Question

URGENT HELP PLEASE!!!
An employer wishes to give his staff a bonus at Christmas time. If the staff receive a median wage of $780 per week with a mean deviation of $70, and the employer gives each person a $150 bonus at Christmas time, what is:
(a) the median amount received at Christmas time, including the bonus?
(b) the mean deviation of the wages at Christmas time, including the bonus?

Answer 1

Suppose we have a data set 10, 40, 50, 60, 100
The median is 50
\[\text{Mean} = \bar{x} = \frac{10+40+50+60+100}{5}=52\]\[\sum|x-\bar{x}|=|10-52|+|40-52|+|50-52|+|60-52|+|100-52| = 112\]\[\text{Mean deviation} = \frac{\sum|x-\bar{x}|}{n}= \frac{112}{5}=22.4\]
Now, if we add a constant 7 to the data set, the new set of data is 17, 47. 57, 67, 107
The new median is 57 = 50 + 7 = original median + constant
\[\text{New mean} = \bar{x} = \frac{17+47+57+67+107}{5}=59=52+9 = \text{original mean + constant}\]\[New \sum|x-\bar{x}|=|17-59|+|47-59|+|57-59|+|67-59|+|107-59| = 112\]\[\text{New mean deviation} = \frac{\sum|x-\bar{x}|}{n}= \frac{112}{5}=22.4 = \text{original mean deviation}\]

From the above calculations, we can conclude that when we add a constant k to each datum in the data set:
1. New median = original median + k
2. New mean = original mean + k
3. New mean deviation = original mean deviation.

I think you can proceed if you get the idea.

Answer 2

Isn't it meant to be 52 + 7 though? Yeah I get it now the median changes with the adding to the scores but the mean deviation does not because the mean and the scores are added by the same constant thanks for the explanation :)

Answer 3

Sorry for the typo there. You're right. It should be 52+7 for the new mean.