## 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 samp

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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2021-09-21T23:26:14+00:00
2021-09-21T23:26:14+00:00 1 Answer
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## Answers ( )

Given:–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,

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 =

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