Question

4.132 Suppose xis a random variable best described by a uniform

probability distribution with c= 3 and d= 7.

a. Find f(x)

b. Find the mean and standard deviation of x.

c. Find$\mathbf{P}\mathbf{(}\mathbf{\mu }\mathbf{-}\mathbf{\sigma }\mathbf{\le }\mathbf{x}\mathbf{\le }\mathbf{\mu }\mathbf{+}\mathbf{\sigma }\mathbf{)}$

Short answer

a. The probability density function is

$f\left(x\right)=\left\{\begin{array}{ll}0.25& 3\le x\le 7\\ 0& ;otherwise\end{array}\right.$

b. The mean is 5 and standard deviation is 1.1547.

c.$\mathrm{P}(\mathrm{\mu }-\mathrm{\sigma }\le \mathrm{x}\le \mathrm{\mu }+\mathrm{\sigma })$

Step-by-step solution

Given Information

Here, x is a uniform random variable with parameters c=3 and d=7.

Finding the pdf of 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=3 and d=7.

So, the pdf of x is:

$$\begin{gathered} f\left(x\right)=\frac{1}{7-3} \\ =\frac{1}{4} \\ =0.25 \end{gathered}$$

Thus, f (x) = 0.25 ; 3 < x < 7

Finding the mean and standard deviation of x.

b.

The mean of the random variable x is given by,

$$\begin{gathered} \mu =\frac{c+d}{2} \\ =\frac{3+7}{2} \\ =\frac{10}{2} \\ =5 \end{gathered}$$

The standard deviation of x is given by,

$$\begin{gathered} \sigma =\frac{d-c}{\sqrt{12}} \\ =\frac{7-3}{\sqrt{12}} \\ =\frac{4}{2\sqrt{3}} \\ =\frac{2}{\sqrt{3}} \\ =1.1547 \end{gathered}$$

Thus, the mean $\mu =5$and standard deviation $\sigma =1.1547$.

Finding the P(μ-σ≤x≤μ+σ)

$$\begin{gathered} c. \\ P(\mu -\sigma \le x\le \mu +\sigma )={\int }_{\mu -\sigma }^{\mu +\sigma }f\left(x\right)dx \\ ={\int }_{\mu -\sigma }^{\mu +\sigma }0.25dx \\ =0.25{\int }_{\mu -\sigma }^{\mu +\sigma }dx \\ =0.25{\left[x\right]}_{\mu -\sigma }^{\mu +\sigma } \\ =0.25\left[\mu +\sigma -\mu +\sigma \right] \\ =0.25\times 2\sigma \\ =0.50\times \frac{2}{\sqrt{3}} \\ =0.5774 \end{gathered}$$

So, the required probability is 0.5774.