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.

1. What is the probability that the sample mean will be within +/- 8 of the population mean (to 4 decimals)? (Round z value in intermediate calculations to 2 decimal places.)
   
   
2. What is the probability that the sample mean will be within +/- 18 of the population mean (to 4 decimals)? (Round z value in intermediate calculations to 2 decimal places.)
   

Answer 1

solution:-

given that mean = 300 , standard deviation = 70 , n = 100

a. p((300-8) < x < (300+8))

=> P(292 < x < 308)

=> P((292-300)/(70/sqrt(100)) < z < (308-300)/(70/sqrt(100)))

=> P(-1.14 < z < 1.14)

=> P(z < 1.14) - P(z < -1.14)

=> 0.8729 - 0.1271

=> 0.7458

b. p((300-18) < x < (300+18))

=> P(282 < x < 318)

=> P((282-300)/(70/sqrt(100)) < z < (318-300)/(70/sqrt(100)))

=> P(-2.57 < z < 2.57)

=> P(z < 2.57) - P(z < -2.57)

=> 0.9949 - 0.0051

=> 0.9898