## 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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2022-02-02T13:44:46+00:00
2022-02-02T13:44:46+00:00 1 Answer
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## Answers ( )

Answer:a)Z = 1.67b)What is the mean insurance cost? $1,650What 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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X – \mu}{\sigma}[/tex]

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:[tex]\mu = 1650[/tex]

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? $1,650What 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 [tex]X = 1800, Z = 0.67[/tex]

So

[tex]Z = \frac{X – \mu}{\sigma}[/tex]

[tex]0.67 = \frac{1800 – 1650}{\sigma}[/tex]

[tex]0.67\sigma = 150[/tex]

[tex]\sigma = \frac{150}{0.67}[/tex]

[tex]\sigma = 223.88[/tex]

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