Question

Suppose a shipment of 400 components contains 68 defective and 332 non-defective computer components. From the shipment you take a random sample of 25. When sampling with replacement (so that the p = probability of success does not change), note that a success in this case is selecting a defective part. In our sample of 25 we reject the entire shipment if X>5. What is the probability of rejecting the entire shipment?

Answer 1

Answer:

0.2425 or 24.25%

Step-by-step explanation:

If the shipment of 400 components contains 68 defective and, we can estimate the probability of selecting a defective components as

68/400 = 0.17

Now, we have a binomial experiment of 25 trials and we want to compute the probability that more than 5 trials are “success” (defective component)

P(X>5) = 1 - P(X≤ 5)

Since  

1 - P(X≤ 5) = 1 - (P(X=0) + P(X=1) + P(X=3) + P(X=4) + P(X=5) =  

$1-(\displaystyle\binom{25}{0}(0.17)^0(1-0.17)^{25}+\displaystyle\binom{25}{1}(0.17)^1(1-0.17)^{24}+\displaystyle\binom{25}{2}(0.17)^2(1-0.17)^{23}+\\\\+\displaystyle\binom{25}{3}(0.17)^3(1-0.17)^{22}+\displaystyle\binom{25}{4}(0.17)^4(1-0.17)^{21}+\displaystyle\binom{25}{5}(0.17)^5(1-0.17)^{20})=\\\\\Rightarrow \boxed{P(X>5)=0.2425}$

Answer 2

Using the binomial distribution, it is found that there is a 0.2425 = 24.25% probability of rejecting the entire shipment.

The components are chosen with replacement, thus, the binomial distribution is used to solve this question.

Binomial distribution:  

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

• x is the number of successes.
• p is the probability of a success on a single trial.
• n is the number of trials.

In this problem:

• Sample of 25, thus $n = 25$
• 68 out of 400 are defective, thus $p = \frac{68}{400} = 0.17$

The probability of rejecting the entire shipment is:

$P(X > 5) = 1 - P(X \leq 5)$

In which:

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

Then

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{25,0}.(0.17)^{0}.(0.83)^{25} = 0.0095$

$P(X = 1) = C_{25,1}.(0.17)^{1}.(0.83)^{24} = 0.0486$

$P(X = 2) = C_{25,2}.(0.17)^{2}.(0.83)^{23} = 0.1193$

$P(X = 3) = C_{25,3}.(0.17)^{3}.(0.83)^{22} = 0.1874$

$P(X = 4) = C_{25,4}.(0.17)^{4}.(0.83)^{21} = 0.2111$

$P(X = 5) = C_{25,5}.(0.17)^{5}.(0.83)^{20} = 0.1816$

Then:

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0095 + 0.0486 + 0.1193 + 0.1874 + 0.2111 + 0.1816 = 0.7575$

$P(X > 5) = 1 - P(X \leq 5) = 1 - 0.7575 = 0.2425$

0.2425 = 24.25% probability of rejecting the entire shipment.

A similar problem is given at https://brainly.com/question/24863377