Question

The mean annual cost of an automotive insurance policy is normally distributed with a mean of $1140 and standard deviation of $310.
a. What is the probability that a random sample of 16 policyholders will have a mean insurance policy cost between $1000 and $1250?



Round your z value(s) to two decimal places. Do not round any other intermediate calculations. Round your answer to four decimal places.



Probability =



b. What is the probability that a random sample of 16 policyholders will have a mean insurance policy cost which exceeds $1250?



Round your z value(s) to two decimal places. Do not round any other intermediate calculations. Round your answer to four decimal places.



Probability =



c. What is the probability that a random sample of 16 policyholders will have a mean insurance policy cost below $1220?



Round your z value(s) to two decimal places. Do not round any other intermediate calculations. Round your answer to four decimal places.



Probability =

Answer 1

Using the normal distribution, it is found that the probabilities are given as follows:

a) 0.8871 = 88.71%.

b) 0.0778 = 7.78%.

c) 0.8485 = 84.85%.

Normal Probability Distribution

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

$Z = \frac{X - \mu}{\sigma}$

• The z-score measures how many standard deviations the measure is above or below the mean.

• Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

• By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

The parameters in this problem are given as follows:

$\mu = 1140, \sigma = 310, n = 16, s = \frac{310}{\sqrt{16}} = 77.5$

Item a:

The probability is the p-value of Z when X = 1250 subtracted by the p-value of Z when X = 1000, hence:

X = 1250:

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{1250 - 1140}{77.5}$

Z = 1.42

Z = 1.42 has a p-value of 0.9222.

X = 1000:

$Z = \frac{X - \mu}{s}$

$Z = \frac{1000 - 1140}{77.5}$

Z = -1.81

Z = -1.81 has a p-value of 0.0351.

0.9222 - 0.0351 = 0.8871 = 88.71% probability.

Item b:

The probability is one subtracted by the p-value of Z when X = 1250, hence:

1 - 0.9222 = 0.0778 = 7.78%.

Item c:

The probability is the p-value of Z when X = 1220, hence:

$Z = \frac{X - \mu}{s}$

$Z = \frac{1220 - 1140}{77.5}$

Z = 1.03

Z = 1.03 has a p-value of 0.8485.

0.8485 = 84.85% probability.

More can be learned about the normal distribution at brainly.com/question/4079902

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