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

What is the probability that the sample mean will be within +/- 3 of the population mean (to 4 decimals)?

What is the probability that the sample mean will be within +/- 12 of the population mean (to 4 decimals)? (Round z value in intermediate calculations to 2 decimal places.)

Answer 1

a) P(297 < < 303)

= P((297 - )/() < ( - )/() < (303 - )/())

= P((297 - 300)/(70/) < Z < (303 - 300)/(70/))

= P(-0.43 < Z < 0.43)

= P(Z < 0.43) - P(Z < -0.43)

= 0.6664 - 0.3336

= 0.3328

b) P(288 < < 312)

= P((288 - )/() < ( - )/() < (312 - )/())

= P((288 - 300)/(70/) < Z < (312 - 300)/(70/))

= P(-1.71 < Z < 1.71)

= P(Z < 1.71) - P(Z < -1.71)

= 0.9564 - 0.0436

= 0.9128