My text book says:
"If a distribution is symmetric, the median plus or minus the semi-interquartile range contains half the scores in the distribution."
I have a simple example set: {1,4,8,12,15}
It is symmetric:
1 _ _ 4 _ _ _ 8 _ _ _ 12 _ _ 15
Median M = Mean = (1 + 4 + 8 + 12 + 15)/5 = 40/5 = 8
Percentile ranks:
p = (r-1)/(n-1), where r for rank, n for size of set
r (25%) = 1 + .25(n-1) = 2
r (50%) = 1 + .5(n-1) = 3
r (75%) = 1 + .75(n-1) = 4
Percentile values:
v (25%) => 4
v (50%) => 8
v (75%) => 12
Semi-interquartile range:
R/2 = 0.5(v (75%) – v (25%)) = 0.5(12 – 4) = 4
Following the text book, then
M + R/2 = 8 + 4 = 12
M – R/2 = 8 – 4 = 4
Half the values would mean to me:
(M + R/2) – (M – R/2) = 12 – 4 = 8
Total range:
TR = 15 – 1 = 14
then,
1/2TR = 7 which is not equal to 8
I believe I do not understand what the text book says.
May I know the mistake in my understanding?
Perhaps an example could help me understand.
Thank you very much.
Best Answer
You have written
$M + R/2 = 8 + 4 = 12$
$M - R/2 = 8 - 4 = 4$
where $M$ is the median and $R$ the semi-interquartile range
Clearly three of the five terms of $\{1,4,8,12,15\}$ are in the interval $[4,12]$
Three is about as close as you can get to half of five when considering integers, and this is what your textbook is saying