The Uniform Distribution

The uniform distribution is a continuous probability distribution and is concerned with events that are equally likely to occur. When working out problems that have a uniform distribution, be careful to note if the data is inclusive or exclusive of endpoints.

The data in [link] are 55 smiling times, in seconds, of an eight-week-old baby.

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 = 11.49 and the sample standard deviation = 6.23.

We will assume that the smiling times, in seconds, follow a uniform distribution between zero and 23 seconds, inclusive. This means that any smiling time from zero to and including 23 seconds is equally likely. The histogram that could be constructed from the sample is an empirical distribution that closely matches the theoretical uniform distribution.

Let X = length, in seconds, of an eight-week-old baby's smile.

The notation for the uniform distribution is

X ~ U(a, b) where a = the lowest value of x and b = the highest value of x.

The probability density function is f(x) = $\frac{1}{b - a}$

for a ≤ x ≤ b.

For this example, X ~ U(0, 23) and f(x) = $\frac{1}{23 - 0}$

for 0 ≤ X ≤ 23.

Formulas for the theoretical mean and standard deviation are

μ = \frac{a + b}{2}

and $σ = \sqrt{\frac{{(b - a)}^{2}}{12}}$

For this problem, the theoretical mean and standard deviation are

μ = $\frac{0 + 23}{2}$

= 11.50 seconds and σ = $\sqrt{\frac{{(23 - 0)}^{2}}{12}}$

= 6.64 seconds.

Notice that the theoretical mean and standard deviation are close to the sample mean and standard deviation in this example.

a. Refer to [link]. 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) $(\frac{1}{23})$

= $\frac{16}{23}$

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

(base) (height) = 0.90

(k - 0) (\frac{1}{23}) = 0.90

k = (23) (0.90) = 20.7

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.

Write a new f(x): f(x) = $\frac{1}{23 - 8}$

= $\frac{1}{15}$

for 8 < x < 23

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

= $\frac{11}{15}$

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 X ~ U (0, 23):

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

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

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

$This diagram shows a horizontal X axis that intersects a vertical F of x axis at the origin. The X axis runs from 0 to 24 while the Y axis only has the fraction one twenty third located about two thirds of the way to the top. A rectangular box extends horizontally from 0 to about 23.7 on the X axis. The box extends vertically up to the fraction one twenty third on the F of x axis. The area of the box between 8 and 12 on the X axis is shaded.$

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?

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) = (base) (height) = (12.5 - 0) (\frac{1}{15}) = 0.8333

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

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 = $\sqrt{\frac{(b - a)^{2}}{12}} = \sqrt{\frac{(15 - 0)^{2}}{12}}$

= 4.3. The Standard deviation is 4.3 minutes. * * *

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

This asks for the 90^th percentile.

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

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

0.90 = (k) (\frac{1}{15})

k = (0.90) (15) = 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.

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.

Suppose the time it takes a nine-year old to eat a donut is 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 ~ U (0.5, 4).* * *

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

a. 0.5714* * *

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.

The second question has a conditional probability. You are asked to find 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. Solve the problem two different ways (see [link]). You must reduce the sample space. First way: Since you know the child has already been eating the donut 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(x) = $\frac{1}{4 - 1.5}$

= $\frac{2}{5}$

for 1.5 ≤ x ≤ 4.

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

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) = (base)(new height) = (4 − 2) $(\frac{2}{5}) = \frac{4}{5}$

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 $\frac{4}{5}$

Second way: Draw the original graph for X ~ U (0.5, 4). Use the conditional formula

P(x > 2\|x > 1.5) = $\frac{P (x > 2 AND x > 1.5)}{P (x > 1 .5)} = \frac{P (x > 2)}{P (x > 1.5)} = \frac{\frac{2}{3.5}}{\frac{2.5}{3.5}} = 0 .8 = \frac{4}{5}$

Ace Heating and Air Conditioning Service finds that the amount of time a repairman 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 ~ U (1.5, 4).

Find the probability that a randomly selected furnace repair requires more than two hours.
Find the probability that a randomly selected furnace repair requires less than three hours.
Find the 30^th percentile of furnace repair times.
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?
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

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; since X ~ U (1.5, 4), x can not be less than 1.5.

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% of repair times are 2.5 hours or less.

{:}

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% of furnace repairs take at least 3.375 hours (3.375 hours or longer). * * *

Note: Since 25% of repair times are 3.375 hours or longer, that means that 75% of repair times are 3.375 hours or less. 3.375 hours is the 75^th percentile of furnace repair times.

e. $μ = \frac{a + b}{2}$

and $σ = \sqrt{\frac{{(b - a)}^{2}}{12}}$

μ \frac{1.5 + 4}{2} 2.75

hours and $σ = \sqrt{\frac{{(4 - 1.5)}^{2}}{12}} = 0.7217$

hours

Chapter Review

If X has a uniform distribution where a < x < b or a ≤ x ≤ b, then X takes on values between a and b (may include a and b). All values x are equally likely. We write X ∼ U(a, b). The mean of X is

μ = \frac{a + b}{2}

for a ≤ x ≤ b. The cumulative distribution function of X is P(X ≤ x) =

\frac{x - a}{b - a}

The probability P(c < X < d) may be found by computing the area under f(x), between c and d. Since the corresponding area is a rectangle, the area may be found simply by multiplying the width and the height.

Formula Review

X = a real number between a and b (in some instances, X can take on the values a and b). a = smallest X; b = largest X

Area Between c and d: P(c < x < d) = (base)(height) = (d – c)

(\frac{1}{b - a})

References

Use the following information to answer the next ten questions. The data that follow are the square footage (in 1,000 feet squared) of 28 homes.

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

The sample mean = 2.50 and the sample standard deviation = 0.8302.

The distribution can be written as X ~ U(1.5, 4.5).

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

Use the following information to answer the next eight exercises. A distribution is given as X ~ U(0, 12).

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).

![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.](/statistics-book/resources/CNX_Stats_C05_M03_item002annoN.jpg){:}

Find P(x > 9).

Find the 40^th percentile.

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.

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.

Sketch the graph of the probability distribution.
Identify the following values:
1. Lowest value for $\bar{x}$
  : _______
2. Highest value for $\bar{x}$
  : _______
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.

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

Sketch the graph, shade the area of interest.
Find the probability. P(x < 4|x < 7.5) = _______

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

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

Sketch the graph, and shade the area of interest.
Find the value k such that P(x < k) = 0.75.
The third quartile is _______

Check student’s solution.
k = 7.25
7.25

Homework

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.

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

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

The probability density function of X is $\frac{1}{25 - 16} = \frac{1}{9}$
.

P(X > 19) = (25 – 19)
$(\frac{1}{9})$
=
$\frac{6}{9}$
=
$\frac{2}{3}$
. {:}
P(19 < X < 22) = (22 – 19) $(\frac{1}{9})$
=
$\frac{3}{9}$
=
$\frac{1}{3}$
. {:}
The area must be 0.25, and 0.25 = (width) $(\frac{1}{9})$
, so width = (0.25)(9) = 2.25. Thus, the value is 25 – 2.25 = 22.75.
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)
( 1 7 )
= 4/7.
- Use the formula: P(x > 21|x > 18) = $\frac{P (x > 21 AND x > 18)}{P (x > 18)}$
=
P(x>21) P(x>18)
=
(25−21) (25−18)
=
4 7
.

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

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

The Uniform Distribution

Chapter Review

Formula Review

References

Homework

Glossary

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