IQR Example
Standard Deviation Example
University | Average SAT Score |
University of Washington | 1760 |
Ohio State | 1775 |
UConn | 1775 |
Rutgers | 1795 |
University of Delaware | 1795 |
University of Texas Austin | 1805 |
DePauw | 1825 |
University of Miami | 1875 |
The College of New Jersey | 1880 |
Lafayette | 1915 |
Georgia Tech | 1930 |
Smith College | 1940 |
Bryn Mawr | 1960 |
Mount Holyoke | 1960 |
Case Western | 1965 |
Colby | 2015 |
Macalester | 2025 |
Barnard | 2055 |
Wesleyan | 2070 |
Vassar | 2075 |
We can use the original SAT scores to calculate the interquartile range and standard deviation.
We already know the median. The middle value is also the value for quartile 2.
Quartile 1 is the middle value of the bottom half of the list. That is the average between
the 5th and 6th values. = 1800 = Q1
Quartile 3 is the middle value of the top half of the list. That is the average between the
15th and 16th values. = 1990 = Q3
The interquartile range is the spread of the middle half of the data. The interquartile
range = Q3 – Q1 = 1990 – 1800 = 190.
There is a 190 point spread in the middle half of the data.
University | Average SAT Score |
||
University of Washington | 1760 | -150 | 22500 |
Ohio State | 1775 | -135 | 18225 |
UConn | 1775 | -135 | 18225 |
Rutgers | 1795 | -115 | 13225 |
University of Delaware | 1795 | -115 | 13225 |
University of Texas Austin | 1805 | -105 | 11025 |
DePauw | 1825 | -85 | 7225 |
University of Miami | 1875 | -35 | 1225 |
The College of New Jersey | 1880 | -30 | 900 |
Lafayette | 1915 | 5 | 25 |
Georgia Tech | 1930 | 20 | 400 |
Smith College | 1940 | 30 | 900 |
Bryn Mawr | 1960 | 50 | 2500 |
Mount Holyoke | 1960 | 50 | 2500 |
Case Western | 1965 | 55 | 3025 |
Colby | 2015 | 105 | 11025 |
Macalester | 2025 | 115 | 13225 |
Barnard | 2055 | 145 | 21025 |
Wesleyan | 2070 | 160 | 25600 |
Vassar | 2075 | 165 | 27225 |
The standard deviation is another way of looking at the spread of the data, or how much the scores vary among themselves. The standard deviation uses the difference between every score and the mean.
The equation for standard deviation is where is the standard deviation, xi is each value, is the mean, and n is the number of values in the set.
Here are the steps.
1. Find the mean. We did that earlier = 1910
2. Subtract the mean from each score. Look at column 3 in the table.
3. Square the values in column 3. and write them in column 4.
4. Add the values in column 4 to find the sum of the squares = 213225
5. Divide the sum of squares by the number of values. n = 20 = 10661.25 is the mean of the sum of squares
6. Find the square root of 10661.25.
This is the standard deviation. = 103.3