16. The average annual cost of automobile insurance is $939 with an estimated po

Question

16. The average annual cost of automobile insurance is $939 with an estimated population standard deviation of $245. a. What is the probability that a simple random sample of size 30 insured automobiles will have a sample mean cost less than $1,000? b. What is the probability that a simple random sample of size 150 insured automobiles will have a sample mean cost less than $1,000? c. What is the probability that a simple random sample of size 50 insured automobiles will have a sample mean within $25 of the population mean?

Explanation / Answer

16)

Given: µ = $ 939, = $245

To find the probability, we need to find the Z scores first.

Z = (X - µ)/ [/n].

(a) n = 30, To calculate P( X < 1000)

Z = (1000 – 939)/[245/30] = 1.36

The required probability from the normal distribution tables is = 0.9137

(b) n = 150, To calculate P( X < 1000)

Z = (1000 – 939)/[245/150] = 3.05

The required probability from the normal distribution tables is = 0.9989

(c) n = 50, and it is within $25 of the population mean. That is 939 - 25 = $914 and 939 + 25 = $964

To calculate P (914< X < 964) = P(X < 964) – P(X < 914)

For P( X < 964)

Z = (964 – 939)/[245/50] = 0.72

The probability for P(X < 964) from the normal distribution tables is = 0.7647

For P( X < 914)

Z = (914 – 939)/[245/50] = -0.72

The probability for P(X < a) from the normal distribution tables is = 0.2353

Therefore the required probability is 0.7247 – 0.2353 = 0.5294

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