Question

CNNBC recently reported that the mean annual cost of auto insurance is 967 dollars. Assume the standard deviation is 291 dollars. You take a simple random sample of 92 auto insurance policies.

A. Find the probability that a single randomly selected value is less than 984 dollars.
P(X < 984) =

Find the probability that a sample of size n=92 is randomly selected with a mean less than 984 dollars.
P(M < 984) =

Enter your answers as numbers accurate to 4 decimal places.

B. A leading magazine (like Barron's) reported at one time that the average number of weeks an individual is unemployed is 23 weeks. Assume that for the population of all unemployed individuals the population mean length of unemployment is 23 weeks and that the population standard deviation is 4 weeks. Suppose you would like to select a random sample of 72 unemployed individuals for a follow-up study.

Find the probability that a single randomly selected value is greater than 24.1.
P(X > 24.1) =  

(Enter your answers as numbers accurate to 4 decimal places.)

Find the probability that a sample of size n=72 is randomly selected with a mean greater than 24.1.
P(M > 24.1) =

(Enter your answers as numbers accurate to 4 decimal places.)

C. Business Weekly conducted a survey of graduates from 30 top MBA programs. On the basis of the survey, assume the mean annual salary for graduates 10 years after graduation is 155000 dollars. Assume the standard deviation is 45000 dollars. Suppose you take a simple random sample of 57 graduates.

Find the probability that a single randomly selected policy has a mean value between 135330.7 and 177649.5 dollars.
P(135330.7 < X < 177649.5) =  

(Enter your answers as numbers accurate to 4 decimal places.)

Find the probability that a random sample of size n=57 has a mean value between 135330.7 and 177649.5 dollars.
P(135330.7 < M < 177649.5) =  

(Enter your answers as numbers accurate to 4 decimal places.)

Answer 1

Solution :

Given that ,

mean = = 967

standard deviation = = 291

A.

P(x < 984 ) = P[(x - ) / < ( 984 - 967) / 291 ]

= P(z < 0.06 )

Using z table,

= 0.5239

Probability = 0.5239

n = 92

M = 967

M = / n = 291 / 92 = 30.3388

P(M < 984) = P((M - M ) / M < ( 984 - 967 ) / 30.3388 )

= P(z < 0.56 )

Using z table

= 0.7123   

Probability = 0.7123

B.

mean = = 23

standard deviation = = 4

P(x > 24.1 ) = 1 - p ( x < 24.1 )

= 1- P[(x - ) / < ( 24.1 - 23) / 4]

= 1- P(z < 0.28 )

Using z table,

= 1 - 0.6103

= 0.3897

Probability = 0.3897

n = 72

M = 23

M = / n = 4 / 72 = 0.4714

P(M > 24.1 ) = 1 - P(M < 24.1 )

= 1 - P[(M - M ) / M < ( 24.1 - 23 ) / 0.4714]

= 1 - P(z < 2.33)

Using z table,    

= 1 - 0.9901

= 0.0099

Probability = 0.0099

C.

mean = = 155000

standard deviation = = 45000

P( 135330.7 < x < 177649.5 )

= P[( 135330.7 - 155000) / 45000 ) < (x - ) /  < ( 177649.5 - 155000) / 45000) ]

= P( -0.44 < z < 0.15 )

= P(z < 0.15 ) - P(z < -0.44 )

Using z table,

= 0.5596 - 0.3300

= 0.2296

Probability = 0.2296

n = 57

M = 155000

M = / n = 45000 / 57 = 5960.3961

P( 135330.7 < M < 177649.5 )

= P[( 135330 - 155000) / 5960.3961 < (M - M) / M < ( 177649.5 - 155000) / 5960.3961)]

= P( -3.30 < Z < 3.80 )

= P(Z < 3.80 ) - P(Z < -3.30 )

Using z table,  

= 0.9999 - 0.0005

= 0.9994

Probability = 0.9994