3. Consider the arrival of customers and sales at a car dealership. (a) Suppose

Question

3. Consider the arrival of customers and sales at a car dealership. (a) Suppose that customer arrivals are a Poisson process with 2.4 customer per 1. What is the probability that less than 5 customers arrive during the next hour? (3 points) next customer exceeds 30 minutes? (2 points) (b) Suppose that successive customers are Bernoulli trials where 20% buy a car. i. What is the probability that 2 of the next 10 customers buy a car? (3 points) i. What is the probability it takes 10 customers to get the 2nd sale? (2 points)

Explanation / Answer

Given = 2.4, n = less than 5, t =1
Probability that less than 5 customer arrive next hour will be

P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4)

= e-2.4(2.4)0/0! + e-2.4(2.4)1/1! + e-2.4(2.4)2/2! + e-2.4(2.4)3/3! + e-2.4(2.4)4/4!

= 0.0907 + 0.2177 + 0.2612 + 0.2090 + 0.1254

= 0.9040

Similarly in 2nd we can also calculate the probability for next customer exceeds 30 minutes by putting the values

(e-k )/k!

b(i) Bernoulli random variable:
It is a variable that has 2 possible outcomes: “success”, or “failure”.
Success occurs with probability p and failure with probability
1 p.

P(X = x) = nCx px(1 p)nx
,P(X=2) = 10C2 (0.2)2(1 - 0.2)10-2

P(X=2) = 45 x 0.04 x (0.8)8

= 1.8 x 0.1677 = 0.3019

(ii) In this question the probability will be same as in above question because 10 customer to get 2nd sale means 2 customers buying the car out of 10.

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