Question

Suppose the average client charge per hour for out-of-court work by lawyers in the state of Iowa is $125. Suppose further that a random telephone sample of 32 lawyers in Iowa is taken and that the sample average charge per hour for out-of-court work is $110. If the population variance is $525
(Round the answers to 4 decimal places.)

what is the probability of getting a sample mean of $110 or larger? P=  

What is the probability of getting a sample mean larger than $135 per hour? P=

What is the probability of getting a sample mean of between $120 and $130 per hour? P=

Answer 1

Using central limit theorem ,

P( < x) = P(Z < ( x - ) / sqrt ( ² / n) )

a)

P( >= 110) = P(Z >= ( 110 - 125) / sqrt ( 525 / 32) )

= P(Z > -3.70)

= P(Z < 3.70)

= 0.9999

b)

P( >= 135) = P(Z >= ( 135 - 125) / sqrt ( 525 / 32) )

= P(Z > 2.47)

= 1 - P(Z < 2.47)

= 1 - 0.9932

= 0.0068

c)

P(120 < < 130) = P( < 130) - P( < 120)

= P(Z < ( 130 - 125) / sqrt ( 525 / 32) ) - P(Z < ( 120 - 125) / sqrt ( 525 / 32) )

= P( Z < 1.23) - P( Z < -1.23)

= 0.8907 - 0.1093

= 0.7814