Question

Customers at a gas station pay with a credit card (A), debit card (B), or cash (C). Assume that successive customers make independent choices with P(A) = 0.5, P(B) = 0.3,and P(C) = 0.2.

(a) Among the next 100 customers, what are the mean and variance of the number who pay with a debit card?

mean	customers
variance	customers2

Explain your reasoning.

Because we are interested in whether or not a debit card was used, we can use the binomial distribution. X = the number of customers at the gas station.Because we are interested in whether or not a debit card was used, we can use the binomial distribution. X = the number of customers who use a debit card.    Because we are interested in whether or not a debit card was used, we can use the binomial distribution. X = the probability that a customer used a debit card.

(b) Answer part (a) for the number among the 100 who don't pay with cash.

mean	customers
variance	customers2

Answer 1

Part a)

Let X be the number of customers who use a debit card. So X has the binomial distribution with trials n ( number of customers )= 100 and success probability p = 0.3 and probability of failure q = 1 - p = 1 - 0.3 = 0.7

Mean of binomial distribution = n*p = 100 *0.3 = 30

Variance of binomial distribution = n*p*q = 100*0.3*0.7  = 21

So Mean = 30 customers.

Variance = 21 .

Part b)

Let Y be the number of customers who don’t pay with cash.

So Y also follows normal distribution with trials n = 100 , and p = 0.5+0.3 = 0.8 , q = 1 - 0.8 = 0.2

Here p is probability that customer don't use cash, that is they either use credit card or debit card.

p is probability that customer use cash.

So Mean = n*p = 100*0.8 = 80 customers

Variance = n*p*q = 100*0.8*0.2 = 16