Answer:
(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 <em>X</em> = Number of defective cartridges.
The sample size is: <em>n</em> = 200.
The probability that the shipment is rejected is, <em>p</em> = 0.02.
The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.
But the sample size is too large, i.e. <em>n</em> = 200 > 30.
Then according to the normal approximation to binomial, if <em>np</em> ≥ 10 or <em>n(1 - p)</em> ≥ 10, the sampling distribution of sample proportion follows a normal distribution with mean
<u>Check</u>:
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.