Hello & Welcome! In this video, I’ll be showing how to use
the “less than” cumulative normal distribution tables to recover x and z scores, given areas
or probabilities.
Let’s look at some examples on finding z-scores.
Example 1: We want to find z if the area to the left
of z is 0.8921. It’s always advisable to draw the curve
in order to visualize the area and the position of the z-score.
I have placed z on the right of the center here because the .8921 is greater than 0.5.
And that tells us that the z-score will be positive.
So we go to the positive z-score section and look for the area closest to 0.8921.
Note that the z-scores are printed in red and the areas are printed in black.
The closest area to 0.8921 we find here is 0.8925, corresponding to a z score of 1.24.
In the second example we want to find z such that 71.9% of the area is to the right of
z. 0.719 is greater than 0.5 here, and it is
to the right of z, so z has to be on the negative side of the center.
So we go to the negative section of the normal tables.
Since the areas in the tables are “less than” areas, we need to find the “less
than” area corresponding to z here. That is, 1 – 0.7190 which gives 0.2810.
Searching the area section, we find 0.2810 exactly here, corresponding to a z score of -0.58.
In the 3rd example, we want to find the z-score corresponding to the 10th percentile.
Since percentile means percent below, we have .10 of the area below z and so it will be
on the negative side. The closest area to .10 we can find on the
table is 0.1003. And that corresponds to a z score of -1.28.
So the z-score corresponding to the 10th percentile is -1.28.
Now recall the standard normal score transformation formula. To recover the score x, we can simply
make x the subject of the formula by first cross multiplying and bringing sigma up, and
then moving the mean to the left. And finally, exchange both sides.
Let’s see the application in the following example:
Suppose we have a normal distribution with a mean of 20 and a standard deviation of 3.
We want to find the score that represents the 85th percentile.
Since percentile represents percent below, we first draw the standard normal curve and
place the 0.85 to the left of z here.
Now, doing reverse lookup on the positive section of the z tables, we see that the closest
probability to 0.85 here is .8508. And that corresponds to a z-score of 1.04.
We can then either substitute the values into the z-formula and solve for x or we can directly
apply the x formula we obtained earlier.
And substituting the values gives x = 23.12. So the 85th percentile is 23.12.
Next example.
In this one, we want to find a score such that only 8% of the values are above that score.
That is, we have 0.08 in the right tail here. Since the areas in our tables are less-than
areas, we do 1 minus .08 to obtain a less-than area of 0.92.
We then search for the closest to 0.92 in the table. And the closest we find here is
.9207 which corresponds to a z-score of 1.41.
Substituting again in the x formula we have x = 24.23.
That is, only 8% of the scores are above 24.23.
Let’s look at one more example. Here we want to find two values (a lower limit,
and an upper limit), that will be boundaries for the middle 75% of the scores.
With.75 in the middle, we have .25 divided equally into the tails. That is, .125 in each
tail. For the lower limit, we go to the negative
section of the tables and search for the closest to .125.
The closest we find here is .1251 which corresponds to a z-score of -1.15.
For the upper limit, the area below is .75 + .125 which gives .875.
Looking that up in the table, the closest area is 0.8749 corresponding to a z-score
of 1.15.
That should be expected because the normal curve is symmetric. That is, we don’t even
need to look up the positive z score. Symmetry should tell us that it will be 1.15.
Let’s now calculate the actual scores. For the lower limit, substituting -1.15 gives
16.55. And for the upper limit, with +1.15, we have
23.45.
In summary, to recover a score for a normal distribution,
Construct the normal curve identifying the area of the required probability.
Then search for the closest less-than area to it in the z tables.
Once you find the closest area, locate its corresponding z-score and, if necessary, substitute
it into the x formula to obtain the required x-value.
And that’s it for this video.
Thanks for watching.