Answer :
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.