Question

CNNBC recently reported that the mean annual cost of auto insurance is 1047 dollars. Assume the standard deviation is 196 dollars. You will use a simple random sample of 102 auto insurance policies.

Find the probability that a single randomly selected policy has a mean value between 1058.6 and 1078.1 dollars.
P(1058.6 < X < 1078.1) =

Find the probability that a random sample of size n=102n=102 has a mean value between 1058.6 and 1078.1 dollars.
P(1058.6 < M < 1078.1) =

Enter your answers as numbers accurate to 4 decimal places.

Answer 1

Solution:

We are given that: CNNBC recently reported that the mean annual cost of auto insurance is 1047 dollars with the standard deviation is 196 dollars.

That is:

   and

A simple random sample of 102 auto insurance policies is selected.

That is: n = 102

Part i) Find the probability that a single randomly selected policy has a mean value between 1058.6 and 1078.1 dollars.
P(1058.6 < X < 1078.1) = .........?

Find z score for x = 1058.6 and for x = 1078.1

z score formula is:

Thus we get:

P(1058.6 < X < 1078.1) = P( 0.06 < Z < 0.16)

P(1058.6 < X < 1078.1) = P( Z < 0.16) - P( Z < 0.06)

Look in z table for z = 0.1 and 0.06 as well as for z = 0.0 and 0.06 and find area.

P(Z < 0.16)= 0.5636 and P( Z < 0.06) = 0.5239

Thus

P(1058.6 < X < 1078.1) = P( Z < 0.16) - P( Z < 0.06)

P(1058.6 < X < 1078.1) = 0.5636 - 0.5239

P(1058.6 < X < 1078.1) = 0.0397

Part ii) Find the probability that a random sample of size n=102 has a mean value between 1058.6 and 1078.1 dollars.
P(1058.6 < M < 1078.1) = ........?

Find z scores for M = 1058.6 and for M = 1078.1

Thus we get:

P(1058.6 < M < 1078.1) = P( 0.60 < Z < 1.60)

P(1058.6 < M < 1078.1) = P( Z < 1.60) - P( Z < 0.60)

Look in z table for z = 0.6 and 0.00 as well as for z = 1.6 and 0.00 and find area

P(Z < 1.60)= 0.9452 and P( Z < 0.60) = 0.7257

P(1058.6 < M < 1078.1) = P( Z < 1.60) - P( Z < 0.60)

P(1058.6 < M < 1078.1) = 0.9452 - 0.7257

P(1058.6 < M < 1078.1) = 0.2195