A manufacturer of computer printers purchases plastic ink cartridges from a vendor. When a large ship- ment is received, a random sample of 200 cartridges is se- lected, and each cartridge is inspected. If the sample pro- portion of defective cartridges 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 car- tridges in the shipment is .05?
b. What is the approximate probability that a shipment will not be returned if the true proportion of defective car- tridges in the shipment is .10?

a. 0.9744

b = 0.00008 (approximately 0)

Step-by-step explanation:

We have the sampling distribution following a normal distribution.

1

the proportion, p^ is 0.05, n = 200

Standard deviation = √p^(1-p^)/n

= √0.05(1-0.05)/200

= √0.0475/200

= √0.0002375

= 0.0154

The probability (p>0.02)

= p(z>0.02-0.05/0.0154)

= P(z>-1.95)

= P(z<1.95)

When we use the standard normal table = 0.9744

So, 0.9744 is the approximate probability that a shipment will be returned if the true proportion of defective car- tridges in the shipment is 0.05

b. P^ = 0.10( true population of defective cartridges = 0.10)

= √0.10(1-0.10)/200

= 0.0212

Probability it will not be returned

= P(p<=0.02)

= P(z<=0.02-0.10/0.0212)

= P(Z<= -3.77)

Using the standard normal table

1-p(z<3.77)

= 1 - 0.99992

= 0.00008

This is approximately 0

Thisis the approximate probability that a shipment will not be returned if the true proportion of defective car- tridges in the shipment is .10

(w – 2.5)(w + 2.5)

(–4v – 9)(–4v + 9)

step-by-step explanation:

replication and blocking

step-by-step explanation: