CHAPTER 5

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CHAPTER 5

SOLUTIONS TO PROBLEMS

The maximum variation for the males can be determined as:

The maximum variation for the females can be determined as:

The observed variation for the males is the sum of 5,372 + 25,596 + 32,390 + 44,872 + 5,508 + 6,970 + 9,656 + 33,210 + 46,008 + 58,220 = 267,802.

The observed variation for the females is the sum of 2,852 + 12,214 + 13,826 + 13,826 + 9,062 + 10,258 + 10,258 + 43,931 + 43,931 + 49,729 = 209,887.

The value of the IQV for the males, therefore, is:

The value of the IQV for the females, is:

The two groups have approximately the same amount of variability.
a. The range is 322 - 42 = 280 executions

b. The interquartile range is found by first rank ordering the number of executions:
```
42		245
83		249
113		251
133		259
138		264
143		273
159		280
167		296
172		297
198		299
209		322
244
```
With 23 observations, the median is in the 23+1/2 or 12^th position (mdn = 244 executions). Then find the quartile positions by the formula (truncated median position + 1)/2. This is equal to (12 +1)/2 = 13/2 = 6.5. To find the fir st quartile (Q₁), then, take the average of the 6^th and 7^th scores from the lowest score. These are 143 and 159, respectively, so the first quartile is equal to (143+159)/2 = 151. To find the third quartile (Q₃) , take the average of the 6^th and 7^th scores from the highest score. These are 273 and 264, respectively. The third quartile is equal to (273+264)/2 = 268.5. The interquartile range can now be determined:

c. We will use the computational formula (formula 5.11) to determine the variance and standard variation. First, sum all of the scores shown above (S=4,836) and square this number [Sx²=(4,836)² = 23,386,896. Then, square each score above (x²) and sum up these squared scores (Sx²=1,147,422). Finally, substitute into formula 5.11:
The range is equal to 14.7 - 9.7 = 5 for the first column of cities, and 80.1 - 9.7 = 70.4 for the second column of cities.

The variance and standard deviation for the first column of cities are equal to:

The standard deviation and variance for the second column of cities are equal to:
.
In order to determine the index of qualitative variation, we need to determine both the observed and maximum heterogeneity for each offense. For property offenders:

For Violent Offenders:

For Drug Offenders:

For Status Offenders:

The greatest variability is found for violent and drug offenders followed by status and property offenders.

The variation ratio for each offense are as follows:

According to the variation ratio, violent and drug offenders evidence the greatest variation. A smaller proportion of their cases fall at the mode than for property and drug offenders.

To calculate the standard deviation and variance, we need to determine the midpoint of each interval. The best way to solve this problem is to make the following table:

Interval        freq       m        fm       m²        fm²        
 0 -  4          76        2       152       4         304
 5 -  9          52        7       364      49        2548
10 - 14          38       12       456     144        5472
15 - 19          21       17       357     289        6069
20 - 24          10       22       220     484        4840
25 - 29           8       27       216     729        5832                        S = 205           S = 1765           S = 25065

Now we can determine the variance and standard deviation with our computing formulas:

To prepare to answer this question, we will begin by rank ordering the rape rates. Since we will need these for our computations, we will also sum the first column (Sx), and square each x value and plac e that in the second column.
```
 x           x²  18.3       334.89
  25.9       670.81
  28.2       795.24
  28.7       823.69
  34.0      1156.00
  34.6      1197.16
  35.4      1253.16
  40.0      1600.00
  40.9      1672.81
  42.3      1789.29
  42.4      1797.76
  44.6      1989.16
  45.6      2079.36
  45.9      2106.81
  47.0      2209.00
  53.4      2851.56
  53.4      2851.56
  58.9      3469.21  
S=719.5   S=30,647.47
```
With this completed, we can now answer our questions.

a. The range is 58.9 - 18.3 = 40.6

b. To calculate the interquartile range, we first find the quartile positions. The quartile positions is equal to . Where 9 is the truncated median position. The first quartile, then, is the 5^th position from the lowest score, and the third quartile is the 5^th position from the highest score. Finding the scores corresponding to these positions, we can now calculate the interquartile range as:

c. With the calculations in the table above, we can use the computing formula for the standard deviation as follows:

d. Knowing what the standard deviation is, we can easily determine the variance:
.

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