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 weighs more than 180 pounds? Assume that we want 96% confidence that the error is no more than 2 percentage points.

A) 923
B) 1267
C) 2628
D) 2001

B) 1267

Step-by-step explanation:

Percentage of people above 180 pounds.

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

This percentage is 1 subtracted by the pvalue of Z when X = 26. So

has a pvalue of 0.8599

1 - 0.8599 = 0.1401

14.01% of people above 180 pounds.

Confidence interval for the proportion:

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 given by:

For this question, we have that:

96% confidence level

So , z is the value of Z that has a pvalue of , so .

We need

A sample size of n.

n is found when M = 0.02. So

So the correct answer is:

B) 1267

i dont know nkw

step-by-step explanation:

kenffledkedncklzkdfeserw

include a picture of the drawing in order for me to

step-by-step explanation: