Question

A population has a mean of 300 and a standard deviation of 70. Suppose a sample of size 100 is selected and  is used to estimate . Use z-table.

A. What is the probability that the sample mean will be within +/- 8 of the population mean (to 4 decimals)? (Round z value in intermediate calculations to 2 decimal places.)

B. What is the probability that the sample mean will be within +/- 10 of the population mean (to 4 decimals)? (Round z value in intermediate calculations to 2 decimal places.)

Answer 1

A population has a mean of = 300 and a standard deviation of =70

Sample size = n = 100

a)

We have to find the probability that the sample mean will be within +/- 8 of the population mean.

I.e P( 300-8 < < 300 + 8 ) = P( 292 < < 308 )

We need to find z score for 292 and 308

z-score for 292 ,

z-score for 308

P( 292 < < 308 )= P( -1.14 < z < 1.14 )

P( -1.14 < z < 1.14 ) = P( z < 1.14 ) - P( z < -1.14 )

Using z score table , P( z < -1.14 ) = 0.1271 and P( z < 1.14 ) = 0.8729

P( -1.14 < z < 1.14 ) = 0.8729 - 0.1271 = 0.7458

The probability that the sample mean will be within +/- 8 of the population mean is 0.7458

b)

We have to find the probability that the sample mean will be within +/- 10 of the population mean.

I.e P( 300-10 < < 300 + 10 ) = P( 290 < < 310 )

We need to find z score for 290 and 310

z-score for 290 ,

z-score for 310

P( 290 < < 310 )= P( -1.43 < z < 1.43 )

P( -1.43 < z < 1.43 ) = P( z < 1.43 ) - P( z < -1.43 )

Using z score table , P( z < -1.43 ) = 0.0764 and P( z < 1.43 ) = 0.9236

P( -1.43 < z < 1.43 ) = 0.9236 - 0.0764= 0.8472

The probability that the sample mean will be within +/- 10 of the population mean is 0.8472