Question

A normal population has a mean of 61 and a standard deviation of 10. You select a random sample of 9. Compute the probability that the sample mean is: (Round your z values to 2 decimal places and final answers to 4 decimal places):

Greater than 64.

Less than 58.

Between 58 and 64.

Answer 1

Solution :

Given ,

mean = = 61

standard deviation = = 10

n = 9

sampling distribution of the sample mean = 61

sampling distribution of the standard deviation(standard error)

= / n = 10 / 9 = 3.33

P( >64 ) = 1 - P( < 64)

= 1 - P[( - ) / < (64-61) /3.33 ]

= 1 - P(z < 0.90)

Using z table

= 1 -0.8159

= 0.1841

probability= 0.1841

b)

P( < 58)

= P[( - ) / < (58-61) /3.33 ]

=P(z < -0.90)

Using z table

probability =0.1841

c)

P(58<    < 64) = P[(58-61) /3.33< ( - ) / < (64-61) /3.33 )

= P( -0.90< Z < 0.90)

= P(Z < 0.90) - P(Z < -0.90)

Using z table

=0.8159-0.1841

probability=0.6318