Question

A population has a mean of 200 and a standard deviation of 70. 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 +/- 11 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 = = 70

n = 100

= = 200

= / n = 70 / 100 = 7

a) 200 -+ 6 = 194, 206

P( 194 < < 206 )

= P[( 194 - 200 ) / 7 < ( - ) / < ( 206 - 200 ) / 7 )]

= P( -0.86 < Z < 0.86 )

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

Using z table

= 0.8051 - 0.1949

= 0.6102

b) 200 -+ 11 = 189, 211

P( 189 < < 211 )

= P[( 189 - 200 ) / 7 < ( - ) / < ( 211 - 200 ) / 7 )]

= P( -1.57 < Z < 1.57 )

= P(Z < 1.57 ) - P(Z < -1.57 )

Using z table

= 0.9418 - 0.0582

= 0.8836