Question

CNNBC recently reported that the mean annual cost of auto insurance is 965 dollars. Assume the standard deviation is 113 dollars. You will use a simple random sample of 57 auto insurance policies.Find the probability that a single randomly selected policy has a mean value between 939.6 and 972.5 dollars.P(939.6 < X < 972.5)

Answer 1

Answer:

P(939.6 < X < 972.5) = 0.6469

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

CNNBC recently reported that the mean annual cost of auto insurance is 965 dollars. Assume the standard deviation is 113 dollars.

This means that $\mu = 965, \sigma = 113$

Sample of 57:

This means that $n = 57, s = \frac{113}{\sqrt{57}} = 14.97$

Find the probability that a single randomly selected policy has a mean value between 939.6 and 972.5 dollars.

This is the pvalue of Z when X = 972.5 subtracted by the pvalue of Z when X = 939.6. So

X = 972.5

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

By the Central Limit Theorem

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

$Z = \frac{972.5 - 965}{14.97}$

$Z = 0.5$

$Z = 0.5$ has a pvalue of 0.6915

X = 939.6

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

$Z = \frac{939.6 - 965}{14.97}$

$Z = -1.7$

$Z = -1.7$ has a pvalue of 0.0446

0.6915 - 0.0446 = 0.6469

So

P(939.6 < X < 972.5) = 0.6469