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4.133 Suppose xis a random variable best described by a uniform

probability distribution with c= 20 and d= 45.

a. Find f(x)

b. Find the mean and standard deviation of x.

c. Graph f (x) and locate μand the interval μ±2σonthe graph. Note that the probability that xassumes avalue within the interval μ±2σis equal to 1.

Short Answer

Expert verified

a. The probability density function is

f(x)=0.0420≤x≤450;otherwise

b. The mean is 32.5 and the standard deviation is 7.2169

c. The lower limit is 18.0662 and the upper limit is 46.9338

Step by step solution

01

Given Information

Here x is a uniform random variable with parameters c=20 and d=45.

02

Finding the f (x)

a.

The probability density function random variable x is given by

f(x)=1d-c;c<x<d

Here, c=20 and d=45

So the pdf of x is:

f(x)=145-20=125=0.04

Thus, f (x0 = 0.04 ; 20 < x <78C8 45

03

Finding the mean and standard deviation of x.

b.

The mean of the random variable x is given by,

μ=c+d2=20+452=652=32.5

The standard deviation of x is given by,

σ=d-c12=45-2012=2523=7.2169

Thus, the mean μ=32.5and standard deviation σ=7.2169.

04

The Graph

c.

Here, the 2σlimit is given by,

μ-2σ=32.5-2×7.2169=18.0662μ+2σ=32.5+2×7.2169=46.9338

Here, the interval limits are outside of the actual range of random variable x.

The following graph depicts the relevant situation.

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