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The data set below gives the points per game averages for the 10 players who had the highest averages (minimum 70 games or
1400 points) during the 2004-2005 National
Basketball Association regular season. Find the interquartile range and list any outliers that may exist.
35, 26, 19, 27, 23, 28, 39, 21, 18, 15

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Absolute Deviation- Is the absolute difference between the elements in the data set and the mean.
Mean Absolute Deviation or Mean Deviation
- is the average of the absolute deviations.
This is the average variation between the elements and the mean.

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Example: Find the Mean Absolute Deviation of the following data set:
10, 8, 5.
First: find the mean of the data set.
10+8+5= 23
23÷3= 7.66
Second: subtract the mean from each element in the data set and take the absolute value ( the absolute deviation).
10-7.66=l2.34l= 2.34
8-7.66=l.34l=.34
5-7.66= l-2.66l=2.66
Third: Find the mean of the resulting numbers.
2.34+.34+2.66=5.34
5.34÷3= 1.78
The mean absolute value is 1.78.

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Find the mean absolute value of the following data set:
100, 120, 115, 110.
Answer:
Find the mean of the set:
100+120+115+110= 445
445÷ 4= 111.25
Find the absolute deviation:
100 – 111.25= l-11.25l= 11.25
120 – 111.25=l8.75l= 8.75
115- 111.25=l3.75l= 3.75
110- 111.25=l-1.25l= 1.25
Find the mean of the absolute deviation values:
11.25+8.75+3.75+1.25= 25
25÷4= 6.25
The mean absolute deviation is 6.25.

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Find the mean absolute deviation of the following data set:
200, 179, 205, 190.
Answer:
Find the mean of the data.
200+179+205+190=774
774÷4= 193.5 The mean of the data set is
193.5.
Find the absolute deviation of each value.
200-193.5=l 6.5l= 6.5
179-193.5=l-14.5l=14.5
205-193.5=l11.5l=11.5
190-193.5= l-3.5l=3.5
Find the mean of the absolute deviation values.
6.5+14.5+11.5+3.5= 36
36÷4= 9 The mean absolute deviation of the data set is 9.

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Varianceis used as a measure of how far a set of numbers are spread out from each other
Example: 4, 3, 5, 7, 2, 8, 1, 10
Find the Variance of the data set.
• Find the mean (x)= 5
• Subtract the mean from each element and square.
4-5 = -1^2 = 1 2-5 = -3^2= 9
3-5 = -2^2 = 4 8- 5 = 3^2 = 9
5-5 = 0 ^2 = 0 1-5 = -4^2 = 16
7-5 = 2^2 = 4 10-5 = 5^2 = 25
Add and divide by the number of elements.
1+4+0+4+9+9+16+25= 68
68÷8 = 8.5
The variance is 8.5

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Find the variance of the data set below.
20, 17, 29, 25, 14
Answer:
X= 21
Variance = 28.8

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Standard Deviations- shows how how much variation or “dispersion” there is from the mean. It is denoted as σ (read as “sigma”). It is the square root of the variance.
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A low standard deviation indicates that the data points tend to be very close to the mean, whereas high standard deviation indicates that the data are spread out over a large range of values.
There is a formula to help you calculate the standard deviation.

Find the standard deviation of the data set:
10, 2, 6, 3, 9, 5, 7.
First: Find the mean of the data set.
42÷ 7= 6
The mean of the data set is 6.
Second: Subtract each element from the mean and square them.
(10-6)^2+(2-6)^2+(6-6)^2+(3-6)^2+(9-6)^2+(5-
6)^2+(7-6)^2. After you square the numbers you should have these numbers
16+16+0+9+9+1+1

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Third: Calculate the mean of these numbers.
16+16+0+9+9+1+1=52
52÷7= 7.429 The mean is 7.429.
Fourth: Square Root the mean.
Sqrt(7.429)=2.725
The Standard deviation is 2.725.

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After you have subtracted the mean from your data points and before you square them.
You will notice something interesting.

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Find the standard deviation of the following data set:
2, 5, 7, 3, 4, 9, 10, 1.
Answer:
Find the mean.
2+5+7+3+4+9+10+1= 41
41÷8= 5.125 The mean of the data set is 5.125
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Subtract the mean from the elements in the set and square them.
(2-5.125)^2+(5-5.125)^2+(7-5.125)^2+(3-5.125)^2+(4-
5.125)^2+(9-5.125)^2+(10-5.125)^2+(1-5.125)^2=
(-3.125)^2+(-.125)^2+(1.875)^2+(-2.125)^2+(-
1.125)^2+(3.875)^2+(4.875)^2+(-4.125)^2=
9.765+.0156+3.515+4.515+1.265+15.016+23.765+17.015=74.872
74.872÷8= 9.359.
Take the square root of the new number:
Sqrt(9.359)= 3.059
The standard deviation of the data set is 3.059.