Measure of Spread

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

Warm-up

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.

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= 2323÷3= 7.66Second: 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=.345-7.66= l-2.66l=2.66Third: Find the mean of the resulting numbers. 2.34+.34+2.66=5.345.34÷3= 1.78 The mean absolute value is 1.78.

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= 445445÷ 4= 111.25 Find the absolute deviation:100 – 111.25= l-11.25l= 11.25 120 – 111.25=l8.75l= 8.75115- 111.25=l3.75l= 3.75110- 111.25=l-1.25l= 1.25 Find the mean of the absolute deviation values:11.25+8.75+3.75+1.25= 2525÷4= 6.25 The mean absolute deviation is 6.25.

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=774774÷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= 6.5179-193.5=l-14.5l=14.5205-193.5=l11.5l=11.5190-193.5= 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.

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, 10Find 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 = 167-5 = 2^2 = 4 10-5 = 5^2 = 25

Add and divide by the number of elements. 1+4+0+4+9+9+16+25= 6868÷8 = 8.5The variance is 8.5

Example Find the variance of the data set below. 20, 17, 29, 25, 14

Answer:X= 21Variance = 28.8

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.

Find the standard deviation of the data set:10, 2, 6, 3, 9, 5, 7.

How to calculate Standard Deviation

First: Find the mean of the data set.42÷ 7= 6The 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

Third: Calculate the mean of these numbers.16+16+0+9+9+1+1=5252÷7= 7.429 The mean is 7.429.

How To Calculate Standard Deviation

Fourth: Square Root the mean.Sqrt(7.429)=2.725

The Standard deviation is 2.725.

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

A way to check your work

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= 4141÷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=74.87274.872÷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.