To test if the data is approximately normal, use the empirical rule for the norm

Question

To test if the data is approximately normal, use the empirical rule for the normal distribution. Find the percentage of male salaries within one, two, and three standard deviation of the mean. Compare the percentages to those predicted by the empirical rule for the normal distribution. Repeat the calculations for the female salaries. Comment on the results.

I just need the information for the calculations for the females salaries.  On the link below you can find the data needed.

http://www.chegg.com/homework-help/questions-and-answers/refer-excel-file-salary-attached-assignment-address-questions-variable-data-set-provided-c-q18506046

Explanation / Answer

the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 52.792
standard Deviation ( sd )= 17.9923
68% OF DATA
About 68% of the area under the normal curve is within one standard deviation of the mean. i.e. (u ± 1s.d)
So to the given normal distribution about 68% of the observations lie in between
= [52.792 ± 17.9923]
= [ 52.792 - 17.9923 , 52.792 + 17.9923]
= [ 34.7997 , 70.7843 ]
95% OF DATA
About 95% of the area under the normal curve is within two standard deviation of the mean. i.e. (u ± 2s.d)
So to the given normal distribution about 95% of the observations lie in between
= [52.792 ± 2 * 17.9923]
= [ 52.792 - 2 * 17.9923 , 52.792 + 2* 17.9923]
= [ 16.8074 , 88.7766 ]
99.7% OF DATA
About 99.7% of the area under the normal curve is within two standard deviation of the mean. i.e. (u ± 3s.d)
So to the given normal distribution about 99.7% of the observations lie in between
= [52.792 ± 3 * 17.9923]
= [ 52.792 - 3 * 17.9923 , 52.792 + 3* 17.9923]
= [ -1.1849 , 106.7689 ]
-----------------------------------------------------------------------------------------------------------------------------------------------

the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 38.348
standard Deviation ( sd )= 17.9923
68% OF DATA
About 68% of the area under the normal curve is within one standard deviation of the mean. i.e. (u ± 1s.d)
So to the given normal distribution about 68% of the observations lie in between
= [38.348 ± 17.9923]
= [ 38.348 - 17.9923 , 38.348 + 17.9923]
= [ 20.3557 , 56.3403 ]
95% OF DATA
About 95% of the area under the normal curve is within two standard deviation of the mean. i.e. (u ± 2s.d)
So to the given normal distribution about 95% of the observations lie in between
= [38.348 ± 2 * 17.9923]
= [ 38.348 - 2 * 17.9923 , 38.348 + 2* 17.9923]
= [ 2.3634 , 74.3326 ]
99.7% OF DATA
About 99.7% of the area under the normal curve is within two standard deviation of the mean. i.e. (u ± 3s.d)
So to the given normal distribution about 99.7% of the observations lie in between
= [38.348 ± 3 * 17.9923]
= [ 38.348 - 3 * 17.9923 , 38.348 + 3* 17.9923]
= [ -15.6289 , 92.3249 ]

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.