4. (a) Suppose that an employee of a company that makes batteries tells you that

Question

4. (a) Suppose that an employee of a company that makes batteries tells you that on average 0.5% of the batteries produced contain voltage that is too low. The employee further tells you that a shipment contains 500 batteries.

i. Find the expected number of low-voltage batteries and the variance in number of low-voltage batteries in shipments of 500.

ii. Find the probability of a shipment of 500 batteries containing 4 or more low-voltage batteries.

(b) Now suppose (without knowing the information in (a)) that a different employee tells you that on average a shipment contains 2.5 low-voltage batteries. This employee does not tell you the size of the shipment or the percentage of low-voltage batteries. Assume it is reasonable to use a Poisson random variable to model the number of low-voltage batteries in a shipment.

i. Find the expected number of low-voltage batteries and the variance in number of low-voltage batteries per shipment.

ii. Find the probability of a shipment of containing 4 or more low-voltage batteries.

(c) Compare your answers in (a) with your answers in (b). Is it possible that both employees are telling you the truth?

Explanation / Answer

Question 4.a.i.

Here, we have to find the expected number of low voltage batteries and variance in number of low voltage batteries in shipments of 500.

Expected number = n*p

We are given n = 500 and p = 0.005

So, expected number = mean = n*p = 500*0.005 = 2.5

Variance = n*p*q

Where, q = 1 – p = 1 – 0.005 = 0.995

Variance = 500*0.005*0.995 = 2.4875

SD = sqrt(2.4875) = 1.577181

Question 4.a.ii.

Here, we have to use normal approximation to binomial distribution.

We have to find P(X4) = 1 – P(X<4)

Z = (X – mean)/SD

Z = (4 – 2.5) / 1.577181

Z = 0.951064

P(Z<0.951064) = 0.829214

P(X4) = 1 – P(X<4) = 1 - 0.829214 = 0.170786

Required probability = 0.170786

Question 4.b.i.

Here, we have to use Poisson approximation to binomial distribution.

Mean for Poisson distribution =

= µ = n*p = 500*0.005 = 2.5

Question 4.b.ii.

We have to find P(X4)

P(X4) = 1 – P(X<4)

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

P(X=x) = ^x*e^(-) / x!

P(X=0) = 2.5^0*e^(-2.5)/0! = 0.082085

P(X=1) = 2.5^1*e^(-2.5)/1! = 0.205212

P(X=2) = 2.5^2*e^(-2.5)/2! = 0.256516

P(X=3) = 2.5^3*e^(-2.5)/3! = 0.213763

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

P(X<4) = 0.757576

P(X4) = 1 – P(X<4)

P(X4) = 1 – 0.757576

P(X4) = 0.242424

Required probability = 0.242424

Question 4.b.

We observed that means in both parts (a) and (b) are same. Required probabilities are different in both parts due to use of normal approximation to binomial distribution and Poisson approximation to binomial distribution. So, it is possible that both employees are telling you the truth.

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