Question

2 (f) If a new grade, Y, is added to the data set, so the data set becomes 12, 78, 24, 28, 42, 60, 94, 62, 68, Y (where Y is the new grade), what is the maximum amount by which the new median could differ from the old median (in 2(a))? (This means, if the new median is mnew and the old median that you found in 2(a) was mold, what is the maximum value that |mnew−mold|can have? Here, | x| denotes the absolute value of x, which is always greater than or equal to 0. For example, if the new median is 56 and the old median was 62, I would say that the new median differs from the old median by | 56 - 62| = 6).

Mnewmold

Answer 1

Here you haven't provided the question 2(a). Possibly in question 2(a), there were 9 numbers 12, 78, 24, 28, 42, 60, 94, 62, 68. If so, then rearranging the data in ascending order we find the observations as follows:

12, 24, 28, 42, 60, 62, 68, 78, 94.

There are 9 observations which is odd in number. So the median would be the (9+1)/2 th observation i.e., the 5th observation. So the median for the old data set is 60, as this is the 5th observation in the data set.

Thus m_old = 60.

Now in the new data set one observation Y has been added. Inclusion of Y makes the count of the observations 10, which is an even number. So for the new data the median would be the arithmetic mean of the 10/2 th and (10/2 + 1) th observations i.e., the 5th and 6th observations.

Now let us consider the different cases:

Case - I: Y <= 42

Then the 5th and 6th observations must be 42 and 60 and so the median in this case is (42+60)/2 = 51

Case - II: 42 < Y <= 60

Then the 5th and 6th observations will be Y and 60 respectively and so the median is (Y+60)/2 which is greater than 51, the median in Case - I, since 42 < Y

Case - III: 60 < Y <= 62

Here the 5th and 6th observations would be 60 and Y and the median would be (60+Y)/2, which is again greater than the medians in Cases - I and II, since Y > 60

Case - IV: 62 < Y

In this case the 5th and 6th observations are 60 and 62 and so the median is (60+62)/2 = 61, which is bigger than all the medians in Cases - I, II and III, since Y > 62

So we can see from the above discussions that the lowest and largest value the new median can have is 51 and 61 respectively. Therefore it is very obvious that the lowest one i.e., 51 will show the maximum difference with the old median, which is 60.

Hence for the maximum difference, m_new = 51 and the maximum value of |m_new - m_old| = |51-60| = 9 [Ans].

*If it clears your doubt, please let me know.