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 

the correct answer is a) they believe they have a "just cause."

islamic terrorists justify their violent actions because they believe they have a "just cause."

every act of terrorism of islamic sects is done “in the name of allah”. that justifies their actions. islam people who commit those terrorist attacks are not afraid to die. that is why the attacks have horrible consequences. they know they are going to die and they are proud to do it because it is in the name of allah, so it is justified and is for a “good cause.”

since the very beginning of the conflict with israel, islam is fighting a “sacred war”, respecting the words stated in the qur’an. they fight against the domination of the western countries and to defend their occupied territories.