Download Normal Probability Distribution: Vartanian Distributional Shapes

Normal Probability Distribution: Vartanian
Distributional Shapes: Skewed distributions.
Skewed right distribution, positively skewed. downward sloping; tail is to the right; mean is greater than
the median; an example is income.
Skewed left: Upward sloping; tail is to the left; mean is less than the median; an example would be an easy
test.
1
Normal Distributions
Properties of the normal distribution.
1.
2.
3.
4.
The highest point of the distribution is the mean.
The mean=median=mode.
Symmetrical
Relatively high standard deviations indicate a more spread out, and flatter curve, while lower standard
deviations indicate a more condensed, higher curve.
5. The total area under the normal curve is 1. We can determine probabilities for particular areas of the
curve.
6. Almost all of the area under a normal curve is within 3 standard deviations from the mean.
Normal Distribution, Mean of 10, SD=2
0.25
Probability
0.2
0.15
Series1
0.1
0.05
0
0
5
10
X values
15
20
2
Calculating Z scores.
x u
z
SD
A z value will indicate how many standard deviation units the x value is from the mean.
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?
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.
z
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.
70  80
10
   1
10
10
. Thus, you are 1 standard deviation below the mean. This again corresponds to
34.14%.
z
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?
70  80
10
   .91 , or you are .91 standard deviations below the mean. This corresponds to 31.86%
11
11
of the distribution. If you subtract this from 50, you get 18.14, or you’re at the 18.14 percentile.
z
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?
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.
z
Example 5: Assume that the mean is 400 and the SD is 50. At what percentile is a score of 475?
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.
z
Example 6: Assume that the mean is 400 and the SD is 50. At what percentile is a score of 300?
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.
z
3
Example 7: Given the above characteristics for the class, what percentage of scores lie between 72 and 92?
72  80
8
   .73
11
11
92  80
12
z
   1.09
11
11
Z=-.73 corresponds to a value of 23.27
Z=1.09 corresponds to a value of 86.21
z
Thus, the percentage of scores between these two is 26.73+36.21=62.94
Example 8: Only those who are at least 2.3 standard deviations below the mean in income are eligible for
assistance programs. If the mean is $50,000 and the standard deviation is 14,000, what is the maximum
income for receiving assistance?
2.3*14000=32200. Assistance ends at incomes of $17,800.
4