A customer entering an electronics store will buy a flat scr

A customer entering an electronics store will buy a flat screen TV with probability 0.3. Sixty percent of the customers buying a flat screen TV will spend $750.00 and 40% will spend $400. Let X denote the smount spent on flat screen TV\'s by two random customers entering the store.

(a) Find the PMF of X.

(b) Find the mean value and variance of X.

Solution

The probability that a customer enters the electronic store is 0.3

given that the probability that he spends $750 on flat TV is 0.6

implies the probability that a customer who enters the electronic store and spends $750 is 0.3*0.6 = 0.18

the probability that he spends $400 on flat TV is 0.4

implies the probability that a customer who enters the electronic store and spends $750 is 0.3*0.4 = 0.12

Hence the PMF is given by

b. The mean value is given by E(X) = 0*0.7+400*0.12+750*0.18 = $183

varince - E(X²) - [E(X)]² = 0²*0.7+400²*0.12+750²*0.18 - 183² = 86961