Question

A population has a mean of 200 and a standard deviation of 90. Suppose a sample of size 125 is selected x_bar and is used to estimate μ Use z-table.

a. What is the probability that the sample mean will be within +/- 7 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 +/- 15 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 = = 90

= / n = 90 / 125 = 8.0498

a.

= P[(193 - 200) / 8.0498< ( - ) / < (207 - 200) / 8.0498)]

= P(-0.87 < Z < 0.87)

= P(Z < 0.87) - P(Z < -0.87)

= 0.8078 - 0.1922

= 0.6156

Probability = 0.6156

b.

= P[(185 - 200) /8.0498 < ( - ) / < (215 - 200) / 8.0498)]

= P(-1.86 < Z < 1.86)

= P(Z < 1.86) - P(Z < -1.86)

= 0.9686 - 0.0314

= 0.9372

probability = 0.9372