Question

The Hawkins Company randomly samples 10 items from every large batch before the batch is packaged and shipped. According to the contract specifications, 5 percent of the items shipped can be defective. If the inspectors find 1 or fewer defects in the sample of 10, they ship the batch without further inspection. If they find 2 or more, the entire batch is inspected.
(A) Based on this sampling plan, the probability that a batch that meets the contract requirements will be shipped without further inspection is approximately___________.

Answer 1

Answer:

The answer is 0.9138

Step-by-step explanation:

This is a question of Binomial Probability Combination and the formula that will be used in this question is:

$P(x) = nCx * p^{x} * q^{n - x}$

Where,

• n ⇒ Finite sample number, here it is equal to 10
• p ⇒ Success, here it is equal to 5% defective or $\frac{5}{100} = 0.05$
• q ⇒ Failure, here it is equal to (1 - p) i.e. $q = 1 - \frac{5}{100} = 0.95$
• $x$ ⇒ Number of defective products

Now, in order to to pass inspection and be shipped $x$ needs to be equal to '0' or '1'.

So,

Putting $x = 0$ in the formula along with the values mentioned above we get

$P(0) = 10C0 * 0.05^{0} * 0.95^{10 - 0}$

$P(0) = 1 * 1 * 0.5987$

$P(0) = 0.5987$

Similarly,

Putting $x = 1$ in the formula along with the values mentioned in the bullets above we get

$P(1) = 10C1 * 0.05^{1} * 0.95^{10 - 1}$

$P(1) = 10 * 0.05 * 0.6302$

$P(1) = 0.3151$

Now, in order to get the actual probability we need to add $P(0)$ and $P(1)$ because there is a chance that either there is no defective product or there is 1 defective in the shipped-batch from which sample was taken. Hence,

$P(0) + P(1) = 0.5987 + 0.3151$

= 0.9138 (Answer)

Answer 2

Answer:

0.9138 = 91.38%

Step-by-step explanation:

For each item, there are only two possible outcomes. Either it is defective, or it is not. The probability of an item being defective is independent of other items. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

10 items

This means that $n = 10$

5 percent of the items shipped can be defective.

This means that $p = 0.05$

Probability that a batch that meets the contract requirements will be shipped without further inspection

Probability of 1 or fewer defects.

So

$P(X \leq 1) = P(X = 0) + P(X = 1)$

In which

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

$P(X = 0) = C_{10,0}.(0.05)^{0}.(0.95)^{10} = 0.5987$

$P(X = 1) = C_{10,1}.(0.05)^{1}.(0.95)^{9} = 0.3151$

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.5987 + 0.3151 = 0.9138$

So the answer is:

0.9138 = 91.38%