Question

Suppose a newspaper article states that the distribution of auto insurance premiums for residents of California is approximately normal with a mean of $1,650. The article also states that 25% of California residents pay more than $1,800.
(a) What is the Z-score that corresponds to the top 25% (or the 75th percentile) of the standard normal distribution? (use the closest value from table B.1)

(b) What is the mean insurance cost? $

What is the cut off for the 75th percentile? $

(c) Identify the standard deviation of insurance premiums in LA.

Answer 1

Answer:

a) Z = 1.67

b) What is the mean insurance cost? $1,650

What is the cut off for the 75th percentile? $1,800

c) The standard deviation of insurance premiums in LA is $223.88.

Step-by-step explanation:

Problems of normally distributed samples 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.

In this problem, we have that:

$\mu = 1650$

The article also states that 25% of California residents pay more than $1,800, which means that Z when X = 1850 has a pvalue of 0.75.

(a) What is the Z-score that corresponds to the top 25% (or the 75th percentile) of the standard normal distribution?

This is the value of Z which has a pvalue of 0.75.

Looking at the ztable, it is Z = 1.67.

(b) What is the mean insurance cost? $1,650

What is the cut off for the 75th percentile? $1,800

Both stated in the problem

(c) Identify the standard deviation of insurance premiums in LA.

We know that when $X = 1800, Z = 0.67$

So

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

$0.67 = \frac{1800 - 1650}{\sigma}$

$0.67\sigma = 150$

$\sigma = \frac{150}{0.67}$

$\sigma = 223.88$

The standard deviation of insurance premiums in LA is $223.88.