Question

CNNBC recently reported that the mean annual cost of auto insurance is 1049 dollars. Assume the standard deviation is 275 dollars. You take a simple random sample of 72 auto insurance policies.

Find the probability that a single randomly selected value is less than 962 dollars.
P(X < 962) =


Find the probability that a sample of size
n
=
72
is randomly selected with a mean less than 962 dollars.
P(M < 962) =


Enter your answers as numbers accurate to 4 decimal places.

Answer 1

Answer:

0.3745 = 37.45% probability that a single randomly selected value is less than 962 dollars.

0.0037 = 0.37% probability that a sample of size 72 is randomly selected with a mean less than 962 dollars.

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 z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes 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 1049 dollars. Assume the standard deviation is 275 dollars.

This means that $\mu = 1049, \sigma = 275$

Find the probability that a single randomly selected value is less than 962 dollars.

This is the p-value of Z when X = 962. So

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

$Z = \frac{962 - 1049}{275}$

$Z = -0.32$

$Z = -0.32$ has a p-value of 0.3745

0.3745 = 37.45% probability that a single randomly selected value is less than 962 dollars.

Sample of 72

This means that $n = 72, s = \frac{275}{\sqrt{72}}$

Probability that the sample mean is les than 962.

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

By the Central Limit Theorem

$Z = \frac{962 - 1049}{\frac{275}{\sqrt{72}}}$

$Z = -2.68$

$Z = -2.68$ has a p-value of 0.0037

0.0037 = 0.37% probability that a sample of size 72 is randomly selected with a mean less than 962 dollars.