Question

A population has a mean of 300 and a standard deviation of 90. Suppose a sample of size 125 is selected and  is used to estimate . Use z-table.

1. 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.)
   
2. What is the probability that the sample mean will be within +/- 13 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 = = 90

n = 125

= = 300

= / n = 90 / 125 = 8.05

a) P(293 < < 307)  

= P[(293 - 300) /8.05 < ( - ) / < (307 - 300) / 8.05)]

= P(-0.87 < Z < 0.87)

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

Using z table,  

= 0.8078 - 0.1922   

= 0.6156

b) P(287 < < 313)  

= P[(287 - 300) /8.05 < ( - ) / < (313 - 300) / 8.05)]

= P(-1.61 < Z < 1.61)

= P(Z < 1.61) - P(Z < -1.61)

Using z table,  

= 0.9463 - 0.0537

= 0.8926