2.  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

3.  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.

4.  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.

5.  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.

6.  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.

7.  Variance- is 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

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

9.  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.  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.

10. 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

11.  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.

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

13.  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. 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.