Question

A factory produces metal links for building long chains used in the shipping and construction industry. The factory’s R&D unit models the length of an individual metal link (in ?m) by a random variable ?? with expectation ?0 m and variance ?. ?4 ??^?. The length of a chain is equal to the sum of the lengths of individual links that comprise the chain. The factory sells ?0 ? long chains as its final product, which it makes by combining ?002 individual metal links together. The factory guarantees that the chain length is not shorter than ?0 ? − if by chance a chain is too short, the customer is reimbursed and a new chain is issued for free.

a. For what percentage of chains would the factory need to reimburse clients and deliver new chains free of cost?

b. The sales team notices that the factory has been handing out a much larger fraction of free chains than expected as per your answer above. After further probing, the research lab reports that the exact expectation is ?. ?? ?m and not ? ?m as was previously reported based on rounding-off approximation. Do you think the research team committed a serious mistake by rounding off to the nearest whole number instead of using the exact value for expectation? Why?

Answer 1

Solution

Let Xi =length of the ith link. Then, length of the chain, say L = X1 + X2 + ……. + X1002.

Back-up Theory

If X1, X2, ……. ,Xn are iid with mean µ and standard deviation σ, and S = Σ(i = 1 to n)Xi, then,

Mean of S = nµ and standard deviation σ√n. …………………………………………………………. (1)

Now to work out the solution,

Given n = 1002, µ = 5cm and σ = 0.2 cm, nµ = 50.1 m and σ√n = 6.3309 cm = 0.0633 m

So, vide (1), mean of S = 50.1 m and s d of S = 0.0633 m

Part (a)

Percentage of chains the factory need to reimburse clients and deliver new chains free of cost

= 100 x P(S < 50 m)

= 100 x P[Z < {(50 – 50.1)/0.0633}], where Z = {(S – 50.1)/0.0633} ~ N(0, 1)

= 100 x P(Z < - 1.5797)

= 5.71% Answer 1

Part (b)

With n = 1002, µ = 4.99 cm and σ = 0.2 cm, nµ = 49.9998 m and σ√n = 6.3309 cm = 0.0633 m

So, mean of S = 49.9998 m and s d of S = 0.0633 m

Then,

Percentage of chains the factory need to reimburse clients and deliver new chains free of cost

= 100 x P(S < 50 m)

= 100 x P[Z < {(50 – 49.9998)/0.0633}], where Z = {(S – 49.9998)/0.0633} ~ N(0, 1)

= 100 x P(Z < 0.0031)

= 50.21% Answer 2

So, if the research team had taken µ = 4.99 cm, the team would have realized the percentage claim would be as high as 50% and then would have taken corrective steps.

Thus, the research team did commit a serious mistake by rounding off to the nearest whole number instead of using the exact value for expectation. Answer 3

DONE