The uniform circulation is a consistent probability distribution and is concerned with events that room equally likely to occur. Once working out troubles that have a uniform distribution, be careful to note if the data is inclusive or exclusive.

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Example 1

The data in the table below are 55 smiling times, in seconds, of one eight-week-old baby.


The sample average = 11.49 and also the sample standard deviation = 6.23.

We will certainly assume the the smiling times, in seconds, follow a uniform distribution in between zero and also 23 seconds, inclusive. This way that any smiling time from zero to and also including 23 seconds is equally likely. The histogram that might be built from the sample is an empirical circulation that carefully matches the theoretical uniform distribution.

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

The notation for the uniform distribution is X ~ U(a, b) where a = the lowest worth of x and b = the highest value the x.

The probability density role is displaystylef(x)=frac1b-a\ because that axb.

For this example, X ~ U(0, 23) and also displaystylef(x)=frac123-0\ for 0 ≤ X ≤ 23.

Formulas because that the theoretical mean and also standard deviation space displaystylemu=fraca+b2quad extandquadsigma=sqrtfrac(b-a)^212\

For this problem, the theoretical mean and standard deviation space displaystylemu=frac0+232=11.50 ext secondsquad extandquadsigma=sqrtfrac(23-0)^212=6.64 ext seconds\

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

Try It

The data that follow space the variety of passengers on 35 various charter fishing boats. The sample typical = 7.9 and the sample standard deviation = 4.33. The data monitor a uniform distribution where all values between and also including zero and 14 room equally likely. State the values of a and also b. Write the distribution in ideal notation, and also calculate the theoretical mean and standard deviation.


a is zero; b is 14; X ~ U (0, 14); μ = 7 passengers; σ = 4.04 passengers

Example 2

Refer to example 1 What is the probability the a randomly favored eight-week-old infant 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 infant smiles an ext than 12 secs knowing that the infant smiles more than eight seconds.


Find P(2 displaystyleP{(2{Ninety percent the the smiling time fall listed below the 90th percentile, k, therefore P(x P(x( extbase)( extheight)=0.90\displaystyle(k-0)(frac123)=0.90\k=(23)(0.90)=20.7\
conditional. You space asked to discover the probability that an eight-week-old baby smiles more than 12 seconds once you already know the baby has smiled for much more than eight seconds.Find P(x > 12|x > 8) There space two methods to execute the problem.For the first way, usage the fact that this is a conditional and changes the sample space. The graph illustrates the new sample space. You already know the infant smiled much more than eight seconds.Write a new f(x):displaystylef(x)=frac123-8=frac115\for 8
For the second way, usage the conditional formula (shown below) through the original circulation X ~ U (0, 23):For this problem, A is (x > 12) and also B is (x > 8).
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A circulation is offered as X ~ U (0, 20). What is P(2

Example 3

The quantity of time, in minutes, the a human must wait because that a bus is uniformly distributed in between zero and 15 minutes, inclusive.

What is the probability the a person waits fewer than 12.5 minutes?On the average, how long should a person wait? find the mean, μ, and the standard deviation, σ.Ninety percent of the time, the moment a person must wait falls listed below what value? This asks because that the 90th percentile.


Let X = the number of minutes a human being must wait for a bus. a = 0 and also b = 15. X~ U(0, 15). Create the probability density function. displaystylef(x)=frac115-0=frac115\ for 0 ≤x ≤ 15.Find P (x displaystyleP(x x > 7) = 0.875

Example 5

Ace Heating and Air Conditioning company finds the the quantity of time a repairman needs to deal with a furnace is uniformly distributed between 1.5 and four hours. Let x = the time required to settle a furnace. Then x ~ U (1.5, 4).

Find the probability the a randomly selected heater repair requires much more than 2 hours.Find the probability that a randomly selected furnace repair requires much less than 3 hours.Find the 30th percentile of heater repair times.The longest 25% of furnace repair times take at least how long? (In other words: uncover the minimum time because that the longest 25% of repair times.) What percentile does this represent?Find the mean and also standard deviation


To find f(x): displaystylef(x)=frac14-1.5=12.5 ext so f(x)=0.4\P(x > 2) = (base)(height) = (4 – 2)(0.4) = 0.8
Uniform Distribution in between 1.5 and four v shaded area in between two and four representing the probability that the fix timex is better than twoP(x x = 1.5 and also x = 3. Note that the shaded area starts at x = 1.5 rather than at x = 0; due to the fact that X ~ U (1.5, 4), x can no be much less than 1.5.
Uniform Distribution between 1.5 and also four through shaded area in between 1.5 and also three representing the probability that the repair time x is less than three
P (x P(x k = 2.25 , obtained by adding 1.5 come both sidesThe 30th percentile the repair times is 2.25 hours. 30% of fix times are 2.5 hrs or less.
Uniform Distribution between 1.5 and also 4 v an area that 0.25 shaded come the right representing the longest 25% of repair times.P(x > k) = 0.25P(x > k) = (base)(height) = (4 – k)(0.4)0.25 = (4 – k)(0.4); deal with for k:0.625 = 4 − k, derived by splitting both sides by 0.4

−3.375 = − k, acquired by subtracting 4 from both sides: k = 3.375

The longest 25% of heating system repairs take it at least 3.375 hrs (3.375 hrs or longer).

Note: due to the fact that 25% of repair times are 3.375 hours or longer, that way that 75% of fix times are 3.375 hrs or less. 3.375 hours is the 75th percentile of heater repair ext and sigma=sqrtfrac(b-a)^212\mu=frac1.5+42=2.75 ext hours and sigma=sqrtfrac(4-1.5)^212= 0.7217 ext hours\Try ItThe quantity of time a service technician needs to adjust the oil in a automobile is uniformly distributed between 11 and also 21 minutes. Let X = the time required to adjust the oil ~ above a car.

Write the arbitrarily variable X in words. X = __________________.Write the distribution.Graph the distribution.Find P (x > 19).Find the 50th percentile.Let X = the time needed to change the oil in a car.X ~ U (11, 21).P (x > 19) = 0.2the 50th percentile is 16 minutes.

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McDougall, john A. The McDougall program for Maximum weight Loss. Plume, 1995.

Concept Review

If X has a uniform distribution where a x)=(b-x)(frac1b-a)\

Area Between c and d: displaystyleP{(c{pdf: displaystylef(x)=frac1b-a\ for a ≤ x ≤ bcdf: P(Xx) = displaystylefracx-ab-a\mean: displaystylemu=fraca+b2\standard deviation: displaystylesigma=sqrtfrac(b-a)^212\P(c