Question

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 $\mathit{\mu }$and the interval $\mathit{\mu }\mathbf{\pm }\mathbf{2}\mathit{\sigma }$onthe graph. Note that the probability that xassumes avalue within the interval $\mathit{\mu }\mathbf{\pm }\mathbf{2}\mathit{\sigma }$is equal to 1.

Short answer

a. The probability density function is

$f\left(x\right)=\left\{\begin{array}{ll}0.04& 20\le x\le 45\\ 0& ;otherwise\end{array}\right.$

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

Given Information

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

Finding the f (x)

a.

The probability density function random variable x is given by

$f\left(x\right)=\frac{1}{d-c};c<x<d$

Here, c=20 and d=45

So the pdf of x is:

$$\begin{gathered} f\left(x\right)=\frac{1}{45-20} \\ =\frac{1}{25} \\ =0.04 \end{gathered}$$

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

Finding the mean and standard deviation of x.

b.

The mean of the random variable x is given by,

$$\begin{gathered} \mu =\frac{c+d}{2} \\ =\frac{20+45}{2} \\ =\frac{65}{2} \\ =32.5 \end{gathered}$$

The standard deviation of x is given by,

$$\begin{gathered} \sigma =\frac{d-c}{\sqrt{12}} \\ =\frac{45-20}{\sqrt{12}} \\ =\frac{25}{2\sqrt{3}} \\ =7.2169 \end{gathered}$$

Thus, the mean $\mu =32.5$and standard deviation $\sigma =7.2169$.

The Graph

c.

Here, the $2\sigma$limit is given by,

$$\begin{gathered} \mu -2\sigma =32.5-2\times 7.2169 \\ =18.0662 \\ \mu +2\sigma =32.5+2\times 7.2169 \\ =46.9338 \end{gathered}$$

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

The following graph depicts the relevant situation.