F.5 maths dispersion

Original Data (in ascending order):
49, 54, 57, 59, 61, 65, 67, 68

Number of data = 8
Median = (59 + 61)/2 = 60
Max = 68
Min = 49
Upper quartile = (65 + 67)/2 = 66
Lower quartile = (54 + 57)/2 = 55.5

Therefore, Range = 68 - 49 = 19;
IQR = 66 - 55.5 = 10.5.

A datum x is added to the data set.
Number of data = 8 + 1 = 9.

If the range remains unchanged, x must not be larger than the max or smaller than the min.
That is, it is required that 49 ≤ x ≤ 68.

Further, it is required that the IQR remains unchanged.

49 (A) 54 (B) 57(C) 59 (D) 61 (E) 65 (F) 67(G) 68

Consider the above 7 possible "places" [(A) to (G)] for x to be inserted to follow the ascending order.

If x < 57, the median = 59. the upper quartile is kept at 66, the lower quartile is changed, and the IQR is changed.

If x > 65, the median = 61. the lower quartile is kept at 55.5, the upper quartile is changed, and the IQR is changed.

If 57 ≤ x ≤ 65,
the upper quartile is still kept at (65 + 67)/2 = 66,
the lower quartile is still kept at (54 + 57)/2 = 55.5,
so the IQR remains unchanged.

Therefore, combining 49 ≤ x ≤ 68 and 57 ≤ x ≤ 65, the possible value(s) of x are: 57 ≤ x ≤ 65