4. The probabilities of weld failures on items from a certain assembly line are:

Question

4. The probabilities of weld failures on items from a certain assembly line are: 0.9:1 in the actual weld metal itself and the remaining 0.09 in the base metal. A random sample of 50 weld failures is examined. If you assume that the probability of a base-metal weld failure is 0.10 and that each weld failure is independent of all other weld failures in the sample, (a) [5 points] what is the probability of exactly 5 base metal failures in the sample of 50? (b) [5 points] what is the probability of fewer than 3 base metal failures? (c) [5 points] what is the probability of 0 base metal failures? (d) [5 points] what is the expected value (mean) of the number of base metal failures? (e) [5 points] what is the variance of the number of base metal failures?

Explanation / Answer

Let X = number of base metal failures in a sample of 50 weld failures. Then,

X ~ B(50, p), where p = probability of a base metal failure, which is given to be 0.1.

Back-up Theory

If X ~ B(n, p). i.e., X has Binomial Distribution with parameters n and p, where n = number of trials and p = probability of one success, then

probability mass function (pmf) of X is given by

p(x) = P(X = x) = (nCx)(px)(1 - p)n – x, x = 0, 1, 2, ……. , n ………………………………..(1)

[The above probability can also be directly obtained using Excel Function of Binomial Distribution] ……………………………………………………………………………….(1a)

Mean (average) of X = E(X) = np…………………………………………………………..(2)

Variance of X = V(X) = np(1 – p)…………………………………………………………..(3)

Part (a)

Probability of exactly 5 base metal failures

= P(X = 5)

= (50C5)(0.15)(0.9)15[vide (1) above]

= 0.1849 [using Excel Function] ANSWER

Part (b)

Probability of fewer than 3 base metal failures

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

= (50C0)(0.10)(0.9)20 + (50C1)(0.11)(0.9)19 + (50C2)(0.12)(0.9)18 + [vide (1) above]

= 0.1117 [using Excel Function] ANSWER

Part (c)

Probability of zero base metal failures

= P(X = 0)

= (50C0)(0.10)(0.9)20 [vide (1) above]

= 0.0052 [using Excel Function] ANSWER

Part (d)

Expected value (mean) of the number of base metal failures in a sample of 50 weld failures

= E(X) = 50 x 0.1 [vide (2) above]

= 5 ANSWER

Part (e)

Variance of the number of base metal failures in a sample of 50 weld failures

= V(X) = 50 x 0.1 x 0.9 [vide (3) above]

= 4.5 ANSWER

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