Question 13
6 pts
1
Details
CNNBC recently reported that the mean annual cost of auto insurance is 1008 dollars. Assume the standard deviation is 100 dollars. You will use a simple random sample of 60 auto insurance policies. You may assume original population is approximatley normally distributed, and round your answers to three decimals.

Find the probability that a single randomly selected policy has a mean value between 977 and 1037.7 dollars.
\[
P(977< X< 1037.7)=
\]

Find the probability that a random sample of size $\boldsymbol{n}=\mathbf{6 0}$ has a mean value between 977 and 1037.7 dollars.
\[
P(977< M< 1037.7)=
\]

Step 1 ：First, we need to find the Z-scores for 977 and 1037.7 dollars. The formula for a Z-score is: \(Z = \frac{X - \mu}{\sigma}\) where \(X\) is the value we are looking at, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.

Step 2 ：For \(X = 977\) dollars, the Z-score is: \(Z = \frac{977 - 1008}{100} = -0.31\)

Step 3 ：For \(X = 1037.7\) dollars, the Z-score is: \(Z = \frac{1037.7 - 1008}{100} = 0.297\)

Step 4 ：Now, we need to find the probability that a single randomly selected policy has a mean value between these Z-scores. We can use a Z-table to find these probabilities. Looking up these Z-scores in the Z-table, we find that: \(P(Z < -0.31) = 0.378\) and \(P(Z < 0.297) = 0.617\)

Step 5 ：So, the probability that a single randomly selected policy has a mean value between 977 and 1037.7 dollars is: \(P(977 < X < 1037.7) = P(Z < 0.297) - P(Z < -0.31) = 0.617 - 0.378 = 0.239\)

Step 6 ：Next, we need to find the probability that a random sample of size \(n = 60\) has a mean value between 977 and 1037.7 dollars. When dealing with a sample mean, the standard deviation is divided by the square root of the sample size. So, the Z-scores for the sample mean are: \(Z = \frac{977 - 1008}{100 / \sqrt{60}} = -2.19\) and \(Z = \frac{1037.7 - 1008}{100 / \sqrt{60}} = 2.05\)

Step 7 ：Looking up these Z-scores in the Z-table, we find that: \(P(Z < -2.19) = 0.014\) and \(P(Z < 2.05) = 0.980\)

Step 8 ：So, the probability that a random sample of size \(n = 60\) has a mean value between 977 and 1037.7 dollars is: \(P(977 < M < 1037.7) = P(Z < 2.05) - P(Z < -2.19) = 0.980 - 0.014 = 0.966\)

Step 9 ：So, the final answers are: \(\boxed{P(977 < X < 1037.7) = 0.239}\) and \(\boxed{P(977 < M < 1037.7) = 0.966}\)