4.134 Refer to Exercise 4.133. Find the following probabilities:

$$\begin{gathered} a.P(20\le x\le 30)b.P(20<x\le 30) \\ c.P(x\ge 30)d.P(x\ge 45) \\ e.(x\le 40)f.(x<40) \\ g.P(15\le x\le 35)h.P(21.5\le x\le 31.5) \end{gathered}$$

Referring to exercise 4.133, we can say that here x is a uniform random variable with parameters c=20 and d=45

The probability density function (PDF) 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:

$$\begin{gathered} f\left(x\right)=\frac{1}{45-20} \\ =\frac{1}{25} \\ =0.04 \\ f\left(x\right)=0.04;02<x<45 \end{gathered}$$

$$\begin{gathered} F\left(x\right)=P(X\le x) \\ ={\int }_{20}^{x}f\left(t\right)dt \\ ={\int }_{20}^{x}0.04dt \\ =0.04{\int }_{20}^{x}dt \\ =0.04{\left[t\right]}_{20}^x \\ =0.04(x-20) \\ F\left(x\right)=0.04x(x-20) \end{gathered}$$

For continuous random variable x

$$\begin{gathered} P(X<x)=P(X\le x) \\ P(20\le x\le 30)=P(x<20) \\ =F\left(30\right)-F\left(20\right) \\ =0.04\times (30-20)-0.04\times (20-20) \\ =0.40 \end{gathered}$$

Thus, the required probability is 0.40.

Thus, the required probability is -0.40.

c.

Thus, the required probability is 0.60.

Thus, the required probability is 0.

Thus, the required probability is 0.80.

Thus, the required probability is 0.80.

$$\begin{gathered} g. \\ P(15\le x\le 35)=P(20\le x\le 35)[Since,x>20] \\ =P(x\le 35)-P(x<20) \\ =F\left(35\right)-F\left(20\right) \\ =0.04\times (35-20)-0.04\times (20-20) \\ =0.60-0 \\ =0.60 \end{gathered}$$

Thus, the required probability is 0.60.

Thus, the required probability is 0.40.