Question

CNNBC recently reported that the mean annual cost of auto insurance is 975 dollars. Assume the standard deviation is 262 dollars, and the cost is normally distributed. You take a simple random sample of 37 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 more than $937?

a simple random sample of 37 auto insurance policies, find the probability that the average cost is more than $937.

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

Answer 1

Solution :

Given that,

mean = = 975

standard deviation = = 262

a.

X N (975 , 262)

b.

n = 37

= 975

= / n = 262 / 37 = 43.0725

N (975 , 43.0725)

c.

P(x > $937) = 1 - P(x < 937)

= 1 - P[(x - ) / < (937 - 975) / 262)

= 1 - P(z < -0.15)

= 1 - 0.4404

= 0.5596

Probability = 0.5596

d.

P( > $937) = 1 - P( < 937)

= 1 - P[( - ) / < (937 - 975) / 43.0725]

= 1 - P(z < -0.88)

= 1 - 0.1894

= 0.8106

Probability = 0.8106

For part d)

Yes