# Questions about measures of central tendency

41 (41 – 28.375) = 12.625 159.391

2 20 (20 – 28.375) = -8.375 70.141

3 35 (35 – 28.375) = 6.625 43.891

4 25 (25 – 28.375) = -3.375 11.391

5 23 (23 – 28.375) = -5.375 28.891

6 30 (30 – 28.375) = 1.625 2.641

7 21 (21 – 28.375) = -7.375 54.391

8 32 (32 – 28.375) = 3.625 13.141

Σ = 227 Σ = 0.000 Σ = 383.878

** **Let’s calculate the mean as our measure of central tendency by adding the individual ages of each officer and dividing by the number of officers. **The calculation is 227/8 = 28.375 year**

It is important to note from this calculation that the sum of the deviations of each score from the mean is equal to zero. When doing hand calculations of the **mean deviation**, **variance**, and **standard deviation** this is an excellent place to check your math. If the sum of the deviations of each score from the mean does not equal zero (or a number very, very close to zero in situations when you are rounding decimal places) then you have made a mathematical error either in your subtractions or your calculation of the **mean**.

Then, subtract the mean from every number to get the list of deviations. Create a list of these numbers. It’s OK to get negative numbers here.

Next, square the resulting list of numbers (read: multiply them with themselves) to get the squared deviation.

Add up all of the resulting squares to get their total sum.

To get the standard deviation, Divide your result by one less than the number of items in the list and just take the square root of the resulting number.

It is important for you to note that the last step in the calculation of the **variance** that I have described requires you to reduce the sample size by 1. This is done because we are using a **sample** rather than the entire population of officers. If our data is the actual entire **population **we would not subtract 1 from N. We would simply divide by the size of the entire **population.**. In this example the eight observations of the age of police officers is a sample from the total population of police officers in this jurisdiction. By subtracting 1 from the sample size (N – 1) we are adjusting the final value of the **variance (s ^{2}) **resulting in a value that is larger than if we were divide by N. When using sample data it is better to overstate the measure of dispersion than to understate it.

**STANDARD DEVIATION – s**

The **standard deviation (s) **is very simple to calculate.

In our example with the sample of 8 police officers and their Age, the **standard deviation** is:

S = **√383.878/7 ** **s = √54.84 s = 7.41years**

The **standard deviation** has another very important advantage over the other measures of dispersion in that we are able to use the **standard deviation** to estimate the number of variable values within certain areas under the curve representative of those values.

*Using the Standard Deviation*

I have posted a pdf file under the “notes” link that outlines the areas falling under/within the **normal curve/bell curve (or normal distribution).** Please refer to that graphical display for the remainder of this discussion.

*Please read: *

Our calculated mean is 28.375 years. When using the **normal curve/bell curve (or normal distribution)** to represent our variable we would place the mean, 28.375 years at the center of the distribution above X bar.

The numbers that correspond to the **“-1s”** and **“+1s”** are **20.965** (mean – standard deviation) and **35.785 **(mean + standard deviation) respectively. These numbers are calculated by adding one standard deviation unit (7.41 years) to the mean of 28.375 years and subtracting one standard deviation unit (7.41 years) from the mean of 28.375 years. This represents the range of ages between which we would expect to find approximately 68.26% of the total population of police officers in this jurisdiction. We would expect approximately 34.13% to have an age between 20.965 years and 28.375 years. Similarly, we would expect approximately 34.13% to have an age between 28.375 years and 35.785 years.

The numbers that correspond to the **“-2s”** and **“+2s”** are **13.555** and **43.195** respectively. These numbers are calculated by adding two standard deviation units (7.41 years x 2 = 14.82 years) to the mean of 28.375 years and subtracting two standard deviation units (7.41 years x 2 = 14.82 years) from the mean of 28.375 years. This represents the range of ages between which we would expect to find approximately 95.44% of the total population of police officers in this jurisdiction. We would expect approximately 47.72% to have an age between 13.55 years and 28.375 years. Similarly, we would expect approximately 47.72% to have an age between 28.375 years and 43.195 years.

The numbers that correspond to the **“-3s”** and **“+3s”** are **6.145** and **50.605** respectively. These numbers are calculated by adding three standard deviation units (7.41 years x 3 = 22.23 years) to the mean of 28.375 years and subtracting three standard deviation units (7.41 years x 3 = 22.23 years) from the mean of 28.375 years. This represents the range of ages between which we would expect to find approximately 99.74% of the total population of police officers in this jurisdiction. We would expect approximately 49.87% to have an age between 6.145 and 28.375 years. Similarly, we would expect 49.87% to have an age between 28.375 years and 50.605 years.

**More examples for practice:**

STEP 1: Find the Mean for the distribution.

X

9 Mean= **ΣX/N**

8 = 30/6 = 5

6

4

2

1

** ΣX = 30**

** **

STEP 2: Subtract the Mean from each raw score to get the DEVIATION.

X (X- Mean)-Deviation

9 (9-5) +4

8 (8-5) +3

6 (6-5) +1

4 (4-5) -1

2 (2-5) -3

1 (1-5) -4

STEP 3: Square each deviation before adding the SQUARED DEVIATIONS TOGETHER.

X (X- Mean)-Deviation (X- Mean) ² – Squared Deviation

9 (9-5) +4 16

8 (8-5) +3 9

6 (6-5) +1 1

4 (4-5) -1 1

2 (2-5) -3 9

1 (1-5) -4 16

**Σ****(X- Mean) ² = 52**

** **

**STEP 4: **Divide by N-1 and get the SQUARE ROOT OF THE RESULT FOR THE STANDARD DEVIATION.

**S= √52/5 s= √10.4 s= 3.22**

More examples:

The following data represent the number of crime calls at “Hot Spots” in a year. Calculate and interpret the standard deviation of crime calls at these hot spots.

Hot Spot Number # of Calls (X- Mean)-Deviation (X- Mean) ² -Squared Deviation

1 2 -19.5 380.25

2 9 -12.5 156.25

3 11 -10.5 110.25

4 13 -8.5 72.25

5 20 -1.5 2.25

6 20 -1.5 2.25

7 20 -1.5 2.25

8 24 2.5 6.25

9 27 5.5 30.25

10 29 7.5 56.25

11 31 9.5 90.25

12 52 30.5 930.25

**Σ=258 Σ=0 Σ= 1839 **

Mean = 258/12= 21.5 crime calls

s= **√1839/11 s= √167.181 s= 12.93 crime calls**