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 +/- 6 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

=   = 128

= / n = 70/ 100 = 7

a ) within 6 = 300  ±6 = 294, 306

P(294< < 306)  

= P[(294 - 300) / 7 < ( - ) / < (306 - 300) /7)]

= P(-0.86 < Z < 0.86)

= P(Z < 0.86) - P(Z < -0.86)

Using z table,  

= 0.8051 - 0.1949

= 0.6102

Probability = 0.6102

a ) within 18 = 300  ± 18 = 294, 318

P(294< < 306)  

= P[(282 - 300) / 7 < ( - ) / < (318 - 300) /7)]

= P(-2.57 < Z < 2.57)

= P(Z < 2.57) - P(Z < -2.57)

Using z table,  

=0.9949 - 0.0051

=0.9898

Probability =0.9898