Question

A population has a mean of 300 and a standard deviation of 80. Suppose a simple random sample of size 100 is selected and is used to estimate ? . 1. What is the probability that the sample mean will be within +/- 7 of the population mean (to 4 decimals)? 2. What is the probability that the sample mean will be within +/- 13 of the population mean (to 4 decimals)?

Answer 1

1) P(293 < < 307)

= P((293 - )/() < ( - )/() < (307 - )/())

= P((293 - 300)/(80/sqrt(100)) < Z < (307 - 300)/(80/sqrt(100)))

= P(-0.88 < Z < 0.88)

= P(Z < 0.88) - P(Z < -0.88)

= 0.8106 - 0.1894

= 0.6212

2) P(287 < < 313)

= P((287 - )/() < ( - )/() < (313 - )/())

= P((287 - 300)/(80/sqrt(100)) < Z < (313 - 300)/(80/sqrt(100)))

= P(-1.63 < Z < 1.63)

= P(Z < 1.63) - P(Z < -1.63)

= 0.9484 - 0.0516

= 0.8968