STATS 60, Spring 2008
April 14, Lecture 08
A note on interpolation
When discussing percentiles, we said, for example, that the
element in an ordered list corre-
sponds to the 1
quartile. What if the elements were [2,12,34,90]? Then the 1
quartile would be found
at the 1.25th position. We did not discuss how one uses interpolation to obtain the desired result, mostly
because I wanted to emphasize the big picture—that the answer was somewhere between the ﬁrst and second
element in the list. But sometimes you need to ﬁnd an actual number, and so you need to interpolate. The
formula for interpolation is
, where in this example
1 = 0
. If we were interested in the 3.56
3 = 0
and so on.
It seems as if quite a few students are not clear on the basics of
-scores and the standard normal distribution.
To help clarify, ﬁrst note that we sometimes speak of a theoretical (perfect) normal distribution (standard
or not). We also speak of the sample distribution, which can be more or less like a normal distribution but
never perfectly equal to one. Samples are messy by nature.
What happens when we have a sample, from any distribution, and convert it to
-scores? One thing we
doing is transforming that sample into one that has a perfect standard normal distribution. What
doing is converting that sample to one that has a mean of zero and SD of one. We then can compare
-scores against a perfect standard normal distribution. In so doing, the resulting
-values can be used
for a variety of purposes.
What is it that we do when we calculate
-scores? It is a two step process: for any given sample value, we
ﬁrst subtract the mean and then divide by the sample distribution. But what are these steps doing to the
sample as a whole? A non-normal sample distribution with mean of 80 and SD of 24 is shown in Figure 1.A.
The ﬁrst step in calculating
-values is to subtract the sample mean. But that does not change the general
shape of the sample distribution, it just shifts it over until it is centered above zero (panel B in the ﬁgure).
In the second step we divide each centered value by the sample SD. But what does that do? As we have
noted in class, multiplying a sample times a constant, such as 4, increases the SD by a factor of 4. Likewise,
dividing by 4 reduces the SD by a factor of 4. Now, we have a sample with
and we divide it by
so the new SD is
. The sample of
-values now has a standard deviation of one.
To summarize, subtracting the mean from a sample and dividing by the sample SD merely shifts the entire
sample so that it is centered over zero, and then spreads out or squeezes the
-axis so that the new SD is
one. Lastly, when we plot a density curve, the area under the curve is always equal to one by design. So