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

step-by-step explanation:

i think you should think about the question and not look online.

total amount of money that he should pay if he has not used his duty free exemption is $98

step-by-step explanation:

given a united states citizen returning to the states declares the following items at the customs office: 3 shirts at $8.50 each 2 dresses at $27.50 each 1 pair of gold cuff links at $17.50 per pair. we have to find how much he pay

cost of 3 shirts if one cost at $8.50 is

cost of 2 dresses if one cost at $27.50 is  

cost of 1 pair of gold cuff at $17.50

so, total amount of money that he should pay if he has not used his duty free exemption is   $25.5+$55+$17.50=$98