Answer:
You must survey 784 air passengers.
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
.
The margin of error is:

95% confidence level
So
, z is the value of Z that has a pvalue of
, so
.
Assume that nothing is known about the percentage of passengers who prefer aisle seats.
This means that
, which is when the largest sample size will be needed.
Within 3.5 percentage points of the true population percentage.
We have to find n for which M = 0.035. So






You must survey 784 air passengers.
Answer:
56.7
Step-by-step explanation:
We know
mean = sum / count, with count being the amount of papers corrected in this case. We want to find the sum of all the papers as well as the count to figure out the mean of all papers.
For Tony's papers,
mean = sum / count
50 = sum / 40
multiply both sides by 40 to isolate the sum
sum = 50 * 40 = 2000
For Alice's papers,
mean = sum / count
70 = sum / 20
multiply both sides by 20 to isolate the sum
70 * 20 = sum = 1400
The total sum of all 60 papers is equal to the sum of 40 papers + the sum of the remaining 20 papers, or 2000 + 1400 = 3400. The mean of the 60 papers is therefore
mean = sum / count
mean = 3400/60 ≈ 56.7
Here’s the answer w explanation :)
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