Problem 16E

A distributor receives a large shipment of components. The distributor would like to accept the shipment if 10% or fewer of the components are defective and to return it if more than 10% of the components are defective. She decides to sample 10 components, and to return the shipment if more than 1 of the 10 is defective.

a. If the proportion of defectives in the batch is in fact 10%, what is the probability that she will return the shipment?

b. If the proportion of defectives in the batch is 20%, what is the probability that she will return the shipment?

c. If the proportion of defectives in the batch is 2%, what is the probability that she will return the shipment?

d. The distributor decides that she will accept the shipment only if none of the sampled items are defective. What is the minimum number of items she should sample if she wants to have a probability no greater than 0.01 of accepting the shipment if 20% of the components in the shipment are defective?

Solution 16E

Step1 of 5:

Let us consider a random variable X and it represents the number of defects in a 10 sampled.

Here the proportion of defectives is p = 10% then X follows binomial distribution with parameters “n and p” that is X B(n, p),

The probability mass function of binomial distribution is given by

, x = 0,1,2,...,n.

Where,

n = sample size

= 10

x = random variable

p = probability of success

= 10%

= 0.10.

q = 1 - p (probability of failure)

= 1 - 0.10

= 0.90

Here our goal is:

a).We need to find the probability that she will return the shipment, when the proportion of defectives in the batch is in fact 10%.

b).We need to find the probability that she will return the shipment, when the proportion of defectives in the batch is 20%.

c).We need to find the probability that she will return the shipment, when the proportion of defectives in the batch is 2%.

d).We need to find the minimum number of items she should sample when she wants to have a probability no greater than 0.01 of accepting the shipment if 20% of the components in the shipment are defective.

Step2 of 5:

a).

We have n = 10 and p = 0.10.

P(The probability that she will return the shipment) = P(X > 1)

P(X > 1) = 1 - P(X1)

= 1 - {P(X = 0)+P(X = 1)}

= 1 - {+}

= 1 - {1(1)(0.3486)+10(0.10)(0.3874)}

= 1 - {0.3486+0.3874}

= 1 - 0.7360

= 0.2640

Hence, when the proportion of defectives in the batch is in fact 10% then,

P(The probability that she will return the shipment) = 0.2640.

Step3 of 5:

b).

We have n = 10 and p = 0.20.

P(the probability that she will return the shipment) = P(X > 1)

P(X > 1) = 1 - P(X1)

= 1 - {P(X = 0)+P(X = 1)}

= 1 - {+}

= 1 - {1(1)(0.1073)+10(0.20)(0.1342)}

= 1 - {0.1073+0.2684}

= 1 - 0.3757

= 0.6243

Hence, when the proportion of defectives in the batch is in fact 20% then,

P(The probability that she will return the shipment) = 0.6243.

Step4 of 5:

c).

We have n = 10 and p = 0.02.

P(the probability that she will return the shipment) = P(X > 1)

P(X > 1) = 1 - P(X1)

= 1 - {P(X = 0)+P(X = 1)}

= 1 - {+}

= 1 - {1(1)(0.8170)+10(0.02)(0.8337)}

= 1 - {0.8170+0.1667}

= 1 - 0.9837

= 0.0163

Hence, when the proportion of defectives in the batch is in fact 2% then,

P(The probability that she will return the shipment) = 0.0163.

Step5 of 5:

d).

Consider P(none of the sampled items are defective) = 0.01

P(X = 0) 0.01

(1)(1)

0.01

Apply log on both side then we get

nlog(0.8) log(0.01)

n

n

n20.6377

Hence, n 20.6377.