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 96% confidence that the error is no more than 4 percentage points.

xBar ± z * sx / sqrt(n)

where xBar is the sample mean

z is the zscore for having α% of the data in the tails, i.e., P( |Z| > z) = α

sx is the sample standard deviation

n is the sample size

For sample size calcualtions we need the following. The width of the interval, from one end point to the center is:

z * sx / sqrt(n) =

w and solving for n gives:

 n = (z * sx / w) ^ 2

remember that n needs to be an integer. Always take the ceiling, i.e., round up. If you round down then the width of the interval will not be correct, it will be too wide. By rounding up, the interval will be more narrow than asked for, but this is a good thing. It means there is more precision in the estimate.

 here we have

z = 2.05

w = 0.04 * 152 = 6.08

n = (z * sx / w) ^ 2 n = (2.05 * 26 / 6.08)^2

n = 76.8506

Round of the answer since n must be an integer.

n = 77 

y=x+2 hope this jdjxhdnzbdjdo

the answer is 62