Question

in 2011; the mean property crime rate per 100,000 people for the 11 western states was about 3331; the standard deviation was 729. Assume distibution and crime rates is unimodal and symmetric.

a) what percent of western states would you expect to have crime rates between 2602 and 4060?

b) what percentage if western states would you expect to have prioerty crune rates between 1873 and 4789?

c) if someone guessed that the property crime rate in one western state was 9000; would you agree that that number was consistent with this data set?

Answer 1

Solution:

We are given that:  the mean property crime rate per 100,000 people for the 11 western states was about 3331; the standard deviation was 729 and the  distibution of crime rates is unimodal and symmetric.

Thus we have : Mean = and the standard deviation =

Thus we use Empirical rule to find following probabilities:

According to Empirical rule:

1) 68% of the data falls within 1 standard deviation from mean

2) 95% of the data falls within 2 standard deviation from mean

3) 99.7% of the data falls within 3 standard deviation from mean

Part a) what percent of western states would you expect to have crime rates between 2602 and 4060?

P( 2602 < X< 4060) =.........?

Find z scores:

thus we get:

P( 2602 < X< 4060) = P( -1.00 < Z < 1.00 )

Thus 2602 and 4060 are 1 standard deviation from mean.

Thus P( 2602 < X< 4060) = 68%

Thus 68% of western states would you expect to have crime rates between 2602 and 4060.

Part b) what percentage of western states would you expect to have property crime rates between 1873 and 4789?

P( 1873 < X< 4789) =.........?

Thus 1873 and 4789 are two standard deviation from mean value.

Thus according to Empirical rule 95% of the data falls within 2 standard deviation from mean.

Thus

P( 1873 < X< 4789) = 95%

Thus 95% of western states would you expect to have property crime rates between 1873 and 4789.

Part c) if someone guessed that the property crime rate in one western state was 9000; would you agree that that number was consistent with this data set?

find z score for x = 9000

Since z score for x= 9000 is 7.78 which more than 2 standard deviation above the mean value,

thus we do not agree with this number x =9000 that number was consistent with this data set.