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).
Answer:
its C!! :) T3, –2(x, y)
Step-by-step explanation:
I just need points I’m sorry :(
Answer:
1/7 divided by 1/98 will give you 14
0=y
so let put the y value
3.8(0) +4/7. x+2
0. +4/7
0 0.57