Statistical measures.

Measures of Central Tendency

These measures identify the central value or typical value in a dataset.

1. mean
2. median
3. mode

Mean (Arithmetic Mean)

refers to the average of all values in a dataset.

Formula:

$\text{Mean}(\bar{x}) = \frac{\sum x}{n}$

$\text{Where:} \\ \sum x = \text{Sum of all values} \\ n = \text{Number of values}$

Example:

Dataset: 10, 20, 30

$\text{Mean} = \frac{10 + 20 + 30}{3} = \frac{60}{3} = 20$

The mean of ungrouped data

This is obtained by diving the sum of value or scores in the range of individual data by the number of observations. It is the sum of all values in the set divided by the number of observations.

Procedures

The following are procedures for calculating mean of ungrouped data.

1. Find the sum of all data values or observation

2. Divide by the total number of observation

Sample mean $\bar{x}$

$\bar{x} = \frac{\sum_{i=1}^{n} x_i}{n}$

$\text{Where:} \quad \bar{x} = \text{Sample mean}, \quad x_i = \text{Each observation}, \quad n = \text{Sample size}$

$\mu = \frac{\sum_{i=1}^{N} x_i}{N}$

$\text{Where:} \quad \mu = \text{Population mean}, \quad x_i = \text{Each observation}, \quad N = \text{Population size}$

Where;

• n = number of data values in the sample population
• N = number of data values in the population
• Σ = means summation of
• xi = score or observations

Using the notation $\sum_{i=1}^{n} x_i$ we can write:

$\sum_{i=1}^{n} x_i = x_1 + x_2 \ldots + x_n$

Example: 1

1. Study the data given in the Table showing passengers transported by the TAZARA in '000 thousands from 2010 to 2016

   Year	2010	2011	2012	2013	2014	2015	2016
   Passengers in '000 thousands	758	414	678	654	287	327	440

2. Compute the sample mean

   solution

   \bar{x} &= \frac{\sum_{i=1}^{n} x_i}{n} \\ \bar{x} &= \frac{758 + 414 + 678 + 654 + 287 + 327 + 440}{7} = \frac{3558}{7} \\ \bar{x} &= 508.2857 \approx 508 \end{aligned}$$

Therefore, this means that on average; the passengers transported by TAZARA per year were 508 000.

For ungrouped data set with frequencies

The mean of ungrouped data with frequencies can be calculated by using similar procedures.

xi	Frequency (f)
1	3
2	4
3	6
4	2
5	5

Procedures

The following are procedures for calculating mean of ungrouped data with frequency

1. Multiply the data (x) against the respective ( f ) to get (fx).
2. Sum the product (fx)
3. Divide by the total frequency ( **∑**f)

Example

use data in Table for street "A" to calculate the mean of ungrouped

$x$	$f$	$fx$
1	3	3
2	4	8
3	6	18
4	2	8
5	5	25
	$\sum f = 20$	$\sum fx = 62$

Solution

A formula for calculating mean for frequency table is

$\bar{x} = \frac{\sum fx}{\sum f}.$

Where;

• f = frequency
• x = score

$\bar{x} = \frac{\sum fx}{\sum f}.$

Therefore the central value for the score of a given sample is 3.1

Mean of grouped data

This is useful when someone has a big range of data which are supposed to be grouped into classes to facilitate further calculations. Mean in the grouped data can be computed by using two methods namely; assumed mean method and direct method (shortcut method).

Example:

Find the mean of the following population data for town "K".

Class interval	Frequency (f)
0-2	2
3-5	5
6-8	6
9-11	2
12-14	2
15-17	4
18-20	6

Procedures

1. Arrange the data in class interval. For example, 0 - 2, 3 - 5, 6 - 8, 9 - 11, 12 - 14, 15 - 17, 18 - 20.

2. Find the midpoint/classmark of each of the class interval (x).

3. Draw a table with required columns for class interval, frequency (f), midpoint xi and fxi .

   Class interval	Frequency (f)
   0-2	2
   3-5	5
   6-8	6
   9-11	2
   12-14	2
   15-17	4
   18-20	6

4. Calculate the mean of grouped data Mean is given by

   $\bar{x} = \frac{\sum_{i=1}^{n} f x_i}{\sum f}$

   $\sum f x_i = 288, \quad \sum f = 27$

   $\bar{x} = \frac{288}{27} = 10.6666 \approx 10.67$

   Therefore, the central value for the score of a given sample is 10.67

Calculating mean by assumed mean method

When the values or scores under observation are extremely large and in fractions, the use of the direct method becomes inconvenient. As a result the use of the assumed mean method can be done. Assumed mean method is also known as the assumed mean method. The method uses the concept of the arbitrary mean or assumed mean.

Example:

Find the mean of school Y students' scores in Geography using the mean by assumed mean method.

Scores	f
0-10	4
10-20	8
20-30	11
30-40	15
40-50	12
50-60	6
60-70	2

Procedures

1. Find the midpoint/class mark of each class interval (x);
2. Assume one value as mean. In the table, the largest f is on intervals 30- 40, which also happens to be almost in the centre of the distribution, 35 is taken as assumed mean;
3. Subtract the arbitrary mean from each value of midpoint and the resultant value is shown in column d;
4. Multiply each d by respective frequencies (f) to get (fd);
5. Find the algebraic sum of the **∑**fd and divide this sum by ∑f
6. Sometime ∑fd will be positive and sometimes negative;
7. Multiply this value by class mark. This gives the correction factor to be applied to the Assumed Mean; and
8. The assumed Mean + the correction factor = the Actual Mean. Correction factor is computed by **∑**f .

Scores	Mid-point	f	d	fd
0-10	5	4	-30	-120
10-20	15	8	-20	-160
20-30	25	11	-10	-110
30-40	35	15	0	0
40-50	45	12	10	120
50-60	55	6	20	120
60-70	65	2	30	60
		Σf = 58		Σfd = -90

The formula for calculation of the mean by the assumed mean method is given below:

$\bar{x} = \mathbb{A} + \frac{\sum{fd}}{\sum{f}}$

Where:

• A = arbitrary or assumed mean
• f = frequency
• d = deviation from the arbitrary or assumed mean
• **∑**f = number of data values in the population

Assumed mean (A) = 35, the correction factor for the difference between the actual mean and the assumed mean (∑fd)

$= -90 \text{ and } \sum f = 58.$

$\overline{x} = 35 + \frac{(-90)}{58}$

$= 35 + (-1.55)$

$= 33.45$

Therefore, the central scores for the school Y students in Geography is 33.45

Calculating mean by direct method

The same data in Table 1.12 can be used to calculate mean by direct method. For example; use data for school "Y" to calculate mean for grouped data by direct method.

Scores	Mid-point	$f$	$f x_i$
0-10	5	4	20
10-20	15	8	120
20-30	25	11	275
30-40	35	15	525
40-50	45	12	540
50-60	55	6	330
60-70	65	2	130
		$\sum f = 58$	$\sum f x_i = 1940$

The image shows the calculation of the mean ($\bar{x}$) for a grouped frequency distribution. The formula used is:

$\bar{x} = \frac{\sum f x_i}{\sum f}$

Where:

• $\bar{x}$ represents the mean
• $\sum f x_i$ is the sum of the products of each class midpoint ($x_i$) and its frequency ($f$)
• $\sum f$ is the total frequency

From the data:

$\sum f x_i = 1940 \\ \sum f = 58$

Substitute into the formula:

$\bar{x} = \frac{1940}{58}$

Calculate the result:

$\bar{x} \approx 33.45$

Therefore, the central score of the school Y students in Geography is 33.45

Median

refers to the middle value in an ordered dataset.

Steps:

1. Arrange the data in ascending order.
2. If n is odd, the median is the middle value.
3. If n is even, the median is the average of the two middle values.

Example:

Dataset: 10, 20, 30, 40

$\text{Median} = \frac{20+30}{2} = 25$

Median for individual data

Procedures

The following are procedures to be observed in calculating the median for ungrouped data.

1. Arrange data in ascending or descending order.
2. If the total number of items is in odd, add 1 to the total number of observations (n) then divide by 2 to acquire the position of median.
3. The value of the median for ungrouped data is given by the expressions below:

$\text{Median} = \text{Value of the } \left(\frac{n+1}{2}\right)^{th} \text{ term in a ranked data set.}$

Note: the expression is purposely for finding the position of the mid value.

Examples

1. Study the data in Table which shows production of cashew nut '000 tonnes in Tanzania from 2012 to 2016, and find the median.

   Year	2012	2013	2014	2015	2016
   Cashew nut production in tonnes	160	128	130	198	155

   Solution The values of production are arranged in ascending order = 128, 130, 155, 160, 198 n = 5, then

2. $\text{Median Position} = \left( \frac{n + 1}{2} \right)^{\text{th}}$

   $\text{Example: If } n = 5$

   $\left( \frac{5 + 1}{2} \right)^{\text{th}} = \left( \frac{6}{2} \right)^{\text{th}} = 3^{\text{rd}}$

   $\text{Step-by-step:}$

   • Add the numbers in the numerator: $5 + 1 = 6$
   • Divide by the denominator: $\frac{6}{2} = 3$
   • Therefore, the median is at the $3^\text{rd}$ position

   this means the mid value is on the 3rd position of the data set

3. That is the median is 155 ('000) tonnes which mean the mid value of production throughout five years of cashew-nut production.

4. When the total of the range (values) is even, median is obtained by adding the two central values and divide by 2.

Example:

Study the data in Table showing the total monthly rainfall recorded in Tanzania in the year 2016 and find the median.

Month	Jan	Feb	Mar	Apr	May	Jun	Jul	Aug	Sep	Oct	Nov	Dec
Monthly total rainfall (mm)	191.8	131.2	14.0	213.6	41.1	9.2	2.2	8.3	14.1	27.8	64.6	66.6

solution

1. Arrange the value of temperature in ascending order: That is 2.2, 8.3, 9.2, 14.0, 14.1, 27.8, 41.1, 64.6, 66.6, 131.2,191.8, 213.6

2. The position of the mid value is given by the expression

   The image contains a mathematical expression representing the position of the median in an ordered dataset. The expression is:

   This calculation shows that the average of 27.8 mm and 41.1 mm is approximately 34.5 mm.

   $\left( \frac{n + 1}{2} \right)^{\text{th}}$

   Where:

   • n represents the total number of data points in the dataset.
   • The entire expression $\frac{n + 1}{2}$ gives the position of the median in the ordered dataset.
   • The superscript ^{\text{th}} indicates that the result refers to the ordinal (position) of the median.This formula is typically used when dealing with a discrete dataset to find the location of the middle value after the data has been arranged in ascending or descending order.

   Example:

   Given: $\frac{12 + 1}{2}$

   Step-by-step Breakdown:

   • Add the numerator: $12 + 1 = 13$

   • Divide by the denominator: $\frac{13}{2} = 6.5$So, the median lies at the 6.5th position. This means we take the average of the 6th and 7th values in the ordered dataset. Suppose:

   • 6th value = 27.8

   • 7th value = 41.1

   Median Calculation:

   $\frac{27.8 + 41.1}{2} = \frac{68.9}{2} = 34.45 \approx 34.5 \text{ mm}$

   Step-by-step:

   • Add the two middle values: $27.8 + 41.1 = 68.9$
   • Divide by 2: $\frac{68.9}{2} = 34.45$
   • Approximate (if needed): $34.45 \approx 34.5 \text{ mm}$

3. Therefore, the median value for rainfall is 34.5 mm which is the mid value of the rainfall in twelve months.

Median for grouped data

Use the given data in Table which show the distribution of students' scores for school "Y" in Tanzania to calculate median of grouped data.

The following are the procedures for calculating the median for grouped data.

1. Construct the cumulative frequency distribution.

2. Decide the class that contain the median. Median class is the first class with the value of cumulative frequency equal to or at least

   The image shows a mathematical expression:

   $\frac{n}{2}$

   This expression commonly appears in statistical formulas, particularly when determining the position of the median for an even-numbered dataset.

   • $n$ represents the total number of data points.
   • $\frac{n}{2}$ gives the position of the first of the two middle values when the dataset is ordered.
   • The median is then calculated as the average of the values at the $\frac{n}{2}^{\text{th}}$ and $\left(\frac{n}{2} + 1\right)^{\text{th}}$ positions.Without further context, $\frac{n}{2}$ simply represents half the value of $n$.

   Median for a Grouped Frequency Distribution:

   $\text{Median} = L_m + \left( \frac{\frac{n}{2} - cf}{f_m} \right) \cdot i$

   Where:

   • $L_m$ = Lower boundary of the median class
   • $n$ = Total frequency (sum of all frequencies)
   • $cf$ = Cumulative frequency before the median class
   • $f_m$ = Frequency of the median class
   • $i$ = Class width (interval size)This formula is used to estimate the median value for data presented in a grouped frequency table.

Where;

• n = the total frequency
• cf = the cumulative frequency before median class
• f m = the frequency of the median class
• i = the class interval or class width

Note: Lower boundary of each class is calculated by subtracting half of the gap value from the class lower limit. For example, 10 - 0.5 = 9.5. 0.5 is used because the gap value is 1.

Class interval	Frequency (f)	Cumulative (cf)
0 - 4	2	2
5 - 9	6	8
10 - 14	10	18
15 - 19	8	26
20 - 24	4	30

$\text{Median} = 9.5 + \left( \frac{\frac{30}{2} - 8}{10} \right) \times 5$

$= 9.5 + \left( \frac{15 - 8}{10} \right) \times 5$

$= 9.5 + \left( \frac{7}{10} \right) \times 5$

$= 9.5 + 3.5$

$= 13$

Therefore, the mid value for the score is 13.

Mode

refers to the value that appears most frequently in a dataset.

Example:

Dataset: 10, 20, 20, 30

Mode = 20

Mode for the ungrouped data

1. Uni-modal occurs when there is one (1) mode value in a distribution. For example; 4, 5, 6, 6, 7, 8, 9, 10, 10, 10. Mode is 10. Because; it has occurred three times.
2. Bimodal is when set of data has two modes, for example, 1, 2, 3, 3, 4, 4, 4, 5, 5, 6, 6, 6 11. The modes are 4 and 6. This is because; they have both occurred three times.
3. Multi-modal exist when data set has more than two modals, for example; 1, 3, 5, 8, 9, 2, 4, 8, 7, 20, 5, 6, 2, 1, 2, 3, 4, 5, 5, 6, 7, 8, 8, 9, 20, 2. The modes are 2, 5 and 8. This is because; these numbers have more occurrences than others.

Mode for the grouped data

In grouped data, mode can be calculated by the following formula:

Calculation of Mode

$\text{Mode} = L_m + \left( \frac{t_1}{t_1 + t_2} \right) \cdot i$

Note: The modal class is the class interval with the highest frequency.

• $L_m$ is the lower class boundary of the modal class.
• $t_1$ is the frequency of the modal class.
• $t_2$ is the frequency of the class after the modal class.
• $i$ is the class interval.

Where;

= the lower limit of the modal class

• t 1= the frequency of the modal class minus frequency before the modal class. That is the difference 21 between the frequency of modal class and frequency of premodal class.
• t2 = frequency of the modal class minus frequency after the modal class. That is the difference between frequency of modal class and frequency of post modal class.
• i = class width or the size of the class interval.

Example:

Study the Table and then find the mode.

Class interval	Frequency
40 - 44	7
45 - 49	8
50 - 54	11
55 - 59	10
60 - 64	7

Solution

Given

$L_m = 49.5,$

$t_1 = \frac{(11 - 8)}{}= 3$

$t_2 = \frac{(11 - 10)}{}= 1$

$i = 5$

Using the formula for the mode:

$\text{Mode} = 49.5 + \left( \frac{3}{3+1} \right) 5$

$= 49.5 + \frac{3}{4} \times 5$

$= 49.5 + 3.75$

$= 53.25$

Therefore, the maximum frequent denoted for the scores is 53.25

Estimating a mode from a histogram

Mode from the grouped data can be derived from the histogram. For example, using score distribution from Table , a mode can be determined from histogram.

Procedures

The following are procedures for estimating mode from histogram

1. Draw a Histogram;
2. Identify the modal class represented by the tallest bar.
3. Draw the cross lines as shown in the diagram;
4. Draw a perpendicular doted line from the intersection of the two lines until it touches the horizontal axis as shown in Figure below and
5. Read the mode at the intersection of the perpendicular line with the horizontal axis.

Estimating a mode from a histogram

Measures of Dispersion

These measures describe the spread or variability of data.

1. range
2. mean deviation
3. standard deviation

Range

The difference between the highest and lowest values.

Formula:

Range=Highest Value−Lowest Value

• Example: Dataset: 10, 20, 30 Range=30 - 10 = 20

Range for an individual data

Range can be calculated by subtracting the smallest value (S) from the largest value (L) in the distribution . Range is the difference between the maximum value and the minimum value of the data set. Range takes into consideration the extreme values in a data set.

Example:

Study the following distribution of data and calculate the range: 2, 3, 5, 7, 14, 15, 24, and 25.

Solution:

From the given data maximum value is 25 and the minimum value is 2.

The range = Maximum value (L) –minimum value (S) = 25 – 2 = 23

Therefore, the range = 23

The range for the grouped data

This is obtained by using the difference between mid-point of largest and smallest class interval. Consider the data in Table below and find the range.

Class interval	Frequency
15 - 19	2
10 - 14	1
5 - 9	1
0 - 4	2

Solution:

1. Range = Highest upper class limit – Lowest upper class limit 19 − 4 = 15 Range = Highest lower class limit – Lowest lower class limit 15− 0 = 15
2. Range = Highest upper real limit – Lowest upper real limit 19.5− 4.5 = 15 Range = Highest lower real limit – Lowest lower real limit 14.5− (−0.5) = 15 = 15

Therefore a b = , the largest class interval is 15-19 and the smallest class interval is 0 - 4. Hence range is obtained first by computing the midpoint of the largest class interval and that of the smallest class interval as follows:

Midpoint of the largest class interval (L)

The image shows the calculation of the range, likely for a set of data or a distribution. Let's break down the steps:

Calculating a value (L):

$L = \frac{15 + 19}{2} = \frac{34}{2} = 17$

This step calculates the midpoint between the values 15 and 19, resulting in 17.

Calculating the midpoint of the smallest class interval (S):

$S = \frac{0 + 4}{2} = \frac{4}{2} = 2$

The midpoint of the smallest class interval (0 to 4) is calculated as 2.

Calculating the Range:

$\text{Range} = L - S = 17 - 2 = 15$

The range is the difference between the largest and smallest midpoints, which is 15 in this case.

Therefore, based on these calculations, the range of the data or distribution is 15.

Coefficient of range is the ratio of difference between the highest and lowest value of frequency to the sum of the highest and lowest value of frequency. It is a relative measure of the distribution based on the range of any given data set.

The coefficient of range is calculated by the formula: . This is the relative measure.

Whereby;

L is the highest value in the frequency distribution

S is the lowest value in the frequency distribution.

In other words;

The image provides the formula for calculating the coefficient of range:

The coefficient of range is given by:

$\text{The coefficient of range} = \frac{\text{Highest value in frequency} - \text{Lowest value in frequency}}{\text{Highest value in frequency} + \text{Lowest value in frequency}}$

This formula is a measure of dispersion, similar to the range, but it expresses the range relative to the sum of the highest and lowest values. It provides a dimensionless value, making it useful for comparing the dispersion of different datasets, even if they have different scales.

To calculate the coefficient of range, you need to identify the highest and lowest frequencies in the distribution and then substitute these values into the formula.

The coefficient of range is more appropriate for the purposes of comparison.

Example

Study Table below with two sets of data and calculate the coefficient of range separately.

Set 1	8	10	20	9	15	10	13	28
Set 2	30	35	42	50	32	49	39	33

Solution:

As shown in calculation, range of the two sets of data are the same:

Set 1 28 – 8 = 20

Set 2 50 – 30 = 20

Coefficient of range in Set 1 is:

The image shows a calculation resulting in the value 0.55:

$\frac{28 - 8}{28 + 8} = 0.55$

Let's break down the steps involved in this calculation:

1. Calculate the numerator: First, subtract 8 from 28: $28 - 8 = 20$

2. Calculate the denominator: Next, add 8 to 28: $28 + 8 = 36$

3. Perform the division: Finally, divide the result of the numerator by the result of the denominator: $\frac{20}{36}$

4. Simplify and approximate: The fraction $\frac{20}{36}$ can be simplified by dividing both the numerator and the denominator by their greatest common divisor, which is 4: $\frac{20 \div 4}{36 \div 4} = \frac{5}{9}$ Converting this fraction to a decimal gives approximately 0.5555... which is rounded to 0.55 in the image.

Therefore, the calculation $\frac{28 - 8}{28 + 8}$ equals approximately 0.55.

Coefficient of range in Set 2 is:

The image displays a calculation resulting in the value 0.25:

$\frac{50 - 30}{50 + 30} = 0.25$

Let's break down the steps of this calculation:

1. Calculate the numerator: First, subtract 30 from 50: $50 - 30 = 20$

2. Calculate the denominator: Next, add 30 to 50: $50 + 30 = 80$

3. Perform the division: Finally, divide the result of the numerator by the result of the denominator: $\frac{20}{80}$

4. Simplify: The fraction $\frac{20}{80}$ can be simplified by dividing both the numerator and the denominator by their greatest common divisor, which is 20: $\frac{20 \div 20}{80 \div 20} = \frac{1}{4}$

5. Convert to decimal: Converting the fraction $\frac{1}{4}$ to a decimal gives 0.25.

Therefore, the calculation $\frac{50 - 30}{50 + 30}$ equals 0.25.

The coefficient of range tells the degree of dispersion in a set of data. The larger the value, the higher the dispersion.

Mean deviation (MD)

The mean deviation is also known as the average deviation. Mean deviation denotes the amount by which individual value deviates from the mean in irrespective of sign (+ or ̶ ). The negative and positive signs are ignored during computation since the deviations are always equal irrespective of their signs. It is the average difference between various measurements and the mean.

Mean deviation for ungrouped data

The formula for computing the mean deviation is given below.

The image shows the formula for calculating the mean deviation:

$\text{Mean deviation} = \frac{\sum |x_i - \bar{x}|}{n}$

In essence, the mean deviation quantifies the average spread or dispersion of the data around the central tendency (mean).

Where:

The image contains a mathematical expression representing the absolute deviation of a data point from the mean:

$|x_i - \bar{x}|$

• n = the total number of observations
• ∑ = summation
• x = observation or values

Procedures

The following are procedures for calculating mean deviation

1. Compute the mean of the given data set;
2. Find the individual deviation by subtracting the mean from each of the given data set as shown in the table below;
3. Sum the absolute deviation; and
4. Divide by the total number of observations or values

Example:

Compute the mean deviation of the given data of a sample; 2, 3, 4, 5, 6

Solution:

$\bar{x} = \frac{2+3+4+5+6}{5} = \frac{20}{5} = 4$

| xi | x | |xi - x| | | --- | --- | --- | | 2 | 4 | 2 | | 3 | 4 | 1 | | 4 | 4 | 0 | | 5 | 4 | 1 | | 6 | 4 | 2 | | n=5 | | 6 |

Thus, mean deviation (MD)

$\frac{\sum|x_i - \overline{x}|}{n} = \frac{6}{5} = 1.2$

Therefore, the mean deviation is 1.2

Mean deviation for grouped data

This is obtained by using the following formula

$MD = \frac{\sum f_i \left| x_i - \bar{x} \right|}{\sum f_i}$

Example

Study the given scores for form five students in Benbella Girls Secondary School as shown in Table below and find the mean deviation (MD).

Scores	Frequency $f$
20-29	2
30-39	3
40-49	10
50-59	13
60-69	3
70-79	2
80-89	1

Solution

Based on the image, the formula for Mean Deviation (MD) is:

$MD = \frac{\sum f|x_i - \bar{x}|}{\sum f}$

| Scores | f | Class mark (x) | fx | $\bar{x}$ | $|x - \bar{x}|$ | $f|x - \bar{x}|$ | | --- | --- | --- | --- | --- | --- | --- | | 20-29 | 2 | 24.5 | 49 | 50.9 | 26.4 | 52.8 | | 30-39 | 3 | 34.5 | 103.5 | 50.9 | 16.4 | 49.2 | | 40-49 | 10 | 44.5 | 445 | 50.9 | 6.4 | 64 | | 50-59 | 13 | 54.5 | 708.5 | 50.9 | 3.6 | 46.8 | | 60-69 | 3 | 64.5 | 193.5 | 50.9 | 13.6 | 40.8 | | 70-79 | 2 | 74.5 | 149 | 50.9 | 23.6 | 47.2 | | 80-89 | 1 | 84.5 | 84.5 | 50.9 | 33.6 | 33.6 | | Total | Σf = 34 | | Σfx = 1733 | | | $\Sigma f|x - \bar{x}| = 334.4$ |

$\bar{x} = \frac{\sum f x_i}{\sum f} = \frac{1733}{34} = 50.9 \quad \bar{x} = 50.9$

$\text{MD} = \frac{\sum f |x_i - \bar{x}|}{\sum f} = \frac{334.4}{34} = 9.84$

Therefore, the Mean deviation of score is 9.84

Variance

Population variance often denoted by s² is the mean of the squares of the differences between each data value and the mean. For the case of the sample, variance often denoted by S² . Variance is also a measure of the spread between given scores or data set. Measures the distance of each number from the mean and from one number and the other. Therefore, the large variance indicates that scores are far from each other as well. It can be calculated for ungrouped and grouped data.

Procedures

The following are procedures for calculating variance

1. Calculate the mean $\bar{x}$.
2. Find the deviation from the mean $(x - \bar{x})$,
3. Square the deviations from the mean

4. Sum all the square deviations and apply the formula.

Example:

The following distributions of scores; 2,7,3,12,9 can be used to calculate variance.

Solution

Based on the image, the calculation of the mean ($\bar{x}$) is:

The formula is:

$\bar{x} = \frac{\sum x_i}{5}$

From the image, the values are:

• $\sum x_i = 2 + 7 + 3 + 12 + 9$

So, the mean is:

$\bar{x} = \frac{2 + 7 + 3 + 12 + 9}{5} = \frac{33}{5} = 6.6$

x	$\bar{x}$	$(x - \bar{x})$	$(x - \bar{x})^2$
2	6.6	-4.62	21.16
7	6.6	0.4	0.16
3	6.6	-3.6	12.96
12	6.6	5.4	29.16
9	6.6	2.4	5.76
			$\sum_{i=1}^{n} (x - \bar{x})^2 = 69.2$

The formula for computing variance is given below;

$S^2 = \frac{\sum (x - \bar{x})^2}{N}$

Where:

• s²= Variance from sample
• $\bar{x}$.= mean of sample
• n = number in sample

From Table 1.20,

• N = 5
• $\sum (x - \bar{x})^2 = 69.20$

Therefore;

$S^2 = \frac{69.20}{5}$

$S^2 = 13.84$

Standard deviation

It is the value which shows how far the scores are spread from the normal or how the numbers are spread in a distribution. It is the most common index of variability. Standard deviation,(SD) can be calculated for individual or grouped data.

Procedures

Procedures involved in calculation of the standard deviation are:

1. Calculate the mean of distribution;
2. Subtract the mean from each score;
3. Square each of these scores;
4. Add all the squares of these scores;
5. Divide the total by the number of observation;
6. The result obtained in (v) above is called the variance; and
7. Take the square root of the variance; this will give the standard deviation(s).

The formula for computing standard deviation is given as:

$S = \sqrt{\frac{\sum_{i=1}^{n} (x_i - \bar{x})^2}{n}}$

• S = standard deviation from sample.
• x = individual score
• = mean of sample
• n = total number of observations in a sample

Standard deviation for individual data (ungrouped data).

Example:

Determine the standard deviation on scores for class "D" 1, 2, 3, 4, 5, 6, 6, 7, 8, 8

Solution

Score (xi)	$\bar{x}$	$x_i - \bar{x}$	$(x - \bar{x})^2$
8	5	3	9
6	5	1	1
3	5	-2	4
7	5	2	4
2	5	-3	9
8	5	3	9
1	5	-4	16
4	5	-1	1
6	5	1	1
5	5	0	0
---	---	---	---
$\sum_{i=1}^{n} x_i = 50$			$\sum (x_i - \bar{x})^2 = 54$

Mean $(\bar{x}) = \frac{\sum_{i=1}^{n} x_i}{n} = \frac{50}{10} = 5$

Then standard deviation is computed by using the formula

$\text{S} = \sqrt{\frac{\sum_{i=n}^{n}(x_i - \bar{x})^2}{n}} = \sqrt{\frac{54}{10}} = \sqrt{5.4} = 2.32$

Therefore, the standard deviation on score for class 'D' is 2.32.

Standard deviation for grouped data formula

$S = \frac{\sum f(x_i - \bar{x})^2}{\sum f} \text{ or}$

$S = \sqrt{\frac{\sum fx^2}{\sum f} - \left(\frac{\sum fx}{\sum f}\right)^2} \text{ or}$

Example:

Study the Table below which shows scores for Form Five students in school Y and find the standard deviation.

Class interval	f
1 - 10	2
11 - 20	6
21 - 30	4
31 - 40	8
41 - 50	6
51 - 60	4
61 - 70	4
71 - 80	2
81 - 90	3
91 - 100	1

Class Interval	Class-mark (x)	f	fx	𝑥̄	(x - 𝑥̄)	(x - 𝑥̄)²	f(x - 𝑥̄)²
1 – 10	5.5	2	11	43.75	-38.25	1463.06	2926.12
11 – 20	15.5	6	93	43.75	-28.25	798.06	4788.36
21 – 30	25.5	4	102	43.75	-18.25	333.06	1332.24
31 – 40	35.5	8	284	43.75	-8.25	68.06	544.48
41 – 50	45.5	6	273	43.75	1.75	3.06	18.36
51 – 60	55.5	4	222	43.75	11.75	138.06	552.24
61 – 70	65.5	4	262	43.75	21.75	473.06	1892.24
71 – 80	75.5	2	151	43.75	31.75	1008.06	2016.12
81 – 90	85.5	3	256.5	43.75	41.75	1743.06	5229.18
91 – 100	95.5	1	95.5	43.75	51.75	2678.06	2678.06
Total		∑f=40	1750				∑f(x - 𝑥̄)² = 21977.4

$\sum f = 40 \quad \sum fx = 1750$

$\bar{x} = \frac{\sum fx}{\sum f} = \frac{1750}{40} = 43.75$

$\sum f(x - \bar{x})^2 = 21977.4$

From the given formula:

&= \sqrt{\frac{\sum f(x - \bar{x})^2}{\sum fx}} \\ &= \sqrt{\frac{21977.4}{40}} \end{aligned}$$ $$\begin{aligned} &= \sqrt{549.44} \\ &= 23.44 \end{aligned}$$ Therefore, the standard deviation of score for form five students in school Y is 23.44 Alternatively, data in Table below can be used to calculate the standard deviation for grouped data using the formula; $$\text{s} = \sqrt{\frac{\sum fx^2}{\sum f} - \left(\frac{\sum fx}{\sum f}\right)^2}$$ The answer for S will be the same for same data. **Solution** | Class Interval | f | x | x^2 | fx^2 | fx | | --- | --- | --- | --- | --- | --- | | 1 - 10 | 2 | 5.5 | 30.25 | 60.5 | 11 | | 11 - 20 | 6 | 15.5 | 240.25 | 1441.5 | 93 | | 21 - 30 | 4 | 25.5 | 650.25 | 2601 | 102 | | 31 - 40 | 8 | 35.5 | 1260.25 | 10082 | 284 | | 41 - 50 | 6 | 45.5 | 2070.25 | 12421.5 | 273 | | 51 - 60 | 4 | 55.5 | 3080.25 | 12321 | 222 | | 61 - 70 | 4 | 65.5 | 4290.25 | 17161 | 262 | | 71 - 80 | 2 | 75.5 | 5700.25 | 11400.5 | 151 | | 81 - 90 | 3 | 85.5 | 7310.25 | 21930.75 | 256.5 | | 91 - 100 | 1 | 95.5 | 9120.25 | 9120.25 | 95.5 | | Total | 40 | | | 98540 | 1750 | $\sum f = 40 \quad \sum fx^2 = 98540 \quad \sum fx = 1750$ $s = \frac{\sum fx^2}{\sum f} - \left(\frac{\sum fx}{\sum f}\right)^2$ $= \frac{98540}{40} - \left(\frac{1750}{40}\right)^2$ $= 2463.5 - (43.75)^2$ $= 2463.5 - 1914.0625$ $$\begin{aligned} &= \sqrt{549.44} \\ &= 23.44 \end{aligned}$$ Therefore, standard deviation of score for form five students in School Y is 23.44