Question

A company is interested in evaluating its current inspection procedure on large shipments of identical items. The procedure is to take a sample of 5 items and pass the shipment if no more than 1 item is found to be defective. It is known that items are defective at a 10% rate overall. a. What is the probability that the inspection procedure will pass the shipment? b. What is the expected number of defectives in this process of inspecting 5 items? c. If items are defective at a 20% rate overall, what is the probability that you will find 4 defectives in a sample of 5?

Answer 1

Answer:

(a) The probability that the inspection procedure will pass the shipment is 0.9185.

(b) The expected number of defectives in this process of inspecting 5 items is 0.50.

(c) The probability that there will be 4 defectives in a sample of 5 is 0.0016.

Step-by-step explanation:

Let X = number of defective items.

The probability of selecting a defective item is, p = 0.10.

A sample of n = 5 items is selected at random.

The random variable X follows a Binomial distribution with parameters n = 5 and p = 0.10.

The probability mass function of X is:

$P(X=x)={5\choose x}0.10^{x}(1-0.10)^{5-x};\ x=0,1,2,3....$

It is provided that the shipment will pass if there are no more than 1 defective items in the selected 5 units.

(a)

Compute the probability that the inspection procedure will pass the shipment as follows:

P (X ≤ 1) = P (X = 0) + P (X = 1)

             $={5\choose 0}0.10^{0}(1-0.10)^{5-0}+{5\choose 1}0.10^{1}(1-0.10)^{5-1}\\=(1\times 1\times 0.59049) + (5\times 0.10\times 0.6561)\\=0.59049+0.32805\\=0.91854\\\approx0.9185$

Thus, the probability that the inspection procedure will pass the shipment is 0.9185.

(b)

The expected value of a Binomial distribution is:

$E(X)=np$

Compute the expected number of defectives in this process of inspecting 5 items as follows:

$E(X)=5\times0.10=0.50$

Thus, the expected number of defectives in this process of inspecting 5 items is 0.50.

(c)

The probability of finding a defective is now changed to 0.20.

Compute the probability that there will be 4 defectives in a sample of 5 as follows;

$P(X=4)={5\choose 4}0.20^{4}(1-0.20)^{5-4}\\=5\times0.0016\times0.20\\=0.0016$

Thus, the probability that there will be 4 defectives in a sample of 5 is 0.0016.

Answer 2

Answer:

(a) Probability that the inspection procedure will pass the shipment = 0.91854

(b) E(X) = $5 \times 0.1$ = 0.5

(c) Probability of 4 defectives in a sample of 5 is 0.0064      

Step-by-step explanation:

We are given that a company is interested in evaluating its current inspection procedure on large shipments of identical items. The procedure is to take a sample of 5 items and pass the shipment if no more than 1 item is found to be defective. It is known that items are defective at a 10% rate overall.

The above situation can be represented through Binomial distribution;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 5 items

            r = number of success

           p = probability of success which in our question is % of items that are

                  defective, i.e., 10%

LET X = Number of defective items

Also, it is given that a sample of 5 items is taken,

So, it means X ~ $Binom(n=5,p=0.10)$

(a) Probability that the inspection procedure will pass the shipment is given by the fact that if no more than 1 item is found to be defective, then only shipment is passed, i.e.;

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

             = $\binom{5}{0}0.1^{0} (1-0.1)^{5-0} + \binom{5}{1}0.1^{1} (1-0.1)^{5-1}$

             = $1 \times 1 \times 0.9^{5} +5 \times 0.1 \times 0.9^{4}$

             = 0.59049 + 0.32805 = 0.91854

(b) The expected number of defectives in this process of inspecting 5 items is given by = E(X) or Mean of X

So, mean of binomial distribution is, E(X) = $n \times p$

      So,  E(X) = $5 \times 0.1$ = 0.5

Therefore, the expected number of defectives in this process of inspecting 5 items is 1 (after rounding to nearest integer).

(c) Now, the probability of success which is % of items that are defective is changed to 20% rate overall, i.e., p = 0.20 now.

So, probability that you will find 4 defectives in a sample of 5 = P(X = 4)

 P(X = 4) = $\binom{5}{4}0.2^{4} (1-0.2)^{5-4}$

               =  $5 \times 0.2^{4} \times 0.8^{1}$ = 0.0064