Question

5.         Wellington Fabrics of New Zealand produces bolts of woolen cloth for export. Each bolt contains 30 yards of fabric. Industry standards call for the average number of defects per fabric bolt to not exceed five. An inspector randomly selected a bolt of cloth, examined the first 3 yards and found 3 defects therein. The company assumes that the defect rate follows the Poisson distribution.

a.          Given the above information, if the company is meeting the industry standards, what is the average number of defects expected in a 3-yard segment of a cloth bolt?

b.         Calculate the probability of finding 3 or more defects in a 3-yard segment of cloth given the above information.

c.          Given your answer to part b., does it appear that the company is meeting the industry standard for quality? Explain briefly.

d.         To verify the findings the inspector examines 15 yards of another bolt of cloth and discovers five defects.

i.          Calculate the average number of defects expected in a 15-yard segment of cloth;

ii.         Calculate the probability of finding five or more defects in the 15-yard cloth segment.

e.          The normal distribution can be used to approximate the Poisson distribution. To calculate a z-score you need the mean and standard deviation. Remember that the mean and variance of a Poisson random variable are equal.

i.          Calculate the mean and standard deviation for the number of defects in a 15-yard segment of cloth.

ii.         Use this mean and standard deviation to calculate the probability of finding five or more defects in a 15-yard segment of cloth using the normal approximation. [Remember to use the continuity correction!]

iii.        How accurate is this approximation?

Answer 1

Solution

Let X = number of defects per fabric bolt i.e., 30 yards of fabric.

Given the defect rate follows the Poisson distribution,

X ~ Poisson (λ), where λ = average number of defects per 30 yards of fabric. …………. (1)

Back-up Theory

If a random variable X ~ Poisson (λ), i.e., X has Poisson Distribution with mean λ then

probability mass function (pmf) of X is given by P(X = x) = e ^{– λ}.λ^x/(x!) ……………..……..(2)

where x = 0, 1, 2, ……. , ∞

Values of p(x) for various values of λ and x can be obtained by using Excel Function,

Statistical, POISSON(x,Mean,Cumulative) ………………………………………………. (2a)

If X = number of times an event occurs during period t, Y = number of times the same

event occurs during period kt, and X ~ Poisson(λ), then Y ~ Poisson (kλ) ……………….. (3)

Now to work out the solution,

Industry standards call for the average number of defects per fabric bolt to not exceed five

=> λ ≤ 5. An inspector randomly selected a bolt of cloth, examined the first 3 yards and found 3 defects therein.

Part (a)

If the company is meeting the industry standards, the average number of defects expected per 30 yards ≤ 5. So, vide (3), average number of defects expected in a 3-yard segment of a cloth bolt

≤ (5 x 3)/30 = 0.5 Answer 1

Part (b)

If Y = number of defects per 3 yards of fabric, then vide (3), Y ~ Poisson (0.5) and so

probability of finding 3 or more defects in a 3-yard segment of cloth

= P(Y ≥ 3)

= 0.0144 Answer 2

Part (c)

Since the probability in Part (b) is very low, it is most unlikely that the company is meeting the industry standards. Answer   3

Part (d)(i)

If the company is meeting the industry standards, the average number of defects expected per 30 yards ≤ 5. So, vide (3), average number of defects expected in a 15-yard segment of a cloth bolt

≤ (5 x 15)/30 = 2.5 Answer 4

Part (d)(ii)

If W = number of defects per 15 yards of fabric, then vide (3), Y ~ Poisson (2.5) and so

probability of finding 5 or more defects in a 15-yard segment of cloth

= P(W ≥ 5)

= 0.1088 Answer 5

Part (e)(i)

Mean and standard deviation for the number of defects in a 15-yard segment of cloth

= Mean and standard deviation of W

= 2.5 and √2.5

Thus, Mean = 2.5   Answer 6

Standard deviation = 1.5811   Answer 7

Part (e)(ii)

Probability of finding 5 or more defects in a 15-yard segment of cloth

= P(W ≥ 5)

= P[z ≥ {(4.5 – 2.5)/1.5811}]

= P(Z ≥ 1.2649)

= 0.103 Answer 7

Part (e)(iii)

Vide Answer 5 and Answer 7, the Normal approximation is quite accurate. Answer 8

DONE