Question

A population has a mean of 200 and a standard deviation of 80. Suppose a sample of size 100 is selected and is used to estimate . Use z-table. What is the probability that the sample mean will be within +/- 4 of the population mean (to 4 decimals)? (Round z value in intermediate calculations to 2 decimal places.) What is the probability that the sample mean will be within +/- 13 of the population mean (to 4 decimals)? (Round z value in intermediate calculations to 2 decimal places.)

Answer 1

Given that mean = 200 and standard deviation (sigma) = 80, sample size n = 100

sample standard deviation

(A) probability that the sample mean will be within +/- 4 of the population mean

So, +/- 4 of the population mean means between 196 and 204

using normalcdf(lower bound, upper bound, mean, standard deviation)

where lower bound = 196, upper bound= 204, mean= 200 and standard deviation = 8

= normalcdf(196,204,200,8)

= 0.3829

(B) probability that the sample mean will be within +/- 13 of the population mean

So, +/- 13 of the population mean means between 187 and 213

using normalcdf(lower bound, upper bound, mean, standard deviation)

where lower bound = 187, upper bound= 213, mean= 200 and standard deviation = 8

= normalcdf(187,213,200,8)

= 0.8958