Answer to Question #147553 in Statistics and Probability for Vijay

Question

Question No. 1 (4 Marks)

A car wholesaler estimates the quantity of new cars retailed that have been returned back a many number of times for the rectification of defects during the warranty. The results are given in the following table. Show complete working.

Number of cars

returns back

0 1 2 3 4

Probability 0.25 0.34 0.22 0.07 0.03

a. Compute the mean of the number of cars returns for rectifications for defects during the warranty time.

b. Calculate the variance of the number of cars returns for rectifications for defects during the warranty time.

Answer 1

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c} Number \ of \\ cars\ returned & 0 & 1 & 2 & 3 & 4 \\ \hdashline Probability & 0.25 & 0.34 & 0.22 & 0.07 & 0.03 \end{array}$$

Check

$$\displaystyle\sum_{i=1}^np(x_i)=0.25+0.34+0.22+0.07+0.03$$

$$=0.91\not=1$$

Since $\displaystyle\sum_{i=1}^np(x_i)=0.91<1,$ this is not a probability distribution of the discrete random variable $X$ .We cannot  compute the mean  and we cannot calculate the variance.

Let

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c} Number \ of \\ cars\ returned & 0 & 1 & 2 & 3 & 4 \\ \hdashline Probability & 0.25 & 0.43 & 0.22 & 0.07 & 0.03 \end{array}$$

Check

$$\displaystyle\sum_{i=1}^np(x_i)=0.25+0.43+0.22+0.07+0.03=1$$

a.

$$mean=E(X)=\displaystyle\sum_{i=1}^nx_ip(x_i)$$

$$=0(0.25)+1(0.43)+2(0.22)+3(0.07)+4(0.03)$$

$$=1.2$$

$mean=1.2$

2.

$$E(X^2)=\displaystyle\sum_{i=1}^nx_i^2p(x_i)$$

$$=0^2(0.25)+1^2(0.43)+2^2(0.22)+3^2(0.07)+4^2(0.03)$$

$$=2.42$$

$$Var(X)=E(X^2)-(E(X))^2$$

$$=2.42-(1.2)^2=0.98$$

$Var(X)=0.98$