac{1}{9}\) from 0.5 to 9.5 and shade everything to the left of \(x = 4\), a strip running from 0.5 to 4.
Part b — compute the area. The base is \(4 - 0.5 = 3.5\) and the height is \(\frac{1}{9}\):
$$P(x < 4) = (4 - 0.5)\left(\frac{1}{9}\right) = \frac{3.5}{9} \approx 0.3889$$Answer: \(P(x < 4) = \frac{3.5}{9} \approx 0.3889\).
Problem 28. Considering only the cars less than 7.5 years old, find the probability that a randomly chosen car in the lot was less than four years old.
a) Sketch the graph, and shade the area of interest.
b) Find the probability. \(P(x < 4 \mid x < 7.5) =\) ____
Solution
Part a — sketch. "Considering only the cars less than 7.5 years old" shrinks the sample space to the interval from 0.5 to 7.5. Redraw the rectangle over that shorter interval, then shade from 0.5 to 4.
Part b — compute the area. The reduced interval has width \(7.5 - 0.5 = 7\), so the new height is \(\frac{1}{7}\):
$$P(x < 4 \mid x < 7.5) = (4 - 0.5)\left(\frac{1}{7}\right) = \frac{3.5}{7} = 0.5$$Answer: \(P(x < 4 \mid x < 7.5) = \frac{3.5}{7} = 0.5\).
Problem 29. What has changed in the previous two problems that made the solutions different?
Solution
Step 1 — Compare the two sample spaces: the first problem used the full range 0.5 to 9.5, a width of 9. The second one conditioned on \(x < 7.5\), which threw away the two oldest years and left a range of width 7.
Step 2 — See what that does to the height: a narrower interval forces a taller rectangle, from \(\frac{1}{9}\) up to \(\frac{1}{7}\). The window we cared about was the same 3.5 years wide both times, but it now covers a bigger share of what is left.
Answer: The condition shrank the sample space. Removing the cars 7.5 years and older raised the density from \(\frac{1}{9}\) to \(\frac{1}{7}\), so the same 3.5-year window went from \(\frac{3.5}{9} \approx 0.3889\) up to \(\frac{3.5}{7} = 0.5\).
Problem 30. Find the third quartile of ages of cars in the lot. This means you will have to find the value such that \(\frac{3}{4}\), or 75%, of the cars are at most (less than or equal to) that age.
a) Sketch the graph, and shade the area of interest.
b) Find the value \(k\) such that \(P(x < k) = 0.75\).
c) The third quartile is ____
Solution
Part a — sketch. Draw the rectangle of height \(\frac{1}{9}\) from 0.5 to 9.5 and shade the left-hand region holding 75% of the area, running from 0.5 up to the unknown value \(k\).
Part b — solve for \(k\).
$$(k - 0.5)\left(\frac{1}{9}\right) = 0.75$$ $$k - 0.5 = (0.75)(9) = 6.75$$ $$k = 7.25$$Part c. The third quartile is 7.25.
Answer: \(k = 7.25\), so the third quartile is 7.25 years. Seventy-five percent of the cars in the lot are 7.25 years old or younger.
For each probability and percentile problem below, draw the picture.
Problem 31. Births are approximately uniformly distributed throughout the year. They can be said to follow a uniform distribution from 0 to 52 (spread of 52 weeks).
a) \(X \sim\) ____
b) Graph the probability distribution.
c) \(f(x) =\) ____
d) \(\mu =\) ____
e) \(\sigma =\) ____
f) Find the probability that a person is born at the exact moment week 19 ends. That is, find \(P(x = 19) =\) ____
g) \(P(2 < x < 31) =\) ____
h) Find the probability that a person is born after week 40.
i) \(P(12 < x \mid x < 28) =\) ____
j) Find the \(70^{th}\) percentile.
k) Find the minimum for the upper quarter.
Solution
Part a. \(X \sim U(0, 52)\), where \(X\) is the week of the year in which a person is born.
Part b. The graph is a rectangle of height \(\frac{1}{52}\) stretching from \(x = 0\) to \(x = 52\), with "week of the year" on the horizontal axis.
Part c. \(f(x) = \frac{1}{52}\) for \(0 \leq x \leq 52\).
Part d.
$$\mu = \frac{0 + 52}{2} = 26$$Part e.
$$\sigma = \sqrt{\frac{(52 - 0)^{2}}{12}} = \sqrt{\frac{2704}{12}} \approx 15.0111$$Part f. A single point has no width, so \(P(x = 19) = 0\).
Part g.
$$P(2 < x < 31) = (31 - 2)\left(\frac{1}{52}\right) = \frac{29}{52} \approx 0.5577$$Part h.
$$P(x > 40) = (52 - 40)\left(\frac{1}{52}\right) = \frac{12}{52} = \frac{3}{13} \approx 0.2308$$Part i. Conditioning on \(x < 28\) shrinks the sample space to the interval from 0 to 28, so the new height is \(\frac{1}{28}\):
$$P(x > 12 \mid x < 28) = (28 - 12)\left(\frac{1}{28}\right) = \frac{16}{28} = \frac{4}{7} \approx 0.5714$$Part j. Let \(k\) be the \(70^{th}\) percentile:
$$k\left(\frac{1}{52}\right) = 0.70 \quad \Rightarrow \quad k = 36.4$$Part k. The minimum for the upper quarter is the \(75^{th}\) percentile:
$$k\left(\frac{1}{52}\right) = 0.75 \quad \Rightarrow \quad k = 39$$Answer: a. \(X \sim U(0, 52)\); c. \(f(x) = \frac{1}{52}\); d. \(\mu = 26\); e. \(\sigma \approx 15.0111\); f. 0; g. \(\frac{29}{52} \approx 0.5577\); h. \(\frac{3}{13} \approx 0.2308\); i. \(\frac{4}{7} \approx 0.5714\); j. 36.4 weeks; k. 39 weeks.
Problem 32. A random number generator picks a number from one to nine in a uniform manner.
a) \(X \sim\) ____
b) Graph the probability distribution.
c) \(f(x) =\) ____
d) \(\mu =\) ____
e) \(\sigma =\) ____
f) \(P(3.5 < x < 7.25) =\) ____
g) \(P(x > 5.67)\)
h) \(P(x > 5 \mid x > 3) =\) ____
i) Find the \(90^{th}\) percentile.
Solution
Part a. \(X \sim U(1, 9)\).
Part b. The graph is a rectangle of height \(\frac{1}{8}\) stretching from \(x = 1\) to \(x = 9\).
Part c. \(f(x) = \frac{1}{8}\) where \(1 \leq x \leq 9\).
Part d.
$$\mu = \frac{1 + 9}{2} = 5$$Part e.
$$\sigma = \sqrt{\frac{(9 - 1)^{2}}{12}} = \sqrt{\frac{64}{12}} \approx 2.3$$Part f.
$$P(3.5 < x < 7.25) = (7.25 - 3.5)\left(\frac{1}{8}\right) = \frac{3.75}{8} = \frac{15}{32} = 0.46875$$Part g.
$$P(x > 5.67) = (9 - 5.67)\left(\frac{1}{8}\right) = \frac{3.33}{8} = \frac{333}{800} = 0.41625$$Part h. Conditioning on \(x > 3\) leaves the interval from 3 to 9, a width of 6:
$$P(x > 5 \mid x > 3) = (9 - 5)\left(\frac{1}{6}\right) = \frac{4}{6} = \frac{2}{3}$$Part i. Let \(k\) be the \(90^{th}\) percentile. The shaded region starts at 1, not 0:
$$(k - 1)\left(\frac{1}{8}\right) = 0.90 \quad \Rightarrow \quad k - 1 = 7.2 \quad \Rightarrow \quad k = 8.2$$Answer: a. \(X \sim U(1, 9)\); c. \(f(x) = \frac{1}{8}\); d. 5; e. about 2.3; f. \(\frac{15}{32}\); g. \(\frac{333}{800}\); h. \(\frac{2}{3}\); i. 8.2.
Problem 33. According to a study by Dr. John McDougall of his live-in weight loss program, the people who follow his program lose between six and 15 pounds a month until they approach trim body weight. Let's suppose that the weight loss is uniformly distributed. We are interested in the weight loss of a randomly selected individual following the program for one month.
a) Define the random variable. \(X\) = ____
b) \(X \sim\) ____
c) Graph the probability distribution.
d) \(f(x) =\) ____
e) \(\mu =\) ____
f) \(\sigma =\) ____
g) Find the probability that the individual lost more than ten pounds in a month.
h) Suppose it is known that the individual lost more than ten pounds in a month. Find the probability that he lost less than 12 pounds in the month.
i) \(P(7 < x < 13 \mid x > 9) =\) ____ . State this in a probability question, similarly to parts g and h, draw the picture, and find the probability.
Solution
Part a. \(X\) = the weight loss, in pounds, of a randomly selected individual who follows the program for one month.
Part b. \(X \sim U(6, 15)\).
Part c. The graph is a rectangle of height \(\frac{1}{9}\) stretching from \(x = 6\) to \(x = 15\), with "pounds lost in one month" on the horizontal axis.
Part d. \(f(x) = \frac{1}{15 - 6} = \frac{1}{9}\) for \(6 \leq x \leq 15\).
Part e.
$$\mu = \frac{6 + 15}{2} = 10.5 \text{ pounds}$$Part f.
$$\sigma = \sqrt{\frac{(15 - 6)^{2}}{12}} = \sqrt{\frac{81}{12}} \approx 2.5981 \text{ pounds}$$Part g.
$$P(x > 10) = (15 - 10)\left(\frac{1}{9}\right) = \frac{5}{9} \approx 0.5556$$Part h. Conditioning on \(x > 10\) leaves the interval from 10 to 15, a width of 5, so the new height is \(\frac{1}{5}\):
$$P(x < 12 \mid x > 10) = (12 - 10)\left(\frac{1}{5}\right) = \frac{2}{5} = 0.4$$Part i. As a question: given that the individual lost more than nine pounds, find the probability that the loss was between seven and 13 pounds. Conditioning on \(x > 9\) leaves the interval from 9 to 15, a width of 6, so the new height is \(\frac{1}{6}\). The part of "between 7 and 13" that survives is the stretch from 9 to 13:
$$P(7 < x < 13 \mid x > 9) = (13 - 9)\left(\frac{1}{6}\right) = \frac{4}{6} = \frac{2}{3} \approx 0.6667$$Answer: b. \(X \sim U(6, 15)\); d. \(f(x) = \frac{1}{9}\); e. 10.5 pounds; f. about 2.5981 pounds; g. \(\frac{5}{9} \approx 0.5556\); h. 0.4; i. \(\frac{2}{3} \approx 0.6667\).
Problem 34. A subway train arrives every eight minutes during rush hour. We are interested in the length of time a commuter must wait for a train to arrive. The time follows a uniform distribution.
a) Define the random variable. \(X\) = ____
b) \(X \sim\) ____
c) Graph the probability distribution.
d) \(f(x) =\) ____
e) \(\mu =\) ____
f) \(\sigma =\) ____
g) Find the probability that the commuter waits less than one minute.
h) Find the probability that the commuter waits between three and four minutes.
i) Sixty percent of commuters wait more than how long for the train? State this in a probability question, similarly to parts g and h, draw the picture, and find the probability.
Solution
Part a. \(X\) = the length of time, in minutes, a commuter must wait for the train to arrive.
Part b. A train comes every eight minutes and the commuter arrives at a random moment, so the wait is uniform on 0 to 8: \(X \sim U(0, 8)\).
Part c. The graph is a rectangle of height \(\frac{1}{8}\) stretching from \(x = 0\) to \(x = 8\).
Part d. \(f(x) = \frac{1}{8}\) where \(0 \leq x \leq 8\).
Part e.
$$\mu = \frac{0 + 8}{2} = 4 \text{ minutes}$$Part f.
$$\sigma = \sqrt{\frac{(8 - 0)^{2}}{12}} = \sqrt{\frac{64}{12}} \approx 2.31 \text{ minutes}$$Part g.
$$P(x < 1) = (1 - 0)\left(\frac{1}{8}\right) = \frac{1}{8} = 0.125$$Part h.
$$P(3 < x < 4) = (4 - 3)\left(\frac{1}{8}\right) = \frac{1}{8} = 0.125$$Part i. As a question: find the value \(k\) for which \(P(x > k) = 0.60\). Shade the right-hand region:
$$(8 - k)\left(\frac{1}{8}\right) = 0.60 \quad \Rightarrow \quad 8 - k = 4.8 \quad \Rightarrow \quad k = 3.2$$Answer: b. \(X \sim U(0, 8)\); d. \(f(x) = \frac{1}{8}\); e. four minutes; f. about 2.31 minutes; g. 0.125; h. 0.125; i. sixty percent of commuters wait more than 3.2 minutes.
Problem 35. The age of a first grader on September 1 at Garden Elementary School is uniformly distributed from 5.8 to 6.8 years. We randomly select one first grader from the class.
a) Define the random variable. \(X\) = ____
b) \(X \sim\) ____
c) Graph the probability distribution.
d) \(f(x) =\) ____
e) \(\mu =\) ____
f) \(\sigma =\) ____
g) Find the probability that she is over 6.5 years old.
h) Find the probability that she is between four and six years old.
i) Find the \(70^{th}\) percentile for the age of first graders on September 1 at Garden Elementary School.
Solution
Part a. \(X\) = the age, in years, of a randomly selected first grader at Garden Elementary School on September 1.
Part b. \(X \sim U(5.8, 6.8)\).
Part c. The graph is a rectangle of height 1 stretching from \(x = 5.8\) to \(x = 6.8\). The interval is only one unit wide, so the height has to be 1 for the area to come out to 1.
Part d. \(f(x) = \frac{1}{6.8 - 5.8} = 1\) for \(5.8 \leq x \leq 6.8\).
Part e.
$$\mu = \frac{5.8 + 6.8}{2} = 6.3 \text{ years}$$Part f.
$$\sigma = \sqrt{\frac{(6.8 - 5.8)^{2}}{12}} = \sqrt{\frac{1}{12}} \approx 0.2887 \text{ years}$$Part g.
$$P(x > 6.5) = (6.8 - 6.5)(1) = 0.3$$Part h. The distribution has no area below 5.8, so only the stretch from 5.8 to 6 counts:
$$P(4 < x < 6) = (6 - 5.8)(1) = 0.2$$Part i. Let \(k\) be the \(70^{th}\) percentile. The shaded region starts at 5.8:
$$(k - 5.8)(1) = 0.70 \quad \Rightarrow \quad k = 6.5 \text{ years}$$Answer: b. \(X \sim U(5.8, 6.8)\); d. \(f(x) = 1\); e. 6.3 years; f. about 0.2887 years; g. 0.3; h. 0.2; i. 6.5 years.
Use the following information to answer the next three exercises. The Sky Train from the terminal to the rental-car and long-term parking center is supposed to arrive every eight minutes. The waiting times for the train are known to follow a uniform distribution.
Problem 36. What is the average waiting time (in minutes)?
a) zero
b) two
c) three
d) four
Solution
Step 1 — Write the distribution: the train is supposed to arrive every eight minutes and you show up at a random moment, so \(X \sim U(0, 8)\).
Step 2 — Use the midpoint formula:
$$\mu = \frac{0 + 8}{2} = 4$$Answer: d. four minutes.
Problem 37. Find the \(30^{th}\) percentile for the waiting times (in minutes).
a) two
b) 2.4
c) 2.75
d) three
Solution
Step 1 — Set the left-hand area equal to 0.30: with \(X \sim U(0, 8)\) the height is \(\frac{1}{8}\).
$$(k - 0)\left(\frac{1}{8}\right) = 0.30$$Step 2 — Solve for \(k\):
$$k = (0.30)(8) = 2.4$$Answer: b. 2.4 minutes.
Problem 38. The probability of waiting more than seven minutes given a person has waited more than four minutes is?
a) 0.125
b) 0.25
c) 0.5
d) 0.75
Solution
Step 1 — Reduce the sample space: knowing the person has waited more than four minutes leaves the interval from 4 to 8, a width of 4, so the new height is \(\frac{1}{4}\).
Step 2 — Find the area above 7 on the new rectangle:
$$P(x > 7 \mid x > 4) = (8 - 7)\left(\frac{1}{4}\right) = 0.25$$Answer: b. 0.25.
Problem 39. The time (in minutes) until the next bus departs a major bus depot follows a distribution with \(f(x) = \frac{1}{20}\) where \(x\) goes from 25 to 45 minutes.
a) Define the random variable. \(X\) = ____
b) \(X \sim\) ____
c) Graph the probability distribution.
d) The distribution is ____ (name of distribution). It is ____ (discrete or continuous).
e) \(\mu =\) ____
f) \(\sigma =\) ____
g) Find the probability that the time is at most 30 minutes. Sketch and label a graph of the distribution. Shade the area of interest. Write the answer in a probability statement.
h) Find the probability that the time is between 30 and 40 minutes. Sketch and label a graph of the distribution. Shade the area of interest. Write the answer in a probability statement.
i) \(P(25 < x < 55) =\) ____ . State this in a probability statement, similarly to parts g and h, draw the picture, and find the probability.
j) Find the \(90^{th}\) percentile. This means that 90% of the time, the time is less than ____ minutes.
k) Find the \(75^{th}\) percentile. In a complete sentence, state what this means. (See part j.)
l) Find the probability that the time is more than 40 minutes given (or knowing that) it is at least 30 minutes.
Solution
Part a. \(X\) = the time, in minutes, until the next bus departs the depot.
Part b. \(X \sim U(25, 45)\). Check: \(\frac{1}{45 - 25} = \frac{1}{20}\), which matches the density the problem gives.
Part c. The graph is a rectangle of height \(\frac{1}{20}\) stretching from \(x = 25\) to \(x = 45\).
Part d. The distribution is uniform, and it is continuous.
Part e.
$$\mu = \frac{25 + 45}{2} = 35 \text{ minutes}$$Part f.
$$\sigma = \sqrt{\frac{(45 - 25)^{2}}{12}} = \sqrt{\frac{400}{12}} \approx 5.7735 \text{ minutes}$$Part g. Shade from 25 to 30:
$$P(x \leq 30) = (30 - 25)\left(\frac{1}{20}\right) = \frac{5}{20} = 0.25$$The probability that the next bus departs within 30 minutes is 0.25.
Part h. Shade from 30 to 40:
$$P(30 < x < 40) = (40 - 30)\left(\frac{1}{20}\right) = \frac{10}{20} = 0.5$$The probability that the wait is between 30 and 40 minutes is 0.5.
Part i. The distribution stops at 45, so nothing beyond 45 contributes any area. The shaded region is the whole rectangle:
$$P(25 < x < 55) = P(25 < x < 45) = (45 - 25)\left(\frac{1}{20}\right) = 1$$Part j. Let \(k\) be the \(90^{th}\) percentile:
$$(k - 25)\left(\frac{1}{20}\right) = 0.90 \quad \Rightarrow \quad k - 25 = 18 \quad \Rightarrow \quad k = 43$$Ninety percent of the time, the time is less than 43 minutes.
Part k. Let \(k\) be the \(75^{th}\) percentile:
$$(k - 25)\left(\frac{1}{20}\right) = 0.75 \quad \Rightarrow \quad k - 25 = 15 \quad \Rightarrow \quad k = 40$$Seventy-five percent of the time, the next bus departs within 40 minutes.
Part l. Conditioning on \(x \geq 30\) leaves the interval from 30 to 45, a width of 15, so the new height is \(\frac{1}{15}\):
$$P(x > 40 \mid x \geq 30) = (45 - 40)\left(\frac{1}{15}\right) = \frac{5}{15} = \frac{1}{3} \approx 0.3333$$Answer: b. \(X \sim U(25, 45)\); d. uniform and continuous; e. 35 minutes; f. about 5.7735 minutes; g. 0.25; h. 0.5; i. 1; j. 43 minutes; k. 40 minutes; l. \(\frac{1}{3} \approx 0.3333\).
Problem 40. Suppose that the value of a stock varies each day from $16 to $25 with a uniform distribution.
a) Find the probability that the value of the stock is more than $19.
b) Find the probability that the value of the stock is between $19 and $22.
c) Find the upper quartile — 25% of all days the stock is above what value? Draw the graph.
d) Given that the stock is greater than $18, find the probability that the stock is more than $21.
Solution
Step 1 — Write the density. The value runs from \(a = 16\) to \(b = 25\), a width of 9, so
$$f(x) = \frac{1}{25 - 16} = \frac{1}{9}$$Part a.
$$P(x > 19) = (25 - 19)\left(\frac{1}{9}\right) = \frac{6}{9} = \frac{2}{3} \approx 0.6667$$Part b.
$$P(19 < x < 22) = (22 - 19)\left(\frac{1}{9}\right) = \frac{3}{9} = \frac{1}{3} \approx 0.3333$$Part c. The upper quartile cuts off the top 25% of the area, so the shaded strip on the right has area 0.25:
$$0.25 = (\text{width})\left(\frac{1}{9}\right) \quad \Rightarrow \quad \text{width} = (0.25)(9) = 2.25$$Starting from the right end, the cutoff is \(25 - 2.25 = 22.75\). On the graph, shade the strip from \(x = 22.75\) to \(x = 25\).
Part d. This is a conditional. Do it either way.
Reduce the sample space: knowing the stock is above 18 leaves the interval from 18 to 25, a width of 7, so the new height is \(\frac{1}{7}\):
$$P(x > 21 \mid x > 18) = (25 - 21)\left(\frac{1}{7}\right) = \frac{4}{7}$$Or use the formula:
$$P(x > 21 \mid x > 18) = \frac{P(x > 21)}{P(x > 18)} = \frac{25 - 21}{25 - 18} = \frac{4}{7}$$Answer: a. \(\frac{2}{3}\); b. \(\frac{1}{3}\); c. $22.75, so 25% of days the stock closes above $22.75; d. \(\frac{4}{7} \approx 0.5714\).
Problem 41. A fireworks show is designed so that the time between fireworks is between one and five seconds, and follows a uniform distribution.
a) Find the average time between fireworks.
b) Find the probability that the time between fireworks is greater than four seconds.
Solution
Step 1 — Write the distribution. The gap between fireworks is uniform on 1 to 5 seconds, so \(X \sim U(1, 5)\) and
$$f(x) = \frac{1}{5 - 1} = \frac{1}{4}$$Part a.
$$\mu = \frac{1 + 5}{2} = 3 \text{ seconds}$$Part b.
$$P(x > 4) = (5 - 4)\left(\frac{1}{4}\right) = 0.25$$Answer: a. three seconds; b. 0.25.
Problem 42. The number of miles driven by a truck driver falls between 300 and 700, and follows a uniform distribution.
a) Find the probability that the truck driver goes more than 650 miles in a day.
b) Find the probability that the truck driver goes between 400 and 650 miles in a day.
c) At least how many miles does the truck driver travel on the furthest 10% of days?
Solution
Step 1 — Write the distribution. Miles driven are uniform on 300 to 700, a width of 400, so \(X \sim U(300, 700)\) and \(f(x) = \frac{1}{400}\).
Part a.
$$P(x > 650) = \frac{700 - 650}{700 - 300} = \frac{50}{400} = \frac{1}{8} = 0.125$$Part b.
$$P(400 < x < 650) = \frac{650 - 400}{700 - 300} = \frac{250}{400} = 0.625$$Part c. The furthest 10% of days sit in the right-hand strip whose area is 0.10:
$$0.10 = \frac{\text{width}}{700 - 300} \quad \Rightarrow \quad \text{width} = (400)(0.10) = 40$$Measuring 40 miles back from the right end gives \(700 - 40 = 660\).
Answer: a. 0.125; b. 0.625; c. the driver travels at least 660 miles on the furthest 10% of days.
Key Terms
uniform distribution — a continuous distribution in which every value in the interval from \(a\) to \(b\) is equally likely; written \(X \sim U(a, b)\).
probability density function — for a uniform distribution, the constant height \(f(x) = \frac{1}{b-a}\) on \(a \leq x \leq b\), whose area over an interval gives the probability of landing in it.
percentile — the value \(k\) with a given percent of the data at or below it, found by setting (base)(height) equal to that proportion and solving for \(k\).
critical value — another name for the value \(k\) that cuts off a specified area of a distribution.