Answer:
Step-by-step explanation:
(9u + w)(9u + w)
f: 9u * 9u = 81 u^2
o: 9u * w
i: 9u* w
l: w * w = w^2
Result
81u^2 + 9uw + 9uw + w^2
81u^2 + 18uw + w^2
Answer:
The 98% confidence interval for the population proportion of disks which are defective is (0.082, 0.118).
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 1536 floppy disks is drawn. Of these disks, 1383 were not defective.
1536 - 1383 = 153
This means that
98% 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 98% confidence interval for the population proportion of disks which are defective is (0.082, 0.118).
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