Tom works the night shift at a gas station. From pastexperience, Tom knows that

Question

Tom works the night shift at a gas station. From pastexperience, Tom knows that 80% of all customers who purchase gas pay at the pump with a credit card. Assume that customers either pay at the pump with acredit card or not, independently of each other.One evening 15 customers arrive at the gas station and Tom counts the number of them that pay with a credit card.

a)(1.5pts) Define a random variable for this problem and state its distribution and parameters.

b)(1pt) Out of 15 customers who purchase gas, how manyshould Tom expect to pay at the pump with a credit card?

c)(1pt) What is the probability that exactly 10customers out of 15 pay at the pump with acredit card?

d)(1.5pts) What is the probability that at least 9customers out of 15 pay at thepump with a credit card?

e)(1.5pts) What is the probability that the number of customers purchasing gas with a credit card is somewhere from 3 to 12 customers?

Explanation / Answer

Solution:

Part a

The random variable for this problem is defined as the number of customers who pay with a credit card at gas station. This random variable follows a binomial distribution with parameters n = 15 and p = 0.80.

X ~ B(n = 15, p = 0.80)

Part b

Expected number of customers who pay with a credit card = n*p

Expected number of customers = 15*0.80 = 12

Expected number of customers who pay with a credit card is 12.

Part c

We have to find P(X=10)

P(X=x) = nCx*p^x*q^(n – x)

Where, q = 1 – p

We are given, n = 15, x = 10, p = 0.80, q = 1 – 0.80 = 0.20

P(X=10) = 15C10*0.80^10*0.20^(15 – 10)

P(X=10) = 3003* 0.107374* 0.00032

P(X=10) = 0.103182

Required probability = 0.103182

Part d

We have to find P(X9)

P(X9) = 1 – P(X8)

P(X8) = 1 - 0.018059

(By using Binomial table/Excel)

P(X8) = 0.981941

Required probability = 0.981941

Part e

We have to find P(3X12)

P(3X12) = P(X12) – P(X<3) = P(X12) – P(X2)

P(X12) = 0.601977

P(X2) = 0.000000057

P(3X12) = 0.601977 – 0.000000057

P(3X12) = 0.601976734

Required probability = 0.601976734

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