Question

CNNBC recently reported that the mean annual cost of auto insurance is 1002 dollars. Assume the standard deviation is 214 dollars, and the cost is normally distributed. You take a simple random sample of 18 auto insurance policies. Round your answers to 4 decimal places.

What is the distribution of XX? XX ~ N(,)

What is the distribution of ¯xx¯? ¯xx¯ ~ N(,)

What is the probability that one randomly selected auto insurance is less than $1042?

a simple random sample of 18 auto insurance policies, find the probability that the average cost is less than $1042.

For part d), is the assumption of normal necessary? NoYes

Answer 1

a)

X ~ ( , ) = (1002 , 2142)

~ ( ​​​​​​ , / n) = ( 1002 , 2142 / 18)

b)

We convert this to standard normal as

P(X < x) = P(Z < ( x - ) / )

So,

P(X < 1042) = P(Z < ( 1042 - 1002) / 214)

= P(Z < 0.19)

= 0.5753 (From Z table)

c)

Using cenral limit theorem,

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

So,

P( < 1042) = P(Z < ( 1042 - 1002) / sqrt ( 2142 / 18) )

= P(Z < 0.79)

= 0.7852 (from Z table)

d )

No

Since original population is normally distributed, the assumption of normal is not necessary.