Question

A population has a mean of 200 and a standard deviation of 60. Suppose a sample of size 100 is selected and is used to estimate . Use z-table.

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.)

What is the probability that the sample mean will be within +/- 17 of the population mean (to 4 decimals)? (Round z value in intermediate calculations to 2 decimal places.)

Answer 1

Solution :

Given that,

mean = = 200

standard deviation = = 60

n = 100

= = 200

= / n = 60 / 10 = 6

1)

P( 194 < < 206) = P((194 -200) /6 <( - ) / < (206 - 200) / 6))

= P( -1.00 < Z < 1.00)

= P(Z < 1.00) - P(Z < -1.00) Using standard normal table,  

= 0.8413 - 0.1587

= 0.6826

Probability = 0.6826  

2)

P( 183 < < 217) = P((183 -200) /6 <( - ) / < (217 - 200) / 6))

= P( -2.83 < Z < 2.83)

= P(Z < 2.83) - P(Z < -2.83) Using standard normal table,  

= 0.9977 - 0.0023

= 0.9954

Probability = 0.9954