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.

-362

step-by-step explanation:

(12)(4)+2(12)-34(12)+2(12)-3

  \/         \/         \/         \/

  48 +   24 -   408 + 24 - 3

      \   /               \   /

          72     -     432 - 3 = -363