An electronics retailer receives a shipment of 6,000CDs to distribute to its stores. A quality control manager inspects a random
2 answers:
Answer: 300
Step-by-step explanation:
Given , An electronics retailer receives a shipment of 6,000CDs to distribute to its stores.
i.e, Total CD's he has now = 6,000
Also, a quality control manager inspects a random sample(S) of 40CD's , the number of defective(D) CD's are 2.
Then , the proportion of defective( D) CD's =
Now , the number of CDs in the shipment are likely to be defective = ( Total CD's) x (proportion of defective CD's)
= 6000 x 0.05
=300
Hence, the number of CDs in the shipment are likely to be defective =300
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Answer:
Step-by-step explanation:
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1.
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Answer:
The 80% confidence interval for the population proportion of disks which are defective is (0.059, 0.079).
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of
, we have the following confidence interval of proportions.
In which
z is the zscore that has a pvalue of .
Suppose a sample of 1067 floppy disks is drawn. Of these disks, 74 were defective.
This means that
80% confidence level
So , z is the value of Z that has a pvalue of
, so
.
The lower limit of this interval is:
The upper limit of this interval is:
The 80% confidence interval for the population proportion of disks which are defective is (0.059, 0.079).