## Suppose a newspaper article states that the distribution of auto insurance premiums for residents of California is approximately normal with

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.

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3 months 2022-02-02T13:44:46+00:00 1 Answer 0 views 0

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.