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
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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) (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.
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