Question

In sampling a large number of parts are manufactured by a machine, the mean number of defective in a sample of 20 is 2,out of 1000 such samples. How many would br expected to contain atleast 3 defective parts?
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Answer 1

Given :- In sampling a large number of parts are manufactured by a machine, the mean number of defective in a sample of 20 is 2,out of 1000 such samples. How many would br expected to contain atleast 3 defective parts ?

Solution :-

As given , this is case of simple binomial. we are given that probability that an item would be defective in a sample of 20 is 2 .

So,

→ P = (2/20) = (1/10) = 0.1

As, we know,

→ Q = 1 - P

So,

→ Q = 1 - 0.1 = 0.9 . { Probability of non defective .}

now we have , n = 20 and N = 1000 .

as we know,

• P(x = X) = n(C)x * P^x * q^(n - x)

So, in order to find 3 expected defective :-

→ P(x = 3) = 20(C)3 * (0.1)³ * (0.9)¹⁷ = 0.323 .

Therefore,

→ The number of samples having at least three defective parts out of 1000 samples = 1000 * 0.323 = 323. (Ans.) .

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