1. DESCRIBING DISTRIBUTION NUMERICALLY
MEASURES OF CENTER:
MIDRANGE = (MAX + MIN) / 2
MEDIAN IS THE MIDDLE VALUE WITH HALF OF THE DATA ABOVE AND HALF BELOW IT.
MEAN = (SUM OF DATA) / (NUMBER OF COUNTS n)
EXAMPLE:
DATA: 45, 46, 49, 35, 76, 80, 89, 94, 37, 61, 62, 64, 68, 56, 57, 57, 59, 71, 72.
SORTED DATA: 35, 37, 45, 46, 49, 56, 57, 59, 61, 62, 64, 68, 71, 72, 76, 80, 89, 94.
MIDRANGE = ( ) / 2 = 64.5
MEDIAN = 61
MEAN = ( … + 94) / 19 = 62
NOTE: FOR SKEWED DISTRIBUTIONS THE MEDIAN IS A BETTER MEASURE OF THE CENTER THAN THE MEAN.

2. MEASURES OF THE SPREAD RANGE = MAX – MIN
INTERQUARTILE RANGE (IQR) = Q3 – Q1
Q3 = UPPER QUARTILE
= MEDIAN OF UPPER HALF OF DATA(INCLUDE MEDIAN IF n IS ODD)
Q1 = LOWER QUARTILE
MEDIAN OF LOWER HALF OF DATA(INCLUDE MEDIAN IF n IS ODD)
VARIANCE (later)
STANDARD DEVIATION (later)

3. Note: Include the median in the calculation of both quartiles
EXAMPLE: (odd number of observations, 19)
Median = 61
UPPER HALF
[ ]
Q3 = (71 +72) / 2 = 71.5
LOWER HALF
[ ]
Q1 = ( ) / 2 = 52.5
IQR = 71.5 – 52.5 = 19
Note: Include the median in the calculation of both quartiles

4. Quartiles EXAMPLE: (even number of observations, 18)
[60] [ ]
60 = Median = (59+61)/2 (Average of the middle two numbers)
UPPER HALF
[60] [ ]
Q3 = 71
LOWER HALF
[ ]
Q1 = 49
IQR = 71 – 49 = 42

5. 5 – NUMBER SUMMARY:
THE 5-NUMBER SUMMARY OF A DISTRIBUTION REPORTS ITS MEDIAN, QUARTILES, AND EXTREMES(MINIMUM AND MAXIMUM)
MAX = 94
Q3 = 71.5
MEDIAN = 61
Q1 = 52.5
MIN=35
OUTLIERS: DATA VALUES WHICH ARE BEYOND FENCES
IQR = Q3 – Q1 = 19
UPPER FENCE = Q IQR = x19 = 100
LOWER FENCE = Q1 – 1.5IQR = 52.5 – 1.5x19 = 24
IN THE EXAMPLE CONSIDERED ABOVE, THERE ARE NO OUTLIERS.

6. BOXPLOTS WHENEVER WE HAVE A 5-NUMBER SUMMARY OF A\
(QUANTITATIVE) VARIABLE, WE CAN DISPLAY THE
INFORMATION IN A BOXPLOT.
THE CENTER OF A BOXPLOT IS A BOX THAT SHOWS THE MIDDLE HALF OF THE DATA, BETWEEN THE QUARTILES.
THE HEIGHT OF THE BOX IS EQUAL TO THE IQR.
IF THE MEDIAN IS ROUGHLY CENTERED BETWEEN THE QUARTILES, THEN THE MIDDLE HALF OF THE DATA IS ROUGHLY SYMMETRIC. IF IT IS NOT CONTERED, THE DISTRIBUTION IS SKEWED.
THE MAIN USE FOR BOXPLOTS IS TO COMPARE GROUPS.

7. BOXPLOTS

8. Examples:
1. Here are costs of 10 electric smoothtop ranges rated very good or excellent by Consumers Reports in August 2002.
Find the following statistics by hand:
a) mean
b) median and quartiles
c) range and IQR

9. VARIANCE = “AVERAGE” SQUARE DEVIATION FROM THE MEAN
DEVIATION = (each data value) – mean
VARIANCE = 4648 / (19 -1) = 258.8
STANDARD DEVIATION = SQUARE ROOT ( VARIANCE)
= 16.1

10. VARIANCE = “AVERAGE” SQUARE DEVIATION FROM THE MEAN
Step 1: Sort Data:
565 Mean =
750 Median =1025
850 Q1=850
900 Q3=1200
Range = 835
IQR= 350
1050
1200
1250
1400

11. VARIANCE = “AVERAGE” SQUARE DEVIATION FROM THE MEAN
Computing the Variance
DEVIATION = (each data value) – mean
Squared Deviation= ((each data value) – mean)^2
Sum all squared deviations
Variance = (sum of all squared deviations)/(n-1), where n = is the number of observations

12. Variance Example: Variance = 147.2/4 = 36.8
Data Squared Deviations
6.76
Mean = 42.4
Variance = 147.2/4 = 36.8
Std Deviation = square root of variance
Std dev = 6.06

13. Some Remarks If the shape is skewed, report the median and IQR.
Mean and median will be very differnet.
You may want to include the mean and std deviation, but you should point out why the mean and the median differ.
If the histogram is symmetric, report the mean and the std deviation and possibly the median and IQR.