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.

What is the probability that the sample mean will be within +/- 6 of the population mean (to 4 decimals)? (Round z value in intermediate calculations to 2 decimal places.)

Answer 1

Explanation:

Sol

Given that ,

mean =300

standard deviation =a =70

n=100

ux=128

a ) within 6 =300 +/-6 : 294, 306

P(294' ' 306)

: P(-0.86 : Z ' 0.86)

: P(Z ' 0.86) - P(Z ' -0.86)

Using z table,

: 0.8051-0.1949

: 0.6102

Probability : 0.6102

a ) within 18 =300 18 : 294, 318

P(294' ' 306)

: P(Z ' 2.57) - P(Z ' -2.57)

Using z table,

:0.9949-0.0051

:0.9898

Probability =0.9898