SOLUTION: The average sentence length in 2018-2019 for all Canadian offenders admitted with a sex offence as their major admitting offence was μ=36 months. Suppose these sentence lengths a

Algebra ->  Probability-and-statistics -> SOLUTION: The average sentence length in 2018-2019 for all Canadian offenders admitted with a sex offence as their major admitting offence was μ=36 months. Suppose these sentence lengths a      Log On


   

Question 1186033: The average sentence length in 2018-2019 for all Canadian offenders admitted with a sex offence as their major admitting offence was μ=36 months. Suppose these sentence lengths are normally distributed with a population standard deviation
σ=4 months.
What is the serving time (months) of the shortest 5.59% sentence lengths?
Enter the z value for your calculation here:

Enter the serving time (months) of the shortest 5.59% sentence lengths:
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
population mean = 36 months
population standard deviation = 4 months.

you want to find the number of months that are less than or equal to 5.59% of the area under the normal distribution curve.

the z-score for that would be equal to the z-score that has .059 of the area under the normal distribution curve to the left of it.

that z-score would be equal to -1.56322 rounded to 5 decimal places.

use the z-score formula to find the raw score.

the z-score formula is:

z = (x - m) / s

z is the z-score
x is the raw score
m is the mean
s is the standard deviation.

the formula becomes:

-1.56322 = (x - 36) / 4.

solve for x to get:

x = -1.56322 * 4 + 36 = 29.74712.

5.59% of the sentences will be less than 29.74712 months.

this is what it looks like on the normal distribution graph.



the very slight difference between the results shown and the ones i gave you has to do with differences in rounding.