Question

CNNBC recently reported that the mean annual cost of auto insurance is 1048 dollars. Assume the standard deviation is 207 dollars. You take a simple random sample of 91 auto insurance policies.

Find the probability that a single randomly selected value is at least 998 dollars.
P(X > 998) =

Find the probability that a sample of size n=91 is randomly selected with a mean that is at least 998 dollars.
P(M > 998) =

Enter your answers as numbers accurate to 4 decimal places.

Answer 1

Part 1

We need to find the probability that a single randomly selected value is at least 998 dollars. Here

X ~ N (1048, 207)

We use Z score

We need to find P(X > 998)

For X = 998, Z = (998-1048)/207 = -0.2415

So, We need to find P(Z>-0.2415)

P(Z>-0.2415) = 1 -P(Z< -0.2415)

= 1 - 0.4046

= 0.5954

Part 2

We need to find the probability that a sample of size n=91 is randomly selected with a mean that is at least 998 dollars.

Now as per CLT, M follows a normal distribution with mean = 1048 and standard deviation = 207/sqrt(91) = 21.6995

So, M ~ Normal(1048, 21.6995)

We need to find P(M > 998)

For M = 998, Z = (998-1048)/21.6995 = -2.3042

So, We need to find P(M>-2.3042)

P(M>-2.3042) = 1 -P(M< -2.3042)

= 1 - 0.0106

= 0.9894

Please let me know if you have any doubts. Happy to help. Please thumbs up if you like the solution. Thanks