A manufacturer of computer printers purchases plastic ink cartridges from a vendor. When a large shipment is received, arandom sample of 200 cartridges is selected, and each cartridge is inspected. If the sample proportion of defectivecartridges is more than .02, the entire shipment is returned to the vendor. a. What is the approximate probability that a shipment will be returned if the true proportion of defective cartridges in theshipment is .05?b. What is the approximate probability that a shipment will not be returned ifthe true proportion of defective cartridges inthe shipment is .10?

(1) The probability that the shipment will be returned  if the true proportion of defective cartridges in the shipment is 0.05 is 0.9744.

(2) The probability that the shipment will not be returned if the true proportion of defective cartridges in the shipment is 0.10 is 0.00008.

Step-by-step explanation:

Let X = Number of defective cartridges.

The sample size is: n = 200.

The probability that the shipment is rejected is, p = 0.02.

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

But the sample size is too large, i.e. n = 200 > 30.

Then according to the normal approximation to binomial, if np ≥ 10 or n(1 - p) ≥ 10, the sampling distribution of sample proportion follows a normal distribution with mean

Check:

Thus, the sampling distribution of sample proportion follows a Normal distribution.

(1)

the mean and standard deviation:

The probability that the shipment will be returned  if the true proportion of defective cartridges in the shipment is 0.05 is:

Use the standard normal table for the probability.

Thus, the probability that the shipment will be returned  if the true proportion of defective cartridges in the shipment is 0.05 is 0.9744.

(2)

Compute the mean and standard deviation:

The probability that the shipment will not be returned if the true proportion of defective cartridges in the shipment is 0.10 is:

Use the standard normal table for the probability.

Thus, the probability that the shipment will not be returned if the true proportion of defective cartridges in the shipment is 0.10 is 0.00008.

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.