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.
Check
"=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
Check
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.
"=0^2(0.25)+1^2(0.43)+2^2(0.22)+3^2(0.07)+4^2(0.03)"
"Var(X)=E(X^2)-(E(X))^2"
"=2.42-(1.2)^2=0.98"
"Var(X)=0.98"
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