Z Scores and the Normal Curve
Vartanian, SW 134
The z score used in these examples gives the area between the value (x) and the mean ( x ).
Example 1.
Mean=10 SD=4
What proportion of cases are between 10 and 12
z=
x − x 12 − 10
=
= .5
s
4
Z==>Deviation from the mean in standard deviation units.
From this Z score, we can look at a z table ==> and see that the value for .50 is 19.15.
If we had gotten a value of 0.51, the area under the curve would have been 19.50.
Example 2:
If we had wanted to know the area between 8 and 12, we would have simply doubled the # that we got.
2*19.15=38.30
The table gives us half the area of the curve. But since the curve is symmetrical, we can determine the entire
area by doubling what is in the table.
Example 3:
The area below 8.
z=
x − x 8 − 10
=
= −.5
s
4
at Z=.50, the Percentage of the area under the curve between 8 and the mean is 19.15. To determine the area to
the left of 8, we would subtract 19.15 from 50.00 (half the area under the curve). Thus our total area = 50.00 19.15 = 30.85
Example 4:
X = 50 SD =6
We want to find the area to the left of 42. What percentage of people had scores below 42?
x − x 42 − 50
z=
=
= −1.33
s
6
The table in the back of the book tells us that a score of -1.33 = 40.83. Thus, between the mean and 42 there is
an area of 40.83. To determine the area below 42, we subtract 40.83 from 50.00 = 9.17.
1
H:\Word 2010\Lect2.mss\Z Scores\Z_Scores_and_the_Normal_Curve.docx
Calculating Z scores.
z=
x−x
Z values indicate how many standard deviation units the x value is from the mean.
s
Example 1: In a class of 22 students, the mean score is 80%. You get a 90%. The standard deviation is 10.
At what percentile are you located?
z=
90 − 80 10
=
=1
10
10
. Thus, you are 1 standard deviation above the mean. This corresponds to 34.13% above the
mean. If you add this to 50% (because you are 34.13% above the mean, and the mean is located at the 50th
percentile), then you are at the 84.13 percentile.
Example 2: Given the same characteristics as above, but you received a 70% on the exam, at what
percentile would you be located? For this problem, you would subtract 34.13 from 50. 50-34.13=15.87
percentile. The 34.13 is indicating how far below the mean you are located so you subtract it from 50.
z=
70 − 80
10
= − = −1
10
10
. Thus, you are 1 standard deviation below the mean. This again corresponds to 34.14%.
Example 3: In a class of 36 students, the mean test score is 80%. You get a 70%. The standard deviation is
11. Where are you on the distribution?
z=
70 − 80
10
= − = −.91 , or you are .91 standard deviations below the mean. This corresponds to 31.86% of the
11
11
distribution. If you subtract this from 50, you get 18.14, or you’re at the 18.14 percentile.
Example 4: Given the same characteristics as example #3 (where the SD =11 and the mean=10), where would you be
in the distribution if you got a score of 90?
z=
90 − 80 10
=
= .91
11
11
. This corresponds to 31.86%. You would add this on to the 50th percentile, and you would
be located at the 81.86 percentile.
Example 5: Assume that the mean is 400 and the SD is 50. At what percentile is a score of 475?
z=
475 − 400 75
=
= 1.5
50
50
. This corresponds to a value of 43.32. We would add this to 50 because we are above the
mean. So, 50+43.32=93.32. A score of 475 is at the 93.32nd percentile.
Example 6: Assume that the mean is 400 and the SD is 50. At what percentile is a score of 300?
z=
300 − 400
100
=−
= −2.0
50
50
. This corresponds to 47.72 in the z table. We would subtract this from 50 to get
a value of 2.28 percentile.
2
H:\Word 2010\Lect2.mss\Z Scores\Z_Scores_and_the_Normal_Curve.docx