Question

CNNBC recently reported that the mean annual cost of auto insurance is 998 dollars. Assume the standard deviation is 227 dollars. You take a simple random sample of 98 auto insurance policies.

Find the probability that a single randomly selected value is less than 966 dollars. P(X < 966) =

Find the probability that a sample of size n=98n=98 is randomly selected with a mean less than 966 dollars.
P(¯xx¯ < 966) =

Answer 1

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The distribution parameters are given as:

Mean, Mu = 998
Xbar = 966
Stdev, Sigma = sqrt(64) = 98
n = 227
alpha = .01

We will these parameters along with the standardization formula to solve the problem. The formula for standardization is : Z = (Xbar-Mean)/(Stdev/sqrt(n)) for sample mean distribution.

We will look up the Z tables to convert Z to cumulative probability or vice-versa. You can also look NORMSDIST(z) Excel funtion to convert Z to cumulative prob.

1.

P(Xbar<966),
Standardizing: using formula: z = (X-Mean)/Stdev
= P(Z< (966-998)/(227)
= P(Z< -0.1410) [used Z table to convert Z to a cumulative prob]
= 0.4439

2.

P(Xbar<966),
Standardizing:
= P(Z< (966-998)/(227/sqrt(98)))
= P(Z< -1.3955) [used Z table to convert Z to a cumulative prob]
= 0.0814