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

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    2022-02-02T13:45:46+00:00

    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 [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? $1,650

    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 [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.

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