Question

4.137 The random variable xis best described by a uniform probability distribution with c= 100 and d= 200. Find the probability that xassumes a value

a. More than 2 standard deviations from

b. Less than 3 standard deviations from

c. Within 2 standard deviations of

Short answer

a. The probability is 0.

b. The probability is 0.

c. The probability is 1.

Step-by-step solution

Given Information

Here, x is a uniform random variable with parameters c=100 and d=200.

Finding the pdf of x

The probability density function random variable x is given by

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

Here, c=100 and d=200

So, the pdf of x is:

$$\begin{gathered} f\left(x\right)=\frac{1}{200-100} \\ =\frac{1}{100} \\ =0.01 \end{gathered}$$

Thus,$f\left(x\right)=\left\{\begin{array}{ll}0.01& ;100<x<200\\ 0& ;otherwise\end{array}\right.$

The mean is given by,

$$\begin{gathered} \mu =\frac{c+d}{2} \\ =\frac{100+200}{2} \\ =150 \end{gathered}$$

The standard deviation is given by,

$$\begin{gathered} \sigma =\frac{d-c}{\sqrt{12}} \\ =\frac{200-100}{\sqrt{12}} \\ =\frac{100}{\sqrt{12}} \\ =28.8675 \end{gathered}$$

Thus, the mean $\mu =150$and standard deviation $\sigma =28.8675$ .

Finding the probability when x assumes a value more than 2 standard deviations from μ

a.

$$\begin{gathered} P\left(x>\mu +2\sigma \right)=P\left(x>150+2\times 28.8675\right) \\ =P\left(x>207.735\right) \\ ={\int }_{207.735}^{\infty }f\left(x\right)dx \\ ={\int }_{207.735}^{\infty }0dx \\ =0 \end{gathered}$$

Thus, the required probability is 0.

Finding the probability when x assumes a value within 2 standard deviations from μ

$$\begin{gathered} P\left(\mu -2\sigma <x<\mu +2\sigma \right)=P\left(150-2\times 28.8675<x<150+2\times 28.8675\right) \\ =P\left(92.265<x<207.735\right) \\ ={\int }_{92.265}^{207.735}f\left(x\right)dx \\ ={\int }_{100}^{200}0.001dx \\ =0.01{\left[x\right]}_{100}^{200} \\ =0.01\times \left(200-100\right) \\ =0.01\times 100 \\ =1 \end{gathered}$$

Thus, the required probability is 1.