1. The weights of 20 year old vegetarian women are normally distributed with a standard deviation σ = 15 pounds. How large of a sample would be need to estimate the mean weight μ of 20 year old vegetarian women with a margin of error of 5 pounds with 95% confidence?

34.5744 ≈ 35

Step-by-step explanation:

The formula for Margin of Error

= z × σ/√n

Where

z = z score of the confidence interval = z score of 95% = 1.96

σ = 15 pounds

n = number of samples = ??

Margin of Error = 5

Hence,

5 = 1.96 × 15/√n

Cross Multiply

5 × √n = 1.96 × 15

Divide both sides 5

√n = 1.96 × 15/5

√n = 1.96 × 3

√ n = 5.88

Square both sides

n = 5.88²

n= 34.5744

Approximately the size = 35

mk./km: lm

step-by-step explanation:

,mlmlm

; mk

lk

step-by-step explanation: