Sampling Distrubtions

Question

CNBC recently reported that the mean annual cost of auto insurance is $1042. Assume the distribution of auto insurance follows a normal distribution with a standard deviation of $287.

Find the probability that a single randomly selected value is less than $968.

P(X < 968) = 

Find the probability that a sample of size 96 is randomly selected with a mean less than $968.

P(¯¯¯XX¯ < 968) = 

Enter your answers as numbers accurate to 4 decimal places.

Answer 1

Great! we have our assumptions. Randomness, Normality, population mean, population standard deviation.

Also our test values (x<968 for a sample from the population, which is the same to be used for testing the sample mean.

Part 1. μ = 1042 σ = 287. We could calculate a Z value and use a standard normal distribution table by why bother? This is the 21st century, why grab a book and look up values unless required. Using the cumulative normal distribution function found in most scientific calculators, Excel, or just on line we find

P(x<968) where X = ℵ(1042,287) = 0.3983

Part 2 P(xbar<968) where X= ℵ(1042,287), sample is 96 and derived standard deviation of the mean (sometimes called standard error in obsolete texts) of 287/√96. or 29.291815

P(xbar<968) where Xbar = ℵ(1042,29.291815) = 0.0058