The Uniform Distribution

a. Refer to element-229. What is the probability that a randomly chosen eight-week-old baby smiles between two and 18 seconds?

P(2 < x < 18) = (base)(height) = (18 – 2)$\left(\frac{1}{23}\right)$ = $\frac{16}{23}$

View original figure (external; image not reproduced in this text edition)

Original figure description: This graph shows a uniform distribution. The horizontal axis ranges from 0 to 15. The distribution is modeled by a rectangle extending from x = 0 to x = 15. A region from x = 2 to x = 18 is shaded inside the rectangle.

b. Find the 90^th percentile for an eight-week-old baby's smiling time.

b. Ninety percent of the smiling times fall below the 90^th percentile, k, so P(x < k) = 0.90.

$P(x<k)=0.90$ $\left(\text{base}\right)\left(\text{height}\right)=0.90$ $\text{(}k-0\text{)}\left(\frac{1}{23}\right)=0.90$$k=\left(23\right)\left(0.90\right)=20.7$

View original figure (external; image not reproduced in this text edition)

Original figure description: This shows the graph of the function f(x) = 1/15. A horiztonal line ranges from the point (0, 1/15) to the point (15, 1/15). A vertical line extends from the x-axis to the end of the line at point (15, 1/15) creating a rectangle. A region is shaded inside the rectangle from x = 0 to x = k. The shaded area represents P(x < k) = 0.90.

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.

c. This probability question is a conditional. You are asked to find the probability that an eight-week-old baby smiles more than 12 seconds when you already know the baby has smiled for more than eight seconds.

Find P(x > 12|x > 8) There are two ways to do the problem. For the first way, use the fact that this is a conditional and changes the sample space. The graph illustrates the new sample space. You already know the baby smiled more than eight seconds.

for 8 < x < 23

Write a new f(x): f(x) = $\frac{1}{23-\text{ 8}}$ = $\frac{1}{15}$ for 8 < x < 23.

P(x > 12|x > 8) = (23 − 12)$\left(\frac{1}{15}\right)$ = $\frac{11}{15}$

View original figure (external; image not reproduced in this text edition)

Original figure description: f(X)=1/15 graph displaying a boxed region consisting of a horizontal line extending to the right from point 1/15 on the y-axis, a vertical upward line from points 8 and 23 on the x-axis, and the x-axis. A shaded region from points 12-23 occurs within this area.

For the second way, use the conditional formula from Probability Topics with the original distribution.

P(A|B) = $\frac{P\left(A\text{ AND }B\right)}{P\left(B\right)}$

For this problem, A is (x > 12) and B is (x > 8).

So, P(x > 12|x > 8) = $\frac{(x>12\text{ AND }x>8)}{P(x>8)}=\frac{P(x>12)}{P(x>8)}=\frac{\frac{11}{23}}{\frac{15}{23}}=\frac{11}{15}$

View original figure (external; image not reproduced in this text edition)

Original figure description: This shows the graph of the function f(x) = 1/23. A horizontal line ranges from the point (0, 1/23) to the point (23, 1/23). A vertical line extends from the x-axis to the end of the line at point (23, 1/23) creating a rectangle. A shaded region occurs with vertical lines that extend from the horizontal axis to the graph at x = 8 and x = 12.

a. What is the probability that a person waits fewer than 12.5 minutes?

a. Let X = the number of minutes a person must wait for a bus. a = 0 and b = 15. X ~ U(0, 15). Write the probability density function. f (x) = $\frac{1}{15-0}$ = $\frac{1}{15}$ for 0 ≤ x ≤ 15.

Find P (x < 12.5). Draw a graph.

$P(x<k)=\left(\text{base}\right)\left(\text{height}\right)=(12.5-0)\left(\frac{1}{15}\right)=0.8333$

The probability a person waits fewer than 12.5 minutes is 0.8333.

View original figure (external; image not reproduced in this text edition)

Original figure description: This shows the graph of the function f(x) = 1/15. A horiztonal line ranges from the point (0, 1/15) to the point (15, 1/15). A vertical line extends from the x-axis to the end of the line at point (15, 1/15) creating a rectangle. A region is shaded inside the rectangle from x = 0 to x = 12.5.

b. On the average, how long must a person wait? Find the mean, μ, and the standard deviation, σ.

b. μ = $\frac{a+b}{2}$ = $\frac{15+0}{2}$ = 7.5. On the average, a person must wait 7.5 minutes.

σ = $\sqrt{\frac{(b-a{)}^2}{12}}=\sqrt{\frac{(\mathrm{15}-0{)}^2}{12}}$ = 4.3. The standard deviation is 4.3 minutes.

c. Ninety percent of the time, the minutes a person must wait falls below what value?

c. Find the 90^th percentile. Draw a graph. Let k = the 90^th percentile.

$P(x<k)=\left(\text{base}\right)\left(\text{height}\right)=(k-0)\left(\frac{1}{15}\right)$

$0.90=\left(k\right)\left(\frac{1}{15}\right)$

$k=\left(0.90\right)\left(15\right)=13.5$

k is sometimes called a critical value.

The 90^th percentile is 13.5 minutes. Ninety percent of the time, a person must wait at most 13.5 minutes.

View original figure (external; image not reproduced in this text edition)

Original figure description: f(X)=1/15 graph displaying a boxed region consisting of a horizontal line extending to the right from point 1/15 on the y-axis, a vertical upward line from an arbitrary point on the x-axis, and the x and y-axes. A shaded region from points 0-k occurs within this area. The area of this probability region is equal to 0.90.

a. The probability that a randomly selected nine-year-old child eats a doughnut in at least two minutes is _______.

a. 0.5714

b. Find the probability that a different nine-year-old child eats a doughnut in more than two minutes given that the child has already been eating the doughnut for more than 1.5 minutes.

The second question has a conditional probability. You are asked to find the probability that a nine-year-old child eats a doughnut in more than two minutes given that the child has already been eating the donut for more than 1.5 minutes. Solve the problem two different ways (see example-170). You must reduce the sample space. First way: Since you know the child has already been eating the doughnut for more than 1.5 minutes, you are no longer starting at a = 0.5 minutes. Your starting point is 1.5 minutes.

Write a new f(x):

$f\left(x\right)\text{ = }\frac{1}{4-1.5}=\frac{2}{5}\text{ for 1}\text{.5 }\le x\le 4.$

Find P(x > 2|x > 1.5). Draw a graph.

View original figure (external; image not reproduced in this text edition)

Original figure description: f(X)=2/5 graph displaying a boxed region consisting of a horizontal line extending to the right from point 2/5 on the y-axis, a vertical upward line from points 1.5 and 4 on the x-axis, and the x-axis. A shaded region from points 2-4 occurs within this area.

$P(x>2|x>1.5)\text{ = (base)(new height) = (4 – 2)}\left(\frac{2}{5}\right)=\frac{4}{5}$

b. $\frac{4}{5}$

1. Find the probability that a randomly selected furnace repair requires more than two hours.
2. Find the probability that a randomly selected furnace repair requires less than three hours.
3. Find the 30^th percentile of furnace repair times.
4. The longest 25 percent of furnace repair times take at least how long? (In other words: find the minimum time for the longest 25 percent of repair times.) What percentile does this represent?
5. Find the mean and standard deviation

a. To find f(x): f (x) = $\frac{1}{4-1.5}$ = $\frac{1}{2.5}$ so f(x) = 0.4

P(x > 2) = (base)(height) = (4 – 2)(0.4) = 0.8

View original figure (external; image not reproduced in this text edition)

Original figure description: This shows the graph of the function f(x) = 0.4. A horiztonal line ranges from the point (1.5, 0.4) to the point (4, 0.4). Vertical lines extend from the x-axis to the graph at x = 1.5 and x = 4 creating a rectangle. A region is shaded inside the rectangle from x = 2 to x = 4.

b. P(x < 3) = (base)(height) = (3 – 1.5)(0.4) = 0.6

The graph of the rectangle showing the entire distribution would remain the same. However the graph should be shaded between x = 1.5 and x = 3. Note that the shaded area starts at x = 1.5 rather than at x = 0. Because X ~ U(1.5, 4), x cannot be less than 1.5.

View original figure (external; image not reproduced in this text edition)

Original figure description: This shows the graph of the function f(x) = 0.4. A horiztonal line ranges from the point (1.5, 0.4) to the point (4, 0.4). Vertical lines extend from the x-axis to the graph at x = 1.5 and x = 4 creating a rectangle. A region is shaded inside the rectangle from x = 1.5 to x = 3.

c.

View original figure (external; image not reproduced in this text edition)

Original figure description: This shows the graph of the function f(x) = 0.4. A horiztonal line ranges from the point (1.5, 0.4) to the point (4, 0.4). Vertical lines extend from the x-axis to the graph at x = 1.5 and x = 4 creating a rectangle. A region is shaded inside the rectangle from x = 1.5 to x = k. The shaded area represents P(x < k) = 0.3.

P (x < k) = 0.30
P(x < k) = (base)(height) = (k – 1.5)(0.4)
0.3 = (k – 1.5) (0.4); Solve to find k:
0.75 = k – 1.5, obtained by dividing both sides by 0.4
k = 2.25 , obtained by adding 1.5 to both sides
The 30^th percentile of repair times is 2.25 hours. 30 percent of repair times are 2.5 hours or less.

d.

View original figure (external; image not reproduced in this text edition)

Original figure description: A graph with an x and an f(x) axis is shown. A box is drawn on the graph between 1.5 and 4 on the x axis, and 0 and 0.4 on the f(x) axis. The box is shaded blue from measurement k to 4. Text above the box says “shaded area represents P(x>k)=0.25.

P(x > k) = 0.25
P(x > k) = (base)(height) = (4 – k)(0.4)
0.25 = (4 – k)(0.4); Solve for k:
0.625 = 4 − k,
obtained by dividing both sides by 0.4
−3.375 = −k,
obtained by subtracting four from both sides: k = 3.375
The longest 25 percent of furnace repairs take at least 3.375 hours (3.375 hours or longer).
Note: Since 25 percent of repair times are 3.375 hours or longer, that means that 75 percent of repair times are 3.375 hours or less. 3.375 hours is the 75^th percentile of furnace repair times.

e. $\mu =\frac{a+b}{2}$ and $\sigma =\sqrt{\frac{{(b-a)}^2}{12}}$
$\mu \frac{1.5+4}{2}2.75$ hours and $\sigma =\sqrt{\frac{{(4–1.5)}^2}{12}}=0.7217$ hours

What type of distribution is this?

In this distribution, outcomes are equally likely. What does this mean?

It means that the value of x is just as likely to be any number between 1.5 and 4.5.

What is the height of f(x) for the continuous probability distribution?

What are the constraints for the values of x?

1.5 ≤ x ≤ 4.5

Graph P(2 < x < 3).

What is P(2 < x < 3)?

0.3333

What is P(x < 3.5| x < 4)?

What is P(x = 1.5)?

zero

What is the 90^th percentile of square footage for homes?

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.

0.6

What is a? What does it represent?

What is b? What does it represent?

b is 12, and it represents the highest value of x.

What is the probability density function?

What is the theoretical mean?

six

What is the theoretical standard deviation?

Draw the graph of the distribution for P(x > 9).

View original figure (external; image not reproduced in this text edition)

Original figure description: This graph shows a uniform distribution. The horizontal axis ranges from 0 to 12. The distribution is modeled by a rectangle extending from x = 0 to x = 12. A region from x = 9 to x = 12 is shaded inside the rectangle.

Find P(x > 9).

Find the 40^th percentile.

4.8

What is being measured here?

In words, define the random variable X.

X = The age (in years) of cars in the staff parking lot

Are the data discrete or continuous?

The interval of values for x is ______.

0.5 to 9.5

The distribution for X is ______.

Write the probability density function.

f(x) = $\frac{1}{9}$ where x is between 0.5 and 9.5, inclusive.

Graph the probability distribution.

1. Sketch the graph of the probability distribution.

   View original figure (external; image not reproduced in this text edition)

   Original figure description: This is a blank graph template. The vertical and horizontal axes are unlabeled.

2. Identify the following values:
   1. Lowest value for $\overline{x}$: _______
   2. Highest value for $\overline{x}$: _______
   3. Height of the rectangle: _______
   4. Label for x-axis (words): _______
   5. Label for y-axis (words): _______

Find the average age of the cars in the lot.

μ = 5

Find the probability that a randomly chosen car in the lot was less than four years old.

1. Sketch the graph, and shade the area of interest.

   View original figure (external; image not reproduced in this text edition)

   Original figure description: Blank graph with vertical and horizontal axes.

2. Find the probability. P(x < 4) = _______

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.

1. Sketch the graph, shade the area of interest.

   View original figure (external; image not reproduced in this text edition)

   Original figure description: This is a blank graph template. The vertical and horizontal axes are unlabeled.

2. Find the probability. P(x < 4|x < 7.5) = _______

1. Check student’s solution.
2. $\frac{3.5}{7}$

What has changed in the previous two problems that made the solutions different?

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 percent, of the cars are at most (less than or equal to) that age.

1. Sketch the graph, and shade the area of interest.

   View original figure (external; image not reproduced in this text edition)

   Original figure description: Blank graph with vertical and horizontal axes.

2. Find the value k such that P(x < k) = 0.75.
3. The third quartile is _______

1. Check student's solution
2. k = 7.25
3. 7.25

Births are approximately uniformly distributed between the 52 weeks of the year. They can be said to follow a uniform distribution from one to 53 (spread of 52 weeks).

1. X ~ _________
2. Graph the probability distribution.
3. f(x) = _________
4. μ = _________
5. σ = _________
6. Find the probability that a person is born at the exact moment week 19 starts. That is, find P(x = 19) = _________
7. P(2 < x < 31) = _________
8. Find the probability that a person is born after week 40.
9. P(12 < x|x < 28) = _________
10. Find the 70^th percentile.
11. Find the minimum for the upper quarter.

A random number generator picks a number from one to nine in a uniform manner.

1. X ~ _________
2. Graph the probability distribution.
3. f(x) = _________
4. μ = _________
5. σ = _________
6. P(3.5 < x < 7.25) = _________
7. P(x > 5.67)
8. P(x > 5|x > 3) = _________
9. Find the 90^th percentile.

1. X ~ U(1, 9)
2. Check student’s solution
3. $f\left(x\right)=\frac{1}{8}$ where $1\le x\le 9$
4. five
5. 2.3
6. $\frac{15}{32}$
7. $\frac{333}{800}$
8. $\frac{2}{3}$
9. 8.2

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.

1. Define the random variable. X = _________
2. X ~ _________
3. Graph the probability distribution.
4. f(x) = _________
5. μ = _________
6. σ = _________
7. Find the probability that the individual lost more than 10 pounds in a month.
8. Suppose it is known that the individual lost more than 10 pounds in a month. Find the probability that he lost less than 12 pounds in the month.
9. P(7 < x < 13|x > 9) = __________. State this result in a probability question, similarly to Parts g and h, draw the picture, and find the probability.

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.

1. Define the random variable. X = _______
2. X ~ _______
3. Graph the probability distribution.
4. f(x) = _______
5. μ = _______
6. σ = _______
7. Find the probability that the commuter waits less than one minute.
8. Find the probability that the commuter waits between three and four minutes.
9. Sixty percent of commuters wait more than how long for the train? State this result in a probability question, similarly to Parts g and h, draw the picture, and find the probability.

1. X represents the length of time a commuter must wait for a train to arrive on the Red Line.
2. X ~ U(0, 8)
3. $f\left(x\right)=\frac{1}{8}$ where ≤ x ≤ 8
4. four
5. 2.31
6. $\frac{1}{8}$
7. $\frac{1}{8}$
8. 3.2

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.

1. Define the random variable. X = _________
2. X ~ _________
3. Graph the probability distribution.
4. f(x) = _________
5. μ = _________
6. σ = _________
7. Find the probability that she is over 6.5 years old.
8. Find the probability that she is between four and six years old.
9. Find the 70^th percentile for the age of first graders on September 1 at Garden Elementary School.

What is the average waiting time (in minutes)?

1. zero
2. two
3. three
4. four

d

Find the 30^th percentile for the waiting times (in minutes).

1. two
2. 2.4
3. 2.75
4. three

The probability of waiting more than seven minutes given a person has waited more than four minutes is?

1. 0.125
2. 0.25
3. 0.5
4. 0.75

b

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.

1. Define the random variable. X = ________
2. X ~ ________
3. Graph the probability distribution.
4. The distribution is ______________ (name of distribution). It is _____________ (discrete or continuous).
5. μ = ________
6. σ = ________
7. 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.
8. 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.
9. P(25 < x < 55) = _________. State this result in a probability statement, similarly to Parts g and h, draw the picture, and find the probability.
10. Find the 90^th percentile. This means that 90 percent of the time, the time is less than _____ minutes.
11. Find the 75^th percentile. In a complete sentence, state what this means. (See Part j.)
12. Find the probability that the time is more than 40 minutes given (or knowing that) it is at least 30 minutes.

Suppose that the value of a stock varies each day from $16 to $25 with a uniform distribution.

1. Find the probability that the value of the stock is more than $19.
2. Find the probability that the value of the stock between $19 and $22.
3. Find the upper quartile — 25 percent of all days the stock is above what value? Draw the graph.
4. Given that the stock is greater than $18, find the probability that the stock is more than $21.

1. The probability density function of X is $\frac{1}{25-16}=\frac{1}{9}$.
   P(X > 19) = (25 – 19) $\left(\frac{1}{9}\right)$ = $\frac{6}{9}$ = $\frac{2}{3}$.

   View original figure (external; image not reproduced in this text edition)

   Original figure description: A graph is shown. The x($) axis shown is measured from 14 to 26 in increments of two. A box is drawn on the graph between the measurements of 16 and 25, and up to 1/9 on the y axis. The box is shaded blue from 19 to 25. Text above the box states “shaded area represents P(x>19)=2/3.

2. P(19 < X < 22) = (22 – 19) $\left(\frac{1}{9}\right)$ = $\frac{3}{9}$ = $\frac{1}{3}$.

   View original figure (external; image not reproduced in this text edition)

   Original figure description: A graph is shown. The x($) axis shown is measured from 14 to 26 in increments of two. A box is drawn on the graph between the measurements of 16 and 25 and up to 1/9 on the y axis. The box is shaded from 23 to 25. Text above the box states “Shaded area represents P(19<x<22)=1/3.

3. The area must be 0.25, and 0.25 = (width)$\left(\frac{1}{9}\right)$, so width = (0.25)(9) = 2.25. Thus, the value is 25 – 2.25 = 22.75.
4. This is a conditional probability question. P(x > 21| x > 18). You can do this two ways:
   • Draw the graph where a is now 18 and b is still 25. The height is $\frac{1}{(25-18)}$ = $\frac{1}{7}$
     So, P(x > 21|x > 18) = (25 – 21)$\left(\frac{1}{7}\right)$ = 4/7.
   • Use the formula: P(x > 21|x > 18) = $\frac{P(x>21\text{ AND }x>18)}{P(x>18)}$
     = $\frac{P(x>21)}{P(x>18)}$ = $\frac{(25-21)}{(25-18)}$ = $\frac{4}{7}$.

A fireworks show is designed so that the time between fireworks is between one and five seconds, and follows a uniform distribution.

1. Find the average time between fireworks.
2. Find the probability that the time between fireworks is greater than four seconds.

The number of miles driven by a truck driver falls between 300 and 700, and follows a uniform distribution.

1. Find the probability that the truck driver goes more than 650 miles in a day.
2. Find the probability that the truck driver goes between 400 and 650 miles in a day.
3. At least how many miles does the truck driver travel on the 10 percent of days with the highest mileage?

1. P(X > 650) = $\frac{700-650}{700-300}=\frac{50}{400}=\frac{1}{8}$ = 0.125.
2. P(400 < X < 650) = $\frac{650-400}{700-300}=\frac{250}{400}$ = 0.625
3. 0.10 = $\frac{\text{width}}{\text{700}-\text{300}}$, so width = 400(0.10) = 40. Since 700 – 40 = 660, the drivers travel at least 660 miles on the farthest 10 percent of days.