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Consider a large shipment of 400 refrigerators, of which 40 have defective compressors. If X is the number among 15 randomly selected refrigerators that have defective compressors, describe a less tedious way to calculate (at least approximately) P(X ≤ 3) than to use the hypergeometric pmf. We can approximate the hypergeometric distribution with the distribution if the population size and the number of successes are large. Here n = and p = M/N = . Approximate P(X ≤ 3) using that method. (Round your answer to three decimal places.) P(X ≤ 3) ≈

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Answer:

The required probability is 0.94

Step-by-step explanation:

Consider the provided information.

There are 400 refrigerators, of which 40 have defective compressors.

Therefore N = 400 and X = 40

The probability of defective compressors is:

[tex]\frac{40}{400}=0.10[/tex]

It is given that If X is the number among 15 randomly selected refrigerators that have defective compressors,

That means n=15

Apply the probability density function.

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

We need to find P(X ≤ 3)

[tex]P(X\leq3) =P(X=0)+P(X=1)+P(X=2)+P(X=3)\\P(X\leq3) =\frac{15!}{15!}(0.1)^0(1-0.1)^{15}+\frac{15!}{14!}(0.1)^1(1-0.1)^{14}+\frac{15!}{13!2!}(0.1)^2(1-0.1)^{13}+\frac{15!}{12!3!}(0.1)^3(1-0.1)^{12}\\[/tex]

[tex]P(X\leq3) =0.944444369992\approx 0.94[/tex]

Hence, the required probability is 0.94

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