A store selling newspapers orders only n = 4 of a certain

Chapter 2, Problem 9E

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QUESTION:

Problem 9E

A store selling newspapers orders only n = 4 of a certain newspaper because the manager does not get many calls for that publication. If the number of requests per day follows a Poisson distribution with mean 3,

(a) What is the expected value of the number sold?

(b) What is the minimum number that the manager should order so that the chance of having more requests than available newspapers is less than 0.05?

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QUESTION:

Problem 9E

A store selling newspapers orders only n = 4 of a certain newspaper because the manager does not get many calls for that publication. If the number of requests per day follows a Poisson distribution with mean 3,

(a) What is the expected value of the number sold?

(b) What is the minimum number that the manager should order so that the chance of having more requests than available newspapers is less than 0.05?

ANSWER:

Solution 9E

Step1 of 3:

We have A store selling newspapers orders only n = 4 of a certain newspaper.

If the number of requests per day follows a Poisson distribution with mean 3.

We need to find,

(a) What is the expected value of the number sold?

(b) What is the minimum number that the manager should order so that the chance of having more requests than available newspapers is less than 0.05?

Step2 of 3:

Let “X” be random variable which follows poisson distribution with parameters .

That is X P()

The probability mass function of binomial distribution is given below

P(X) = , x =0,1,2,3,...,n.

Where,

X = random variable

n = sample size

p = probability of success(or proportion).

e = a constant it is approximately 2.7182.

a).

Here3 and n = 4  

Let number of papers requested = X

Let number of papers sold = Y

That is Y = {0, 1, 2, 3, 4}

Then,

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