Question

A population has a mean of 200 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 +/- 3 of the population mean What is the probability that the sample mean will be within +/- 17 of the population mean (to 4 decimals)? (Round z value in intermediate calculations to 2 decimal places.)

Answer 1

Solution :

Given that,

mean = = 200

standard deviation = = 70

= / n = 70 / 100 = 7

= P[(-0.03) / 7< ( - ) / < (0.03) / 7)]

= P(-0.00 < Z < 0.00)

= P(Z < 0.00) - P(Z < -0.00)

= 0.5 - 0.5

= 0

Probability = 0

= P[(-0.17) / 7< ( - ) / < (0.17) / 7)]

= P(-0.02 < Z < 0.02)

= P(Z < 0.02) - P(Z < -0.02)

= 0.508 - 0.492

= 0.016

Probability = 0.0160