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Teacher raises Refer to Exercise 20. If each teacher receives a 5% raise instead of a flat \(1000 raise, the amount of the raise will vary from \)1400 to $3000, depending on the present salary.

(a) What will this do to the mean salary? To the median salary? Explain your answers.

(b) Will a 5% raise increase the IQR? Will it increase the standard deviation? Explain your answers.

Short Answer

Expert verified

Part (a) The new median of the salary is 1.05timesof the old median of the salary and the new mean salary is 1.05timesof the old mean salary.

Part (b) The new inter-quartile range is 1.05times the old inter-quartile range, and the new wage's standard deviation is 1.05times that of the old salary.

Step by step solution

01

Part (a) Step 1. Given

Instead of a flat $1000raise, teachers will receive a 5% raise. The money will be between $1400 and $3000

02

Part (a) Step 2. Concept

The formula used: z=xmeanstandarddeviation

03

Part (a) Step 3. Calculation

Assuming that each instructor is given a 5% raise, the new mean and median are calculated as follows:Mean(new)=(salaryofeachteacher+salaryofeachteacher0.05)numberofteacher=(salaryofeachteacher(1+0.05))numberofteachers=1.05salaryofeachteachernumberofteachers=1.05Mean(old)

As a result, the new average salary is 1.05 times the former average income.

Changes in scale do not affect the median position, as we know. As a result, the new median income is equal to the value of the old median wage plus 5%. That is to say,Median(new)=Median(old)+0.05Median(old)=1.05Median(old)

Therefore the new median of the salary is 1.05times of the old median of the salary. As a result, the new pay median is 1.05times the old salary median, and the new mean income is 1.05times the old mean salary.

04

Part (b) Step 1. Explanation

Assuming that each instructor is given a 5%raise, the new standard deviation is calculated as follows: std.dev.(new)=[(1.05salaryofeachteacher)(1.05Mean(old))]2n-1=1.052[(salaryofeachteacher)(Mean(old))]2n-1=1.05[(salaryofeachteacher)(Mean(old))]2n-1=1.05std.dev.(old)

As a result, the new salary's standard deviation is 1.05 times the old salary's standard deviation. The first and third quartiles are still the first and second half of the data's medians; the scale shift has no effect on the median position.

Thus Q1(new)=Q1(old)+0.05Q1(old)=1.05Q1(old)Q3(new)=Q3(old)+0.05Q3(old)=1.05Q3(old)

As a result, the revised interquartile range is as follows: IQR(new)=Q3(new)Q1(new)=1.05Q3(old)1.05Q1(old)=1.05(Q3(old)Q1(old))=1.05IQR(old)

As a result, the new inter-quartile range is 1.05times greater than the old inter-quartile range, and the new salary's standard deviation is 1.05times more than the old salary's standard deviation.

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