Question

A population has a mean of 300 and a standard deviation of 70. Suppose a sample of size 100 is selected x and is used to estimate m. Use z-table.
a. What is the probability that the sample mean will be within +/- 3 of the population mean (to 4 decimals)? (Round z value in intermediate calculations to 2 decimal places.)
b. 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 = = 70

n = 100

=   = 300

= / n = 70 / 100 = 7

a )within 3 = 300 ± 3 = 297, 303

P(297< < 313)  

= P[(297- 300) / 7 < ( - ) / < (313 - 300) / 7)]

= P(-0.43 < Z < 0.43)

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

Using z table,  

= 0.6664 - 0.3336

= 0.3328

Probability =0.3328

= P(-1.86< Z < 1.86)

= P(Z < 1.86) - P(Z < -1.86)

Using z table,  

= 0.9686 - 0.0314

= 0.9372

Probability =0.9371

b ) within 13= 300 ± 13 = 287, 313

P(287< < 313)  

= P[(287- 300) / 7 < ( - ) / < (303 - 300) / 7)]

= P(-1.86< Z < 1.86)

= P(Z < 1.86) - P(Z < -1.86)

Using z table,  

= 0.9686 - 0.0314

= 0.9372

Probability =0.9372