In a certain population, body weights are normally distributed with a mean of 152 pounds 30) 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.

We are given the following variables:

μ = the sample mean = 152 pounds

σ = the standard deviation = 26 pounds

x = the sample value we want to test = 180 pounds

n = the sample size = unknown

MOE = margin of error = 4% = 0.04

Confidence level = 96%

The first thing we can do is to find for the value of z using the formula:

z = (x – μ) / σ

z = (180 – 152) / 26

z = 1.0769 = 1.08

Since we are looking for the people who weigh more than 180 pounds, therefore this is a right tailed z test. The p value is:

p = 0.1401

Then we can use the formula below to solve for n:

n = z^2 * p * (1 – p) / (MOE)^2

n = 1.08^2 * 0.1401 * (1 – 0.1401) / (0.04)^2

n = 87.82 = 88

Therefore around 88 people must be surveyed.

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step-by-step explanation:

step-by-step explanation:

two pairs of adjacent sides are of equal length. this implies that one diagonal divides the kite into congruent triangles, and so the angles between the two pairs of equal sides are equal in measure. it also implies that the diagonals are perpendicular.