CNNBC recently reported that the mean annual cost of auto insurance is 1031 dollars. Assume the standard deviation is 177 dollars. You will use a simple random sample of 102 auto insurance policies.

Question

Find the probability that a single randomly selected policy has a mean value between 1039.8 and 1083.6 dollars.

P(1039.8 < X < 1083.6) =

Find the probability that a random sample of size n = 102

has a mean value between 1039.8 and 1083.6 dollars.

P(1039.8 < (sample mean) < 1083.6) =

Answer 1

Hi Madison,

You will need to compute two z-scores and compute the respective difference in probabilities. Given that this is a sample, formula is:

z = (xbar - mu) / (s/sqrt(n))

Again, you have two values here, so I will refer to lower z and xbar as z1 and xbar1, higher z and xbar as z2, xbar2.

xbar1 = lower sample mean = 1039.8

mu = pop. mean = 1031

s = sample standard deviation = 177

n=102

xbar2 = upper sample mean = 1083.6

z1 = (1039.8 - 1031) / (177/sqrt(102)) = 0.50

z2 = (1083.6 - 1031)/ (177/sqrt(102)) = 3.00

From z-table:

P (Z < 0.50) = 0.6915

P(Z < 3.00) = 0.9987

P = 0.9987 - 0.6915 = 0.3072

I hope this helps.