Chapter 2: Q 22. (page 107)

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 $1.05 t i m e s$ of the old median of the salary and the new mean salary is $1.05 t i m e s$ of the old mean salary.

Part (b) The new inter-quartile range is $1.05 t i m e s$ the old inter-quartile range, and the new wage's standard deviation is $1.05 t i m e s$ that of the old salary.

Step by step solution

Part (a) Step 1. Given

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

Part (a) Step 2. Concept

The formula used: $z = \frac{x - m e a n}{s \tan d a r d d e v i a t i o n}$

Part (a) Step 3. Calculation

Assuming that each instructor is given a 5% raise, the new mean and median are calculated as follows: $M e a n (n e w) = \frac{Σ (s a l a r y o f e a c h t e a c h e r + s a l a r y o f e a c h t e a c h e r \times 0.05)}{n u m b e r o f t e a c h e r} = \frac{Σ (s a l a r y o f e a c h t e a c h e r \times (1 + 0.05))}{n u m b e r o f t e a c h e r s} = 1.05 (\frac{Σ s a l a r y o f e a c h t e a c h e r}{n u m b e r o f t e a c h e r s}) = 1.05 \times M e a n (o l d)$

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, $M e d i a n (n e w) = M e d i a n (o l d) + 0.05 \times M e d i a n (o l d) = 1.05 \times M e d i a n (o l d)$

Therefore the new median of the salary is $1.05 t i m e s$ of the old median of the salary. As a result, the new pay median is $1.05 t i m e s$ the old salary median, and the new mean income is $1.05 t i m e s$ the old mean salary.

Part (b) Step 1. Explanation

Assuming that each instructor is given a $5 %$ raise, the new standard deviation is calculated as follows: $s t d . d e v . (n e w) = \sqrt{\frac{{Σ [(1.05 \times s a l a r y o f e a c h t e a c h e r) - (1.05 \times M e a n (o l d))]}^{2}}{n - 1}} = \sqrt{\frac{{1.05}^{2} Σ {[(s a l a r y o f e a c h t e a c h e r) - (M e a n (o l d))]}^{2}}{n - 1}} = 1.05 \times \sqrt{\frac{Σ {[(s a l a r y o f e a c h t e a c h e r) - (M e a n (o l d))]}^{2}}{n - 1}} = 1.05 \times s t d . d e v . (o l d)$

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 $Q_{1} (n e w) = Q_{1} (o l d) + 0.05 \times Q_{1} (o l d) = 1.05 \times Q_{1} (o l d) Q_{3} (n e w) = Q_{3} (o l d) + 0.05 \times Q_{3} (o l d) = 1.05 \times Q_{3} (o l d)$

As a result, the revised interquartile range is as follows: $I Q R (n e w) = Q_{3} (n e w) - Q_{1} (n e w) = 1.05 \times Q_{3} (o l d) - 1.05 \times Q_{1} (o l d) = 1.05 \times (Q_{3} (o l d) - Q_{1} (o l d)) = 1.05 \times I Q R (o l d)$

As a result, the new inter-quartile range is $1.05$ $t i m e s$ greater than the old inter-quartile range, and the new salary's standard deviation is $1.05 t i m e s$ more than the old salary's standard deviation.