Question

A population has a mean of 400 and a standard deviation of 60. 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 +/- 4 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 +/- 15 of the population mean (to 4 decimals)? (Round z value in intermediate calculations to 2 decimal places.)

Answer 1

Solution :

Given that,

mean = = 400

standard deviation = = 60

n = 125

= = 400

= / n = 60 / 125 = 5.3667

a)

P( 396 < < 404) = P((396 - 400) /5.3667 <( - ) / < (404 - 400) / 5.3667))

= P(-0.75 < Z < 0.75)

= P(Z < 0.75) - P(Z < -0.75) Using z table,

= 0.7734 - 0.2266

= 0.5468

Probability = 0.5468

b)

P( 385 < < 415) = P((385 - 400) / 5.3667<( - ) / < (415 - 400) / 5.3667))

= P(-2.80 < Z < 2.80)

= P(Z < 2.80) - P(Z < -2.80) Using z table,

= 0.9974 - 0.0026

= 0.9948

Probability = 0.9948