Dell Computers receives large shipments of microprocessors (chips) from Intel Corp. It must try to ensure that the proportion of defective chips is small and has contracted to keep the proportion of defective chips at 5% or less. Suppose that Dell tests 25 chips (with replacement) out of a shipment of thousands.
a) What is the probability of obtaining at least three defects in the sample, if the entire shipment has 5% defectives? If a sample results in at least three defects, does it appear that Intel has provided a shipment which is consistent with the contract? Briefly explain, using the calculated probability.
b) For the shipment described in part a, find the expected number of sampled chips that are defective and the standard deviation of the number of sampled chips that are defective.
c) Suppose that 20% of the entire shipment is defective. What is the probability that more than 17 chips in the sample are good (not defective)?​

Using the binomial distribution, it is found that:

a) 0.1271 = 12.71% probability of obtaining at least three defects in the sample, if the entire shipment has 5% defectives. 5% defective is between 1 and 2 defects, but since this probability is more than the unlikely threshold 5%, more than 2 defects could still mean that Intel has provided a shipment which is consistent with the contract.

b) The mean is of 1.25 defective chips and the standard deviation is of 1.09 defective chips.

c) 0.8908 = 89.08% probability that more than 17 chips in the sample are good.

For each chip, there are only two possible outcomes, either it is defective, or it is not. The probability of a chip being defective is independent of any other chip, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

In this problem:

Item a:

Supposing 5% defective, we have that [tex]p = 0.05[/tex].

The probability is:

[tex]P(X \geq 3) = 1 - P(X < 3)[/tex]

In which

[tex]P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]

Then

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{25,0}.(0.05)^{0}.(0.95)^{25} = 0.2774[/tex]

[tex]P(X = 1) = C_{25,1}.(0.05)^{1}.(0.95)^{24} = 0.3650[/tex]

[tex]P(X = 2) = C_{25,2}.(0.05)^{2}.(0.95)^{23} = 0.2305[/tex]

[tex]P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.2774 + 0.3650 + 0.2305 = 0.8729[/tex]

[tex]P(X \geq 3) = 1 - P(X < 3) = 1 - 0.8729 = 0.1271[/tex]

0.1271 = 12.71% probability of obtaining at least three defects in the sample, if the entire shipment has 5% defectives. 5% defective is between 1 and 2 defects, but since this probability is more than the unlikely threshold 5%, more than 2 defects could still mean that Intel has provided a shipment which is consistent with the contract.

Item b:

The mean is:

[tex]E(X) = np = 25(0.05) = 1.25[/tex]

The standard deviation is:

[tex]\sqrt{V(X)} = \sqrt{np(1 - p)} = \sqrt{25(0.05)(0.95)} = 1.09[/tex]

The mean is of 1.25 defective chips and the standard deviation is of 1.09 defective chips.

Item c:

20% are defective, so 80% are good, which means that [tex]p = 0.8[/tex]

The probability is:

[tex]P(X > 17) = P(X = 18) + P(X = 19) + P(X = 20) + P(X = 21) + P(X = 22) + P(X = 23) + P(X = 24) + P(X = 25)[/tex]

Then

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 18) = C_{25,18}.(0.8)^{18}.(0.2)^{7} = 0.1108[/tex]

[tex]P(X = 19) = C_{25,19}.(0.8)^{19}.(0.2)^{6} = 0.1633[/tex]

[tex]P(X = 20) = C_{25,20}.(0.8)^{20}.(0.2)^{5} = 0.1960[/tex]

[tex]P(X = 21) = C_{25,21}.(0.8)^{21}.(0.2)^{4} = 0.1867[/tex]

[tex]P(X = 22) = C_{25,22}.(0.8)^{22}.(0.2)^{3} = 0.1358[/tex]

[tex]P(X = 23) = C_{25,23}.(0.8)^{23}.(0.2)^{2} = 0.0708[/tex]

[tex]P(X = 24) = C_{25,24}.(0.8)^{24}.(0.2)^{1} = 0.0236[/tex]

[tex]P(X = 25) = C_{25,25}.(0.8)^{25}.(0.2)^{0} = 0.0038[/tex]

Then:

[tex]P(X > 17) = P(X = 18) + P(X = 19) + P(X = 20) + P(X = 21) + P(X = 22) + P(X = 23) + P(X = 24) + P(X = 25) = 0.1108 + 0.1633 + 0.1960 + 0.1867 + 0.1358 + 0.0708 + 0.0236 + 0.0038 = 0.8908[/tex]

0.8908 = 89.08% probability that more than 17 chips in the sample are good.

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