The Uniform Distribution
Learning Outcomes
- Recognize the uniform probability distribution and apply it appropriately
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.
Example
The data in the table below 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[latex]=11.49[/latex] and the sample standard deviation[latex]=6.23[/latex].
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 [latex]X=[/latex] length, in seconds, of an eight-week-old baby’s smile.
The notation for the uniform distribution is [latex]X{\sim}U(a,b)[/latex] where [latex]a=[/latex] the lowest value of x and [latex]b=[/latex] the highest value of x.
The probability density function is [latex]{f{{({x})}}}=\frac{{1}}{{{b}-{a}}}[/latex] for [latex]a{\leq}x{\leq}b[/latex].
For this example, [latex]X{\sim}U(0,23)[/latex] and [latex]{f{{({x})}}}=\frac{{1}}{{{23}-{0}}}[/latex] for [latex]0{\leq}X{\leq}23[/latex].
Formulas for the theoretical mean and standard deviation are [latex]{\mu}=\frac{{{a}+{b}}}{{2}}{\quad\text{and}\quad}{\sigma}=\sqrt{{\frac{{{({b}-{a})}^{{2}}}}{{12}}}}[/latex]
For this problem, the theoretical mean and standard deviation are [latex]{\mu}=\frac{{{0}+{23}}}{{2}}={11.50} \text{ seconds}{\quad\text{and}\quad}{\sigma}=\sqrt{{\frac{{{({23}-{0})}^{{2}}}}{{12}}}}={6.64} \text{ seconds}[/latex]
Notice that the theoretical mean and standard deviation are close to the sample mean and standard deviation in this example.
Try It
The data that follow are the number of passengers on 35 different charter fishing boats. The sample mean [latex]=7.9[/latex] and the sample standard deviation [latex]=4.33[/latex]. The data follow a uniform distribution where all values between and including zero and 14 are equally likely. 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 |
[reveal-answer q=”810841″]Show Solution[/reveal-answer]
[hidden-answer a=”810841″]
a is zero; b is 14; [latex]X{\sim}U(0,14)[/latex]; [latex]\mu=7[/latex] passengers; [latex]\sigma=4.04[/latex] passengers
[/hidden-answer]
Example
- Refer to the previous example. What is the probability that a randomly chosen eight-week-old baby smiles between two and 18 seconds?
- Find the 90th percentile for an eight-week-old baby’s smiling time.
- Find the probability that a random eight-week-old baby smiles more than 12 seconds knowing that the baby smiles more than eight seconds.
[reveal-answer q=”96316″]Show Solution[/reveal-answer]
[hidden-answer a=”96316″]
- Find P(2 < x < 18).
[latex]{P}{({2}{<}{x}{<}{18})}={(\text{base})}{(\text{height})}={({18}-{2})}{(\frac{{1}}{{23}})}={(\frac{{16}}{{23}})}.[/latex] - Ninety percent of the smiling times fall below the 90th percentile, k, so P(x <k) = 0.90
[latex]P(x - 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):[latex]{f{{({x})}}}=\frac{{1}}{{{23}-{8}}}=\frac{{1}}{{15}}[/latex]for 8 < x < 23
- For the second way, use the conditional formula (shown below) with the original distribution X ~ U (0, 23):For this problem, A is (x > 12) and B is (x > 8).
- 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):[latex]{f{{({x})}}}=\frac{{1}}{{{23}-{8}}}=\frac{{1}}{{15}}[/latex]for 8 < x < 23
[/hidden-answer]
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A distribution is given as [latex]X{\sim}U(0,20)[/latex]. What is [latex]P(2 The amount of time, in minutes, that a person must wait for a bus is uniformly distributed between zero and 15 minutes, inclusive. The total duration of baseball games in the major league in the 2011 season is uniformly distributed between 447 hours and 521 hours inclusive. 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). Suppose the time it takes a student to finish a 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 ~ 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 she has already taken more than seven minutes. P (x > 8) = 0.7778 P (x > 8 | x > 7) = 0.875 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). −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 75th percentile of furnace repair times. 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. McDougall, John A. The McDougall Program for Maximum Weight Loss. Plume, 1995. 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 [latex]{\mu}=\frac{{{a}+{b}}}{{2}}[/latex]. X is continuous. 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. 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 X ~ U (a, b) The mean is [latex]\mu=\frac{{{a}+{b}}}{{2}}[/latex] The standard deviation is [latex]\sigma=\sqrt{{\frac{{({b}-{a})}^{{2}}}{{12}}}}[/latex] Probability density function: [latex]{f{{({x})}}}=\frac{{1}}{{{b}-{a}}} \text{ for } {a}\leq{X}\leq{b}[/latex] Area to the Left of x: [latex]{P}{({X}{<}{x})}={({x}-{a})}{(\frac{{1}}{{{b}-{a}}})}[/latex]
Area to the Right of x: [latex]{P}{({X}{>}{x})}={({b}-{x})}{(\frac{{1}}{{{b}-{a}}})}[/latex] Area Between c and d: [latex]{P}{({c}{<}{x}{<}{d})}={(\text{base})}{(\text{height})}={({d}-{c})}{(\frac{{1}}{{{b}-{a}}})}[/latex]
Uniform: X ~ U(a, b) where a < x < bExample
Solution
[latex]{P}{({x}{<}{k})}={(\text{base})}{(\text{height})}={({12.5}-{0})}{(\frac{{1}}{{15}})}={0.8333}[/latex]
The probability a person waits less than 12.5 minutes is 0.8333.
[latex]{P}(x
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Example 4
Solution
[latex]{P}{({x}{>}{2}|{x}{>}{1.5})}={(\text{base})}{(\text{new height})}={({4}-{2})}{(\frac{{2}}{{5}})}=\frac{{4}}{{5}}[/latex]
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 [latex]\frac{{4}}{{5}}[/latex].
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Example 5
Solution
P(x > 2) = (base)(height) = (4 – 2)(0.4) = 0.8
Uniform Distribution between 1.5 and four with shaded area between two and four representing the probability that the repair time
x is greater than two
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.
Uniform Distribution between 1.5 and four with shaded area between 1.5 and three representing the probability that the repair time x is less than three
Uniform Distribution between 1.5 and 4 with an area of 0.30 shaded to the left, representing the shortest 30% of repair times.
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 sidesThe 30th percentile of repair times is 2.25 hours. 30% of repair times are 2.5 hours or less.
Uniform Distribution between 1.5 and 4 with an area of 0.25 shaded to the right representing the longest 25% of repair times.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
[latex]{\mu}=\frac{1.5+4}{2}=2.75\text{ hours and }{\sigma}=\sqrt{\frac{(4-1.5)^2}{12}}= 0.7217 \text{ hours}[/latex]
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References
Concept Review
Formula Review