Probability of randomly selected

Question

Fox News recently reported that the mean annual cost of auto insurance is $1000. Assume the standard deviation is $200, and the cost is normally distributed. Take a random sample of 13 auto insurance policies.

1. What is the distribution of X? X~N(___,___)
2. What is the distribution of x-bar? x-bar ~N (___,___)
3. What is the probability that one randomly selected auto insurance is more than $950?_____
4. A simple random sample of 13 auto insurance policies, what is the probability that its average cost more than $950?
5. 
   

Answer 1

X is N(1000,200)

Xbar is N(1000, 200/sqrt(13))

P(x > 950) = P(z > (950-1000)/200) = P(z > -0.25) = 1 - P(z < -0.25)

P(xbar > 950) = P(z > (950 - 1000)/(200/sqrt(13)) = P(z > -0.9) = 1 - P(z < -0.9)