Question

1. A population of values has a normal distribution with μ=182.1 and σ=28.9. You intend to draw a random sample of size n=117.

Find the probability that a single randomly selected value is less than 187.7.
P(X < 187.7) =

Find the probability that a sample of size n=117is randomly selected with a mean less than 187.7.
P(¯x < 187.7) =  

2.
CNNBC recently reported that the mean annual cost of auto insurance is 1045 dollars. Assume the standard deviation is 211 dollars. You take a simple random sample of 69 auto insurance policies.

Find the probability that a single randomly selected value is less than 977 dollars.
P(X < 977) =

Find the probability that a sample of size n=69= is randomly selected with a mean less than 977 dollars.
P(¯xx¯ < 977) =

Answer 1

Solution:

Question 1)

A population of values has a normal distribution with μ = 182.1 and σ = 28.9.

Sample size = n = 117.

Part a) P(X < 187.7) =............?

Find z score for x = 187.7

Thus we get:

P(X < 187.7) = P( Z < 0.19)

Look in z table for z = 0.1 and 0.09 and find corresponding area.

P( Z< 0.19) = 0.5753

thus

P(X < 187.7) = P( Z < 0.19)

Part b) Find:

Find z score for

Thus

Look in z table for z = 2.1 and 0.00 and find corresponding area.

P( Z < 2.10) = 0.9821

Thus

Question 2)

CNNBC recently reported that the mean annual cost of auto insurance is 1045 dollars and the standard deviation is 211 dollars.

Sample size = n = 69

Part a) Find:

P(X < 977) =............?

Find z score for x = 977

Thus we get:

P(X < 977) = P( Z < -0.32 )

Look in z table for z = -0.3 and 0.02 and find corresponding area.

P( Z< -0.32)= 0.3745

Thus

P(X < 977) = P( Z < -0.32 )

Part b)

Find z score for

Thus

Look in z table for z = -2.6 and 0.08 and find corresponding area.

P( Z < -2.68 ) = 0.0037

Thus