In a certain population, body weights are normally distributed with a mean of 15
ID: 3132517 • Letter: I
Question
In a certain population, body weights are normally distributed with a mean of 152 pounds and a standard deviation of 26 pounds. How many people must be surveyed if we want to estimate the percentage who weigh more than 180 pounds? Assume that we want 95% confidence that the error is no more than 4 percentage points.
Note: Use the percentage of people expected to weigh more than 180 pounds in the distribution as an estimate of the population proportion when calculating your sample size. That is, use the normal distribution to determine the probability of P( X > 180) as the estimate for the population proportion when determining the sample size.
For the answer to the question, you must determine the sample size need to get an estimate for a population proportion at 96% confidence with a margin of error of no more than 4%.
601
291
501
232
601
291
501
232
Explanation / Answer
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 180
u = mean = 152
s = standard deviation = 26
Thus,
z = (x - u) / s = 1.076923077
Thus, using a table/technology, the right tailed area of this is
P(z > 1.076923077 ) = 0.140757316
Note that
n = z(alpha/2)^2 p (1 - p) / E^2
where
alpha/2 = 0.025
Using a table/technology,
z(alpha/2) = 1.959963985
Also,
E = 0.04
p = 0.140757316
Thus,
n = 290.3775385
Rounding up,
n = 291 [ANSWER, B]