Chapter 15.5 - Standard Deviation of Continuous Frequency Distribution

In the previous section, we saw some solved examples on variance and standard deviation. In this section, we will see how they are related to continuous frequency distribution.

Calculating variance and standard deviation for continuous frequency distribution

• We know that, when the data is large, we arrange the observations into various groups. It is called continuous frequency distribution.
• We know the method to calculate mean in such cases. Also, in a previous section 15.2, we have seen the method to calculate the following items:
   ♦ Mean deviation about mean for a continuous frequency distribution.
   ♦ Mean deviation about median for a continuous frequency distribution.
• So now we will see the method to calculate the following items:
   ♦ Variance for a continuous frequency distribution.
   ♦ Standard deviation for a continuous frequency distribution.
• The method is similar to what we saw in the solved examples 15.9 and 15.10 of the previous section. There we dealt with discrete values. But here we will be dealing with groups of values. These groups are called class intervals.
• So for each class interval, we choose a value to represent that class interval. Usually we choose the midpoint as the representative value.
• For example, if the class interval is 30-35, then the representative value will be the midpoint, which is:
$\frac{30 + 35}{2}~=~\frac{65}{2}~=~32.5$
• Once the representative value is fixed, the procedure is the same as before.    
• The following solved example will demonstrate the method:

Solved Example 15.12
Find the variance and standard deviation for the following data:

Table15.24

Solution:
1. We are asked to find the variance.
• So our first aim is to find the mean $(\bar{x})$. For that, we want $f_i x_i$.
• For calculating $f_i x_i$, we want $x_i$.
• $x_i$ values are the midpoint values. They are calculated in the third column of the table 15.25 below:

Table 15.25

• $f_i x_i$ is calculated in the column IV. So we can write:
$\bar{x}~=~\frac{\sum{f_i x_i}}{\sum{f_i}}~=~\frac{3100}{50}~=~62$
2. Now we can calculate the variance. For that, we can use columns II, V and VI.
• We have:
$\sigma^2~=~\frac{\sum{f_i (x_i - \bar{x})^2}}{\sum{f_i}}~=~\frac{10050}{50}~=~201$
3. Finally, we can calculate the standard deviation.
$\sigma~=~\sqrt{\sigma^2}~=~\sqrt{201}~=~14.18$

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Shortcut method for calculating variance and standard deviation

• In a previous section, we used a shortcut method (step-deviation method) for finding $\bar{x}$.
• If $\bar{x}$ obtained is large, our present calculations of variance will be tedious and lengthy.
• So we need a method to reduce the size of $\bar{x}$. It can be written in 5 steps:

1. In the step-deviation method, we first reduce $x_i$ into $d_i$.
• $d_i = x_i - a$
   ♦ Where 'a' is the assumed mean.
2. Next we reduce $d_i$ to $u_i$.
• $u_i = \frac{d_i}{h}$
   ♦ Where 'h' is the width of class.
3. We can combine the above two results as follows:
$u_i = \frac{x_i - a}{h}$
• From this, we get:
$x_i = a + h u_i$
4. Once we calculate the $u_i$ values, we find $\bar{u}$.
• Using $\bar{u}$, we find $\bar{x}$ using the formula:
$\bar{x} = a + h \bar{u}$
5. Let us use the above results in (3) and (4), to modify variance. It can be done as follows:

$\begin{array}{ll} {}&{\sigma^2} & {~=~}& {\frac{1}{\sum{f_i}}\left[\sum{f_i \left[x_i - \bar{x} \right]^2} \right]} &{} \\ {}&{} & {~=~}& {\frac{1}{\sum{f_i}}\left[\sum{f_i \left[(a + h u_i) - (a + h \bar{u}) \right]^2} \right]~\color{green}{\text{- - - I}}} &{} \\ {}&{} & {~=~}& {\frac{1}{\sum{f_i}}\left[\sum{f_i \left[a + h u_i - a - h \bar{u} \right]^2} \right]} &{} \\ {}&{} & {~=~}& {\frac{1}{\sum{f_i}}\left[\sum{f_i \left[h u_i - h \bar{u} \right]^2} \right]} &{} \\ {}&{} & {~=~}& {\frac{1}{\sum{f_i}}\left[\sum{f_i \left[h( u_i - \bar{u}) \right]^2} \right]} &{} \\ {}&{} & {~=~}& {\frac{1}{\sum{f_i}}\left[\sum{f_i h^2 ( u_i - \bar{u})^2} \right]} &{} \\ {}&{} & {~=~}& {\frac{h^2}{\sum{f_i}}\left[\sum{f_i (u_i - \bar{u})^2} \right]} &{} \\ \end{array}$

◼ Remarks:
• Line marked as I:
In this line, we replace $x_i$ and $\bar{x}$ using the results in (3) and (4).

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• So now we have three formulas for calculating variance:
(i) $\sigma^2~=~\frac{\sum{f_i (x_i - \bar{x})^2}}{\sum{f_i}}$

(ii) $\sigma^2~=~\left(\frac{1}{\sum{f_i}}\right)^2 \left[\left(\sum{f_i}\right) \left(\sum{f_i x^2 _i} \right)  - \left(\sum{f_i x_i} \right)^2 \right]$

(iii) $\sigma^2~=~\frac{h^2}{\sum{f_i}}\left[\sum{f_i (u_i - \bar{u})^2} \right]$

Note:
Formula (i) is the basic formula. Formula (iii) is used for the shortcut method. Both (i) and (iii) follow the same pattern.

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Let us apply the shortcut method for solved example 15.12 above. It will be our next solved example.

Solved example 15.13
Find the variance and standard deviation for the data in table 15.24 above.
Solution:
1. We are asked to find the variance.
• So our first aim is to find the mean $(\bar{x})$. For that, we want $f_i x_i$.
• For calculating $f_i x_i$, we want $x_i$.
• $x_i$ values are the midpoint values. They are calculated in the third column of the table 15.26 below:

Table 15.26

• Now we reduce the sizes of all $x_i$ values. For that, we consider a middle value as the assumed mean. For our present case, we consider 65 as the assumed mean (a). Then we subtract 'a' from all $x_i$ values. The values obtained after subtraction are denoted as $d_i$. This is calculated in column IV.
• Next, we reduce the size of $d_i$. This is achieved by dividing with a common factor (h). For our present case, 10 is the common factor. It is the class width. The values obtained after division are denoted as $u_i$. This is calculated in column V. 
• Finally, $f_i u_i$ is calculated in the column VI. So we can write:
$\bar{u}~=~\frac{\sum{f_i u_i}}{\sum{f_i}}~=~\frac{-15}{50}~=~-0.3$
2. Now we can calculate ($u_i - \bar{u}$). It is done in column VII.
• Using ($u_i - \bar{u}$), we can calculate $f_i(u_i - \bar{u})^2$. It is done in column VIII.
3. Now we can substitute the values.
• We have:
$\sigma^2~=~\frac{h^2}{\sum{f_i}}\left[\sum{f_i (u_i - \bar{u})^2} \right]~=~\frac{10^2}{50}\left[100.5 \right]~=~201$
4. Finally, we can calculate the standard deviation.
$\sigma~=~\sqrt{\sigma^2}~=~\sqrt{201}~=~14.18$

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Link to a few more solved examples is given below:

Exercise 15.2

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In the next section, we will see Analysis of frequency distributions.

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Chapter 15.4 - Solved Examples on Standard Deviation

In the previous section, we saw variance and standard deviation. In this section, we will see some solved examples.

Solved example 15.9
Find the variance and standard deviation of the following data:
6,8,10,12,14,16,18,20,22,24
Solution:
1. We are asked to find the variance.
• So our first aim is to find the mean $(\bar{x})$. For that, we can use the columns I, II  and III of table 15.19 below:

Table 15.19

• Based on those columns, we can write:
$\bar{x}~=~\frac{\sum{f_i x_i}}{\sum{f_i}}~=~\frac{150}{10}~=~15$
2. Now we can calculate the variance. For that, we can use columns II, IV and V of the table.
• We have:
$\sigma^2~=~\frac{\sum{(x_i - \bar{x})^2}}{\sum{f_i}}~=~\frac{330}{10}~=~33$
3. Finally, we can calculate the standard deviation.
$\sigma~=~\sqrt{\sigma^2}~=~\sqrt{33}~=~5.74$

Solved example 15.10
Find the variance and standard deviation of the following data:

Table 15.20

Solution:
1. We are asked to find the variance.
• So our first aim is to find the mean $(\bar{x})$. For that, we can use the columns I, II  and III of table 15.21 below:

Table 15.21

• Based on those columns, we can write:
$\bar{x}~=~\frac{\sum{f_i x_i}}{\sum{f_i}}~=~\frac{420}{30}~=~14$
2. Now we can calculate the variance. For that, we can use columns II, IV and V of the table.
• We have:
$\sigma^2~=~\frac{\sum{f_i (x_i - \bar{x})^2}}{\sum{f_i}}~=~\frac{1374}{30}~=~45.8$
3. Finally, we can calculate the standard deviation.
$\sigma~=~\sqrt{\sigma^2}~=~\sqrt{45.8}~=~6.77$
4. Note:
In this problem, we have to take frequencies also into account. This can be explained using an example. It can be written in 3 steps:
(i) Frequency of the observation 11 is nine.
• That means, the observation 11 occurs nine times.
(ii) So we have to include "$(11 - \bar{x})^2$" nine times.
• Instead of writing nine times, we can directly calculate "$9 \times (11 - \bar{x})^2$"
(iii) So we calculate "$f_i \times (x_i - \bar{x})^2$" in the column V.

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Another formula for standard deviation

This can be written in 2 steps:
1. We can derive a new formula for variance as follows:

$\begin{array}{ll} {}&{\sigma^2} & {~=~}& {\frac{1}{\sum{f_i}}\left[\sum{f_i \left(x_i - \bar{x} \right)^2} \right]} &{} \\ {}&{} & {~=~}& {\frac{1}{\sum{f_i}}\left[\sum{f_i \left(x^2 _i + \bar{x}^2 - 2 x_i \bar{x} \right)} \right]} &{} \\ {}&{} & {~=~}& {\frac{1}{\sum{f_i}}\left[\sum{f_i x^2 _i} + \sum{f_i \bar{x}^2} - \sum{2 f_i x_i \bar{x} } \right]} &{} \\ {}&{} & {~=~}& {\frac{1}{\sum{f_i}}\left[\sum{f_i x^2 _i} + \bar{x}^2 \sum{f_i } - 2 \bar{x} \sum{f_i x_i } \right]~\color{green}{\text{- - - I}}} &{} \\ {}&{} & {~=~}& {\frac{1}{\sum{f_i}}\left(\sum{f_i x^2 _i} \right) + \frac{1}{\sum{f_i}} \left(\bar{x}^2 \sum{f_i } \right) - \frac{1}{\sum{f_i}} \left(2 \bar{x} \sum{f_i x_i } \right)} &{} \\ {}&{} & {~=~}& {\frac{1}{\sum{f_i}}\left(\sum{f_i x^2 _i} \right) + \left(\bar{x}^2 \right) - \left(2 \bar{x}^2 \right)~\color{green}{\text{- - - II}}} &{} \\ {}&{} & {~=~}& {\frac{1}{\sum{f_i}}\left(\sum{f_i x^2 _i} \right) - \left(\bar{x}^2 \right)} &{} \\ {}&{} & {~=~}& {\frac{1}{\sum{f_i}}\left(\sum{f_i x^2 _i} \right) - \left[\frac{1}{\sum{f_i}}\left(\sum{f_i x_i} \right) \right]^2 ~\color{green}{\text{- - - III}}} &{} \\ {}&{} & {~=~}& {\frac{1}{\sum{f_i}}\left(\sum{f_i x^2 _i} \right) - \left[\left(\frac{1}{\sum{f_i}}\right)^2\left(\sum{f_i x_i} \right)^2 \right]} &{} \\ {}&{} & {~=~}& {\left(\frac{1}{\sum{f_i}}\right)^2 \left[\left(\sum{f_i}\right) \left(\sum{f_i x^2 _i} \right) - \left(\sum{f_i x_i} \right)^2 \right]} &{} \\ \end{array}$

◼ Remarks:
(i) Line marked as I
• $\bar{x}$ is a constant. So we take it outside summations.
(ii) Line marked as II
• In the third term, $\frac{\sum{f_i x_i}}{\sum{f_i}}$ is $\bar{x}$
(iii) Line marked as III
• In this line, we write $\frac{\sum{f_i x_i}}{\sum{f_i}}$ in the place of $\bar{x}$.

2. Once we obtain the formula for variance, we can easily write the formula for standard deviation.

$\sigma~=~\sqrt{\left(\frac{1}{\sum{f_i}}\right)^2 \left[\left(\sum{f_i}\right) \left(\sum{f_i x^2 _i} \right) - \left(\sum{f_i x_i} \right)^2 \right]}$ 

⇒ $\sigma~=~\frac{1}{\sum{f_i}}\sqrt{\left(\sum{f_i}\right) \left(\sum{f_i x^2 _i} \right) - \left(\sum{f_i x_i} \right)^2}$

Solved example 15.11
Find the variance and standard deviation of the following data:

Table15.22

Solution:
1. For calculating variance, we have:
$\sigma^2~=~\left(\frac{1}{\sum{f_i}}\right)^2 \left[\left(\sum{f_i}\right) \left(\sum{f_i x^2 _i} \right) - \left(\sum{f_i x_i} \right)^2 \right]$
2. For calculating standard deviation, we have:
$\sigma~=~\frac{1}{\sum{f_i}}\sqrt{\left(\sum{f_i}\right) \left(\sum{f_i x^2 _i} \right) - \left(\sum{f_i x_i} \right)^2}$
3. For applying the formulas, we need the following items:
(i) $\sum{f_i}$
• This is calculated in column II of table 15.23 below:

Table 15.23

(ii) $\sum{f_i x_i}$
• This is calculated in column III of table 15.23.
(iii) $\sum{f_i (x_i)^2}$
• This is calculated in column V of table 15.23.
4. Substituting the known values in (1), we get:
$\sigma^2~=~\left(\frac{1}{48}\right)^2 \left[\left(48\right) \left(9652 \right) - \left(614 \right)^2 \right]~=~37.456$
5. So we can write:
$\sigma~=~\sqrt{37.456}~=~6.12$

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In the next section, we will see variance and standard deviation for continuous frequency distribution.

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Chapter 15.3 - Standard Deviation

In the previous section, we completed a discussion on mean deviation. In this section, we will see standard deviation.

• In the previous sections,we have studied two items:
   ♦ Mean deviation about Mean.
   ♦ Mean deviation about Median.
• Both have some limitations. Those limitations can be explained in 4 steps:
1. Consider the mean deviation about mean.
• We take the absolute values of the deviations. So the -ve signs are ignored.
• When a result is obtained by ignoring the signs, that result cannot be used for any further algebraic calculations.
2. Consider the mean deviation about median.
• Here also, we take the absolute values of the deviations. So the -ve signs are ignored.
• So this result is also not helpful for any further algebraic calculations.
3. Consider the two sums:
   ♦ Sum of absolute deviations about mean
   ♦ Sum of absolute deviations about median
• For any problem that we consider, the first sum mentioned above, will be greater than the second sum.
• But we divide both sums by the same quantity, which is “total number of observations”. So a confusion arises, as to which sum gives the accurate picture.
4. Consider the two series below:
   ♦ 35, 40, 47, 50, 54, 60, 65, 68, 72, 79
   ♦ 5, 8, 47, 50, 54, 60, 65, 68, 72, 79
• In the first series, the observations are more or less evenly distributed about the median.
• In the second series, the median will not represent the true nature because, two extreme values (5 and 8) are present. In such a series, if we use the deviations about median, we will get inaccurate results.

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• Based on the limitations mentioned above, we can write:
An improved method is necessary to measure dispersion.
• Standard deviation is such an improved method, which is widely used in science, engineering, economics, business studies and sociology.

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Variance and Standard deviation

This can be explained in 22 steps:
1. Recall that, for calculating mean deviation, we used the absolute values of the deviations. This was necessary to prevent the cancellation between +ve and -ve values.
• By taking the absolute values, we could ensure that only non-negative values come into the calculations.
2. There is another method to ensure the presence of only non-negative values:
Taking squares of the deviations.
3. Consider a series with n observations.
   ♦ Let $x_1, x_2, x_3, x_4,~.~.~.~, x_n$ be those n observations.
   ♦ Let $\bar{x}$ be the mean of those n observations.
• Then the deviations will be:
$(x_1 - \bar{x}), (x_2 - \bar{x}), (x_3 - \bar{x}), (x_4 - \bar{x}),~.~.~.~, (x_n - \bar{x})$
• So the squares of the deviations will be:
$(x_1 - \bar{x})^2, (x_2 - \bar{x})^2, (x_3 - \bar{x})^2, (x_4 - \bar{x})^2,~.~.~.~, (x_n - \bar{x})^2$
4. Now we can write the sum of those squares:
$(x_1 - \bar{x})^2~+~(x_2 - \bar{x})^2~+~(x_3 - \bar{x})^2~+~(x_4 - \bar{x})^2~+~.~.~.~+~ (x_n - \bar{x})^2~=~\sum{(x_i - \bar{x})^2}$
5. Consider the sum in the above step (4).
• No term in the L.H.S can become -ve. This is because, all terms are squares.
• Each term in the L.H.S will be either zero or +ve.
• Then, if the R.H.S is zero, it means that each term in the L.H.S is zero.
6. Suppose that, the R.H.S is zero.
Then we can write:

$\begin{array}{ll} {\text{(i)}}&{(x_1 - \bar{x})^2} & {~=~}& {0} &{} \\ {\Rightarrow}&{x_1 - \bar{x}} & {~=~}& {0} &{} \\ {\Rightarrow}&{x_1} & {~=~}& {\bar{x}} &{} \\ {\text{(ii)}}&{(x_2 - \bar{x})^2} & {~=~}& {0} &{} \\ {\Rightarrow}&{x_2 - \bar{x}} & {~=~}& {0} &{} \\ {\Rightarrow}&{x_2} & {~=~}& {\bar{x}} &{} \\ {\text{(iii)}}&{(x_3 - \bar{x})^2} & {~=~}& {0} &{} \\ {\Rightarrow}&{x_3 - \bar{x}} & {~=~}& {0} &{} \\ {\Rightarrow}&{x_3} & {~=~}& {\bar{x}} &{} \\ \end{array}$

- - -
- - -
so on . . .

7. Based on the above step (6), we can write:
• If the sum of the squares of the deviations is zero, then each observation will be equal to $\bar{x}$
   ♦ That means, every observation is the same.
   ♦ That means, there is no dispersion.
• So we can write:
If the sum of the squares of the deviations is zero, then the series has no dispersion.
8. Consider again, the result in (4).
• If the deviations are small, then sum of the squares of the deviations will be small.
• If the deviations are small, then the dispersion of the series is said to be small.
• So we can write:
If the sum of the squares of the deviations is small, then the series has small dispersion.
9. Consider again, the result in (4).
• If the deviations are large, then sum of the squares of the deviations will be large.
• If the deviations are large, then the dispersion of the series is said to be large.
• So we can write:
If the sum of the squares of the deviations is large, then the series has large dispersion.
10. Let us compile the results in the above steps (7), (8) and (9):
(i) If the sum of squares is zero, then there is no dispersion.
(ii) If the sum of squares is small, then the dispersion is small.
(iii) If the sum of squares is large, then the dispersion is large.
11. Based on the above step (10), we are tempted to conclude that:
Sum of squares of the deviations is a good measure of dispersion.
• But before making such a conclusion, we will look at some examples.
12. As our first example, let us consider the following series of 6 observations:
5, 15, 25, 35, 45, 55
• The table 15.17 below shows the calculations:

Table 15.17

• We can write:
   ♦ $\bar{x}~=~\frac{\sum{f_i x_i}}{\sum{f_i}}~=~\frac{180}{6}~=~30$
   ♦ $\sum{(x_i - 30)^2~=~1750}$
13. As our second example, let us consider the following series of 31 observations:
15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45
• The table 15.18 below shows the calculations:

Table 15.18

• We can write:
   ♦ $\bar{x}~=~\frac{\sum{f_i x_i}}{\sum{f_i}}~=~\frac{930}{31}~=~30$
   ♦ $\sum{(x_i - 30)^2~=~2480}$

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Note:
• The sum of all items in the column VI is:
(-15)² + (15)² + (-14)² + (14)² + (-13)² + (13)² + . . . + (-1)² + (1)² +
• This can be written as:
2[15² + 14² + 13² + . . .  + (1)²]
• In the above result, the portion inside square brackets is:
Sum of the squares of first n natural numbers, where n = 15.
• We have a formula for calculating that sum (details here):
$\sum\limits_{k\,=\,1}^{k\,=\,n}{k^2}~=~\frac{n(2n + 1) (n+1)}{6}$
• Substituting n = 15, we get:
$\sum\limits_{k\,=\,1}^{k\,=\,15}{k^2}~=~\frac{15(30 + 1) (15+1)}{6}~=~\frac{15 × 31  × 16}{6}~=~1240$
• So the sum in column VI = (2 × 1240) = 2480

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14. Comparing the results in steps (12) and (13), and based on the knowledge that we acquired so far, we are forced to write this:
Sum of squares is larger for the second example. So the series in the second example has greater dispersion.
15. However, let us plot both the series on the same graph. It is shown in fig.15.3 below:

Fig.15.3

• The green circles denote the six observations in the first example.
• The red diamonds denote the thirty one observations in the second example.
• We see that:
   ♦ The mean in both cases is 30.
   ♦ The six observations of example 1, are dispersed along a wider length of the number line.
   ♦ The thirty one observations of example 2, are dispersed along a narrower length of the number line.
• So we can write:
Example 2 has a lesser dispersion.
16. It is clear that, what we wrote in step (14) is wrong.
◼ How did it become wrong?
• We can write a brief answer in 3 steps:
(i) In the second example, individual deviations are small.
(ii) But compared to the first example, there are a large number of observations.
(iii) So the sum of squares became larger.
17. We must not use sum of squares to measure dispersion.
• Instead, we must modify it in accordance to the number of observations.
• For applying the modification, we must divide the sum by the number of observations ($\sum{f_i}$).
• When we divide the sum by $\sum{f_i}$, we get this:
   ♦ Mean
   ♦ of the
   ♦ squares
   ♦ of the
   ♦ deviations.
18. This mean is called variance. It is denoted by $\sigma^2$. It is read as sigma square.
• We can write:
$\text{Variance}~(\sigma^2) ~=~ \frac{\sum{(x_i- \bar{x})^2}}{\sum{f_i}}$
19. Let us calculate the variance for our examples:
• The variance of our first example will be:
$\frac{1750}{6}~=~291.67$
• The variance of our second example will be:
$\frac{2480}{31}~=~80$
20. We see that, the second example has a smaller variance.
• Based on this, we can write:
Data in the second example has a smaller dispersion.
21. While dealing with variance, we encounter a difficulty related to units. It can be explained using some examples.
Example 1:
This can be written in 2 steps:
(i) Suppose that the observations in a series, are the heights of students in a class.
(ii) Then we will be using two units:
   ♦ Each observation ($x_i$) will be expressed in cm.
   ♦ The mean ($\bar{x}$) will also be expressed in cm.
   ♦ The variance ($\sigma^2$) will be expressed in cm²
Example 2:
This can be written in 2 steps:
(i) Suppose that the observations in a series, are the weights of fruits from a field.
(ii) Then we will be using two units:
   ♦ Each observation ($x_i$) will be expressed in gm.
   ♦ The mean ($\bar{x}$) will also be expressed in gm.
   ♦ The variance ($\sigma^2$) will be expressed in gm²
22. In order to avoid the usage of a unit and it's square, we take the square root of the variance.
• Remember that, there will be a +ve square root and a -ve square root. We must discard the -ve square root.
• The +ve square root of the variance is called standard deviation. It is denoted by $\sigma$.
• So we can write:
$\sigma~=~\sqrt{\frac{\sum{(x_i- \bar{x})^2}}{\sum{f_i}}}$

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In the next section, we will see some solved examples.

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Chapter 15.2 - Mean Deviation in the case of Continuous Frequency Distribution

In the previous section, we saw the basics about mean deviation. We saw some solved examples also. In this section, we will see a few more solved examples. Later in this section, we will see mean deviation in the case of continuous frequency distribution.

Solved Example 15.4
Find the mean deviation about the mean for the following data:

Table 15.8

Solution:
1. We are asked to find the mean deviation about the mean.
• So our first aim is to find the mean $(\bar{x})$. For that, we can use the columns I, II and III of table 15.9 below:

Table 15.9

• Based on the table, we can write:
$\bar{x}~=~\frac{\sum{f_i x_i}}{\sum{f_i}}~=~\frac{300}{40}~=~7.5$
2. Now we can calculate the mean deviation. For that, we can use columns II, IV and V of the table.

• We have:
$\text{M.D.}(\bar{x})~=~\frac{\sum{(f_i |x_i - \bar{x}|)}}{\sum{f_i}}$
• Substituting the values, we get:
$\text{M.D.}(\bar{x})~=~\frac{92}{40}~=~2.3$

Solved Example 15.5
Find the mean deviation about the median for the following data:

Table 15.10

Solution:
1. We are asked to find the mean deviation about the median.
• So our first aim is to find the median (M). For that, we can use the columns I, II  and III of the table 15.11 below. The observations are arranged in ascending order.

Table 15.11

• Based on the table, we can write:
Total number of observations = n = ∑f_i = 30
• "30" is an even number. So the median will be the average of the values at:
$\frac{n}{2}~\text{and}~\left(\frac{n}{2} + 1 \right)$
   ♦ That is., the average of the values at:
$\frac{30}{2}~\text{and}~\left(\frac{30}{2} + 1 \right)$
   ♦ That is., the average of the values at:
15 and 16,
• From the column for cumulative frequency, we see that:
   ♦ the value at the 15^th position is 13.
   ♦ the value at the 16^th position is also 13.
• So we get: M = average of 13 and 13 = 13 
2. Now we can calculate the mean deviation. For that, we can use columns II, IV and V of the table.

• We have:
$\text{M.D.(M)}~=~\frac{\sum{(f_i |x_i - M|)}}{\sum{f_i}}$
• Substituting the values, we get:
$\text{M.D.(M)}~=~\frac{149}{30}~=~4.97$

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Calculating mean deviation about mean for continuous frequency distribution

• We know that, when the data is large, we arrange the observations into various groups. It is called continuous frequency distribution.
• We have seen the method to calculate mean and median in such cases. So now we will see the method to calculate the following items:
   ♦ mean deviation about mean for a continuous frequency distribution.
   ♦ mean deviation about median for a continuous frequency distribution.
• The method is similar to what we saw in the previous section. There we dealt with discrete values. But here we will be dealing with groups of values. These groups are called class intervals.
• So for each class interval, we choose a value to represent that class interval. Usually we choose the midpoint as the representative value.
• For example, if the class interval is 30-35, then the representative value will be the midpoint, which is:
$\frac{30 + 35}{2}~=~\frac{65}{2}~=~32.5$
• Once the representative value is fixed, the procedure is the same as before.    
• The following solved example will demonstrate the method:

Solved Example 15.6
Find the mean deviation about the mean for the following data:

Table 15.12

Solution:
1. We are asked to find the mean deviation about the mean.
• So our first aim is to find the mean $(\bar{x})$. For that, we want $f_i x_i$.
• For calculating $f_i x_i$, we want $x_i$.
• $x_i$ values are the midpoint values. They are calculated in the third column of the table 15.13 below:

Table 15.13

• $f_i x_i$ is calculated in the column IV. So we can write:
$\bar{x}~=~\frac{\sum{f_i x_i}}{\sum{f_i}}~=~\frac{1800}{40}~=~45$
2. Now we can calculate the mean deviation. For that, we can use columns II, V and VI.

• We have:
$\text{M.D.}(\bar{x})~=~\frac{\sum{(f_i |x_i - \bar{x}|)}}{\sum{f_i}}$
• Substituting the values, we get:
$\text{M.D.}(\bar{x})~=~\frac{400}{40}~=~10$

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Shortcut method for calculating mean deviation about $\bar{x}$

• In our earlier classes, we have seen a shortcut method to find $\bar{x}$. It is called assumed mean method (details here).
• We have also seen the modification of that method. It is called step-deviation method (details here). Using that method, we calculated $\bar{x}$ very easily.
    ♦ If $\bar{x}$ can be calculated very easily,
    ♦ it means that,
    ♦ mean deviation about $\bar{x}$ can also be calculated very easily.
• So let us apply the step-deviation method to the solved example 6 that we saw above. It will be our next solved example:

Solved Example 15.7
Find the mean deviation about the mean for the data in table 15.12.
Solution:
1. We are asked to find the mean deviation about the mean.
• So our first aim is to find the mean $(\bar{x})$. For that, we want $f_i x_i$.
• For calculating $f_i x_i$, we want $x_i$.
• $x_i$ values are the midpoint values. They are calculated in the third column of the table 15.14 below:

Table 15.14

• Now we reduce the sizes of all $x_i$ values. For that, we consider a middle value as the assumed mean. For our present case, we consider 45 as the assumed mean (a). Then we subtract 'a' from all $x_i$ values. The values obtained after subtraction are denoted as $d_i$. This is calculated in column IV.
• Next, we reduce the size of $d_i$. This is achieved by dividing with a common factor (h). For our present case, 10 is the common factor. The values obtained after division are denoted as $u_i$. This is calculated in column V. 
• Finally, $f_i u_i$ is calculated in the column VI. So we can write:
$\bar{u}~=~\frac{\sum{f_i u_i}}{\sum{f_i}}~=~\frac{0}{40}~=~0$
• Using $\bar{u}$, we can calculate $\bar{x}$.
$\bar{x} = a + h \bar{u} = (45 + 10 \times 0) = 45$
2. Now we can calculate the mean deviation. For that, we can use columns II, VII and VIII.

• We have:
$\text{M.D.}(\bar{x})~=~\frac{\sum{(f_i |x_i - \bar{x}|)}}{\sum{f_i}}$
• Substituting the values, we get:
$\text{M.D.}(\bar{x})~=~\frac{400}{40}~=~10$

---

While using the step-deviation method for calculating the “mean deviation about the mean”, we must keep an important fact in our minds. It can be written in 2 steps:
1. Calculation of the “mean deviation about the mean” involves two parts.
• In the first part, we calculate the mean ($\bar{x}$).
• In the second part, we calculate the mean deviation about that mean.
2. The step-deviation is a shortcut method for the first part only.
• For the second part, we need to follow the usual procedure.

---

Calculating mean deviation about median for continuous frequency distribution

• We know how to find the median (M) of a continuous frequency distribution (details here).
• Once we find the M, we can find the required mean deviation as usual.
• The following solved example demonstrates the procedure.

Solved Example 15.8
Find the mean deviation about the median for the following data:

Table 15.15

Solution:
1. We are asked to find the mean deviation about the median.
• So our first aim is to find the median (M). It can be done in steps:
(i) In the column II of table 15.16 below, we see that, $\sum{f_i}$ is 50. That means, there are a total of 50 observations. So we can write: n = 50.
(ii) '50' is an even number. So the median will be the average of the following two values:
    ♦ The value at the $\frac{n}{2}$ position.
    ♦ The value at the $\frac{n}{2} + 1$ position.
(iii) Let us calculate those positions:
    ♦ $\frac{n}{2}~=~\frac{50}{2}~=~25$
    ♦ $\frac{n}{2} + 1~=~(25 + 1)~=~26$
(iv) In the table 15.16 below, consider the classes with cumulative frequencies 13 and 28.

Table 15.16

• It is clear that, positions 25 and 26 occur inside the class 20-30.
• So 20-30 is the median class.
(v) But we do not know the actual values inside the classes. So we do not know the exact values at 25 and 26 positions. To find the median in such a situation , we use the formula:
$M~=~l + \left[\frac{\frac{n}{2}~-~cf}{f} \right]h$

• Where:
    ♦ l = lower limit of the median class
    ♦ n = number of observations
    ♦ cf = cumulative frequency of the class preceding the median class
    ♦ f = frequency of the median class
    ♦ h = width of the class interval (assuming all classes are of the same width)
(vi) In our present case:
    ♦ The median class is 20-30. So l = 20
    ♦ n = total number of observations = 50
    ♦ cf = cumulative frequency of class 10-20 = 13
    ♦ f = frequency of class 20-30 = 15
    ♦ h = width of class intervals = 10
• Substituting these values, we get:
$M~=~20 + \left[\frac{\frac{50}{2}~-~13}{15} \right]10 ~=~28$
2. Now we can calculate the mean deviation. For that, we can use columns II, IV, V and VI of the table.

• We have:
$\text{M.D.(M)}~=~\frac{\sum{(f_i |x_i - M|)}}{\sum{f_i}}$
• Substituting the values, we get:
$\text{M.D.(M)}~=~\frac{508}{50}~=~10.16$

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Link to a few more solved examples is given below:

Exercise 15.1

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In the next section, we will see Variance and Standard deviation.

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Chapter 15.1 - Mean Deviation

In the previous section, we saw range and it's limitations. In this section, we will see mean deviation.

Mean deviation

This can be written in steps:
1. Mean deviation is: “mean (average) of the deviations”.
   ♦ So first we have to calculate the deviations.
   ♦ Then we calculate the mean of those deviations.
2. How do we calculate the deviations?
The answer can be written in 3 steps:
(i) First write down the value of the central tendency.
• The central tendency to be used, can be either mean or median.
• That central tendency number is represented by the letter ‘a’.
(ii) The values in the data can be represented using the letter ‘x’
• Pick the first value (x₁) in the data. Calculate (x₁ - a) for that value.
• This (x₁ - a) is the deviation of the first value.
(iii) In this way, we must calculate the deviation for all values in the data.    
3. Once all the deviations are calculated, we must find the sum of those deviations.
• This sum, when divided by the number of observations (n = ∑f_i) will give the mean deviation.
4. Let us calculate the mean deviation for the batsmen A  and B.
• Let us use the mean as the central tendency. So a = 53 for both A and B.
   ♦ We have to calculate (x-a), which is (x – 53).
   ♦ We have to calculate this (x-53) for all values.
• This is shown in table 15.4 (a) and (b) below:

5. Now we encounter a problem. It can be explained in two steps:
(i) We know that, mean deviation = $\frac{\sum{(x_i - 53)}}{\sum{f_i}}$
   ♦ Where (n = ∑f_i) is the number of observations.
(ii) But the sum in the numerator works out to zero for both A and B.
• So the mean deviation will be zero for both A and B.
6. This situation is not unexpected. If we calculate the deviations in this way, the sum will be zero in all problems. Not just for A and B. The reason can be explained in 4 steps:
(i) In the fig.15.2(a) below, the red bars are the observations in a data. The mean of those observations is the one with the arrow mark.

Fig.15.2

• We see that:
   ♦ Some observations are smaller than the mean.
   ♦ Some observations are larger than the mean.
(ii) In fig.b, the yellow bars are the deviations of those observations which are smaller than the mean. These deviations will be -ve.    
(iii) Also in fig.b, the green bars are the deviations of those observations which are larger than the mean. These deviations will be +ve.
(iv) Now we calculate the sum of deviations:
   ♦ The sum of yellow bars
   ♦ will be equal to
   ♦ The sum of green bars.
• The yellow bars are -ve and green bars are +ve. So the total sum of all deviations will be zero.
7. So we have to apply a modification to the deviation. It can be written in 3 steps:
(i) By the term “deviation”, we are referring to the “difference from the central value”.
(ii) In fig.15.2(b) above, the lengths of the yellow bars are -ve deviations.
• But “lengths” are “distances”. They do not have signs.
   ♦ It is the magnitude of those lengths that matters.    ♦ There is no need to put the -ve signs.
(iii) So the modification can be applied by taking the absolute values.
• We can write:
   ♦ The sum used in the numerator
   ♦ must be
   ♦ The sum of absolute values of the deviations.
8. Based on this, we can write the formulas for calculating mean deviation (M.D):
(i) $\text{M.D.(a)}~=~\frac{\sum{(|x_i - a|)}}{\sum{f_i}}$
• If any observation x_i is present more than once, it's frequency is greater than 1. So it is better to apply frequency to all observations in general. So the formula becomes:
$\text{M.D.(a)}~=~\frac{\sum{(f_i |x_i - a|)}}{\sum{f_i}}$
• In this formula,
   ♦ "M.D.(a)" indicates that, the mean deviation is taken about the central tendency value "a".
   ♦ ∑f_i is the sum of frequencies, which will give the total number of observations.
(ii) If we decide to use the "mean" as the central tendency value, then "a" will become "$\bar{x}$".
• So the formula becomes:
$\text{M.D.}(\bar{x})~=~\frac{\sum{(f_i |x_i - \bar{x}|)}}{\sum{f_i}}$
(iii) If we decide to use the "median" as the central tendency value, then "a" will become "M".
• So the formula becomes:
$\text{M.D.(M)}~=~\frac{\sum{(f_i |x_i - \text{M}|)}}{\sum{f_i}}$

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Now we will see some solved examples:
Solved Example 15.1
Find the mean deviation about the mean for the following data:
6,7,10,12,13,4,8,12
Solution:
1. We are asked to find the mean deviation about the mean.
• So our first aim is to find the mean $(\bar{x})$. For that, we can use the first, second and third columns of table 15.5 below:

Table 15.5

• Based on those columns, we can write:
$\bar{x}~=~\frac{\sum{f_i x_i}}{\sum{f_i}}~=~\frac{72}{8}~=~9$
2. Now we can calculate the mean deviation. For that, we can use second, fourth and fifth columns of the table.
• We have:
$\text{M.D.}(\bar{x})~=~\frac{\sum{(f_i |x_i - \bar{x}|)}}{\sum{f_i}}$
• Substituting the values, we get:
$\text{M.D.}(\bar{x})~=~\frac{22}{8}~=~2.75$

Solved Example 15.2
Find the mean deviation about the mean for the following data:
12,3,18,17,4,9,17,19,20,15,8,17,2,3,16,11,3,1,0,5
Solution:
1. We are asked to find the mean deviation about the mean.
• So our first aim is to find the mean $(\bar{x})$. For that, we can use the first, second and third columns of table 15.6 below:

Table.15.6

• Based on the table, we can write:
$\bar{x}~=~\frac{\sum{f_i x_i}}{\sum{f_i}}~=~\frac{200}{20}~=~10$
2. Now we can calculate the mean deviation. For that, we can use second, fourth and fifth columns of the table.
• We have:
$\text{M.D.}(\bar{x})~=~\frac{\sum{(f_i |x_i - \bar{x}|)}}{\sum{f_i}}$
• Substituting the values, we get:
$\text{M.D.}(\bar{x})~=~\frac{124}{20}~=~6.2$

Solved Example 15.3
Find the mean deviation about the median for the following data:
3,9,5,3,12,10,18,4,7,19,21
Solution:
1. We are asked to find the mean deviation about the median.
• So our first aim is to find the median M. For that, we can use the first, second and third columns of the table 15.7 below. The observations are arranged in ascending order.

Table 15.7

• Based on the table, we can write:
Total number of observations = n = ∑f_i = 11
• "11" is an odd number. So the position of the median will be:
$\frac{n+1}{2}~=~\frac{11+1}{2}~=~\frac{12}{2}~=~6$
• From the column for cumulative frequency, we see that, the observation at the sixth position is 9.
• So we get: M = 9 
2. Now we can calculate the mean deviation. For that, we can use second, fourth and fifth columns of the table.

• We have:
$\text{M.D.(M)}~=~\frac{\sum{(f_i |x_i - M|)}}{\sum{f_i}}$
• Substituting the values, we get:
$\text{M.D.(M)}~=~\frac{58}{11}~=~5.27$

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In the next section, we will see a few more solved examples.

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Chapter 15 - Statistics

In the previous section, we completed a discussion on mathematical reasoning. In this section, we will see statistics.

In our earlier classes, we have seen the basics of statistics. The links to those notes are given below:

    ♦ Statistics part I consists of chapters 1, 1.1, 1.2, . . . up to 1.5
    ♦ Statistics part II consists of chapters 25, 25.1, . . . up to 25.10
    ♦ Statistics part III consists of chapters 37, 37.1, . . . up to 37.7

The reader must have a thorough knowledge on the above three parts. In our present discussion, we will see more details about central tendency.

We know that, mean, median and mode are three measures of central tendency. But in many cases, these three items may not be able to give us the actual nature of the data. Let us see an example. It can be written in 10 steps:
1. The following table 15.1 gives the runs scored by two batsmen in the last ten matches.

Table 15.1

• Let us find the mean and median of A and B.
2. First we will find the mean and median of A.
• To find the mean:
    ♦ The table 15.2 below shows the calculations.
    ♦ Observations are arranged in ascending order.

Table 15.2

We get: $\text{Mean}~(\bar x)~=~\frac{\sum{f_i x_i}}{\sum f_i}~=~\frac{530}{10}~=~53$
• To find the median:
(i) Total number of observations n = 10
This ‘10’ is an even number.
So the median is the mean of $\left(\frac{n}{2} \right)^{\text{th}}$ and $\left(\frac{n}{2} + 1 \right)^{\text{th}}$ values.
(ii) Calculating the positions:
   ♦ $\left(\frac{n}{2} \right)~=~\frac{10}{2}~=~5$ 
   ♦ $\left(\frac{n}{2} + 1 \right)~=~\frac{10}{2} + 1~=~6$
(iii) From the cumulative frequency column, we get:
    ♦ fifth value = 42      
    ♦ sixth value = 64      
(iv) So median = $\frac{42 + 64}{2}~=~\frac{106}{2}~=~53$
3. Now we will find the mean and median of B.
• To find the mean:
    ♦ The table 15.3 below shows the calculations.
    ♦ Observations are arranged in ascending order.

Table 15.3

We get: $\text{Mean}~(\bar x)~=~\frac{\sum{f_i x_i}}{\sum f_i}~=~\frac{530}{10}~=~53$
• To find the median:
(i) Total number of observations n = 10
This ‘10’ is an even number.
So the median is the mean of $\left(\frac{n}{2} \right)^{\text{th}}$ and $\left(\frac{n}{2} + 1 \right)^{\text{th}}$ values.
(ii) Calculating the positions:
   ♦ $\left(\frac{n}{2} \right)~=~\frac{10}{2}~=~5$ 
   ♦ $\left(\frac{n}{2} + 1 \right)~=~\frac{10}{2} + 1~=~6$
(iii) From the cumulative frequency column, we get:
    ♦ fifth value = 53      
    ♦ sixth value = 53      
(iv) So median = $\frac{53 + 53}{2}~=~\frac{106}{2}~=~53$
4. Now we can write a comparison:
• Comparing the means:
    ♦ Mean of A = 53
    ♦ Mean of B = 53
• Comparing the medians:
    ♦ Median of A = 53
    ♦ Median of B = 53
5. We see that, both mean and median are same for A and B. This gives us the impression that, both batsmen give similar performances.
6. But if we look at the observations carefully, we will see that, the performances are not similar.
• Batsman A scores low runs like 0 and 5. He scores high runs like 91 and 117 also.
• Batsman B always scores runs which are near or equal to 53.
7. So it is clear that, we cannot completely depend upon mean and median for taking decisions.
8. Fig.15.1 below shows the plot of the scores.

Fig.15.1

   ♦ Green circles denote the scores of batsman A.
   ♦ Red diamonds denote scores of batsman B.
• We see that:
   ♦ The ten red diamonds are close together.
   ♦ The ten green circles are scattered.
• We can say this in any one of the three ways written below:
   ♦ The green circles are more scattered.  
   ♦ The green circles are more spread out.  
   ♦ The green circles are more dispersed.
9. We want a method to measure this dispersion.
• Using that method, we must get a number. Once we get such a number, we will be able to say this:
   ♦ If the number is large, then the dispersion is high.
   ♦ If the number is small, then the dispersion is low.
◼ This number is called measure of dispersion.
10. There are four methods for finding the measure of dispersion:
(i) Range  (ii) Quartile deviation  (iii) Mean deviation  (iv) Standard deviation.
• In this chapter, we will be discussing all the above four methods except quartile deviation.

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Range

This can be explained in 4 steps:
1. Consider the example of the two batsmen A and B that we saw above.
• For batsman A,
   ♦ Maximum value is 117.
   ♦ Minimum value is 0.
• Maximum value – Minimum value = (117 – 0) = 117
• This number 117 is the range of the data of A. 
2. For batsman B,
   ♦ Maximum value is 60
   ♦ Minimum value is 46.
• Maximum value – Minimum value = (60 – 46) = 14
• This number 14 is the range of the data of B.
3. We see that:
Range of A > Range of B
• So we can write:
When compared to B, the dispersion of A is higher.
4. Difference of maximum and minimum values in a data is called the range of that data.
• The range gives us an idea about dispersion. Higher the range, higher is the dispersion.

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Limitations of range

This can be written in 2 steps:
1. Range gives us only a rough idea about dispersion.
2. We know that, mean and median are two important measures of central tendency.
• While calculating the range, mean and median are not taken into account.
• So the range is not able to give us a relation between dispersion and central tendency.

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In the next section, we will see mean deviation.

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Chapter 14.5 - Miscellaneous Exercise

In the previous section, we saw validating statements. In this section, we will see some miscellaneous examples.

Solved example 14.14
Check whether “Or” used in the following compound statement is exclusive or inclusive. Write the component statements of the compound statement and use them to check whether the compound statement is true or not. Justify your answer.
t: you are wet when it rains or you are in a river.
Solution:
1. We have:
t: you are wet when it rains or you are in a river.
2. The component statements are:
p: you are wet when it rains.
q: you are wet when you are in a river.
   ♦ p and q are connected by “or”.
3. Let us analyze the component statements:
• During rain, you are wet.
• When in a river, you are wet.
• When in a river, if it rains, you are wet.
• So the “or” used in this case is inclusive “or”.
4. Both the component statements are true. So t is true.

Solved example 14.15
Write the negation of the following statements
(i) p: For every real number x, x² > x.
(ii) q: There exists a rational number x such that x² = 2.
(iii) r: All birds have wings.
(iv) s: All students study mathematics at the elementary level.
Solution:
Part (i)
1. Consider the statement:
p: For every real number x, x² > x.
2. We want ~p.
• p tells us that: For all real numbers, the given property is valid.
• ~p must tell us that "For all" is false.
3. This ~p can be achieved by saying that there is at least one real number for which the property is not valid.
• So we get:
~p: There exists one real number x for which x² ≤ x.

Part (ii)
1. Consider the statement:
q: There exists a rational number x such that x² = 2.
2. We want ~q.
• q tells us that: There is at least one rational number for which the given property is valid.
• ~q must tell us that "at least one" is false.
3. This ~q can be achieved by saying that for all rational numbers, the property is not valid.
• So we get:
~p: For all rational numbers x, x² ≠2.

Part (iii)
1. Consider the statement:
r: All birds have wings.
2. We want ~r.
• r tells us that: For all birds, the given property is valid.
• ~r must tell us that "For all" is false.
3. This ~r can be achieved by saying that there is at least one bird for which the property is not valid.
• So we get:
~r: There exists one bird which do not have wings.

Part (iv)
1. Consider the statement:
s: All students study mathematics at the elementary level.
2. We want ~s.
• s tells us that: For all students, the given property is valid.
• ~s must tell us that "For all" is false.
3. This ~r can be achieved by saying that there is at least one student for which the property is not valid.
• So we get:
~r: There exists one student who does not study mathematics at the elementary level.

Solved example 14.16
Using the words “necessary and sufficient”, rewrite the statement “The integer n is odd if and only if n² is odd”. Also check whether the statement is true.
Solution:
Part (i): Rewriting the statement
Using the words “necessary and sufficient”, we can write:
“The integer n is odd” is necessary and sufficient condition for “n² to be odd” and vice versa. 

Part (ii): Checking the validity.
• The given compound statement is in the form “p if and only if q”.
The component statements are:
p: The integer n is odd.
q: n² is odd.
• In such cases, we know that:
    ♦ Whenever p is true, q is also true.
    ♦ Whenever q is true, p is also true.
• So we need to show two items:
Case (i) If p is true, then q is true.
Case (ii) If q is true, then p is true.

Case (i): If p is true, then q is true.
    ♦ First, we assume that p is true.
    ♦ Based on this assumption, we check q.
1. Assuming p to be true:
• Let us assume that, integer n is indeed odd.
2. Since it is an odd integer, we can write:
    ♦ n = 2m + 1
        ✰ Where m is some integer.
3. Now we calculate n². We get:
$\begin{array}{ll} {}&{n^2} & {~=~}& {(2m+1)^2} &{} \\ {}&{} & {~=~}& {4m^2 + 4m + 1} &{} \\ {}&{} & {~=~}& {2(2m^2 + 2m) + 1} &{} \\ \end{array}$

4. Consider the result in (3):
• m is an integer.
⇒ m² will be an integer.
⇒ (2m² + 2m) will be even.
⇒ 2(2m² + 2m) will be even.
⇒ 2(2m² + 2m) + 1 will be odd.

5. Based on (3) and (4), we can write:
n² is odd.
• So the statement q is true.
6. We obtained the truth value of q as T.
• We obtained this by assuming that the truth value of p is T.
• So the truth value of “if p then q” is T.

Case (ii): If q is true, then p is true.
Here we will use the contrapositive method. That is., if ~p then ~q.
    ♦ First, we assume that ~p is true.
    ♦ Based on this assumption, we check ~q.
The statements are:
~p: The integer n is even.
~q: n² is even.
1. Assuming ~p to be true:
• Let us assume that, n is an even integer and n² is indeed even.
2. Since it is an even integer, we can write:
    ♦ n = 2m
        ✰ Where m is some integer.
3. Now we calculate n². We get:
$\begin{array}{ll} {}&{n^2} & {~=~}& {(2m)^2} &{} \\ {}&{} & {~=~}& {4m^2} &{} \\ {}&{} & {~=~}& {2(2m^2)} &{} \\   \end{array}$

4. Consider the result in (3):
• m is an integer.
⇒ m² will be an integer.
⇒ 2m² will be even.
⇒ 2(2m²) will also be even.

5. Based on (3) and (4), we can write:
n² is even.
• So the statement ~q is true.
6. We obtained the truth value of ~q as T.
• We obtained this by assuming that the truth value of ~p is T.
• So the truth value of “if ~p then ~q” is T.
• So the truth value of “if q then p” is T.

◼ From case (i), we obtained:
• The truth value of “if p then q” is T.
◼ From case (ii), we obtained:
• The truth value of “if q then p” is T.
◼ So we can write: The given "p if and only if q" is true.
• That is.,
The statement “The integer n is odd if and only if n² is odd” is true.

Solved example 14.17
For the given statement, identify the necessary and sufficient conditions
t: If you drive over 80 km per hour, then you will get a fine. 
Solution:
1. Consider the statement:
t: If you drive over 80 km per hour, then you will get a fine.
2. The component statements are:
p: You drive over 80 km per hour.
q: You will get a fine.
3. This is a statement with "if p then q".
• So we can write:
   ♦ p is a sufficient condition for q.
   ♦ q is a necessary condition for p.
(see example 2 at the beginning of section 14.3)
4. Thus we get:
• The sufficient condition is:
Driving over 80 km per hour.
(This is sufficient to get a fine)
• The necessary condition is:
Getting a fine.
(This is necessary if the speed is above 80 km per hour)

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The link below gives some more solved examples:

Miscellaneous Exercise

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In the next chapter, we will see statistics.

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Chapter 14.4 - Validating Statements

In the previous section, we saw Implications. In this section, we will see validating statements.

• Truth value of a statement tells us whether a statement is true or false.
• If a statement is true, then its truth value is true.
    ♦ We write it as T.
• If a statement is false, then its truth value is false.
    ♦ We write it as F.
• Validating statements is the process of finding the truth value of a statement.

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• Validating simple statements is easy. But for compound statements, we have to follow certain rules.
• The rules depend on the connectives/implications present in the compound statement.

Rule 1: Statements with “and”.
• If the connective in a compound statement is “and”, then the truth value of each of the component statements must be T. Then only we get T for the compound state.
• So when the component statements are p and q, we can use the following two steps:
Step 1: Show that p is true.
Step 2: Show that q is true. 

Rule 2: Statements with “or”.
• If the connective in a compound statement is “or”, then the truth value of at least one component statement must be T. Then only we get T for the compound state.
• So when the component statements are p and q, we can use the following three cases:
Case 1: p is false but q is true.
Case 2: p is true but q is false.
Case 3: Both p and q are true.
• If we get any of the above three cases, then the truth value of the compound statement is T.  

Rule 3: Statements with “if-then”.
• When the compound statement is in the form “if p then q”, we know that:
Whenever p is true, q is also true.
• So we can think of a method to validate such statements:
    ♦ First, we assume that p is true.
    ♦ Based on this assumption, we check q.
    ♦ If q is true, then the compound statement can be given the truth value T
• This method is known as direct method.

• When the compound statement is in the form “if p then q”, we know that:
if ~q then ~p is applicable.
• So we can think of another method to validate such statements:
    ♦ First, we assume that ~q is true.
    ♦ Based on this assumption, we check ~p.
    ♦ If ~p is true, then the compound statement can be given the truth value T
• This method is known as contrapositive method.
• So for validating “if p then q” statements, we have two methods:
(i) Direct method  (ii) Contrapositive method.

Rule 4: Statements with “if and only if”.
• When the compound statement is in the form “p if and only if q”, we know that:
    ♦ Whenever p is true, q is also true.
    ♦ Whenever q is true, p is also true.
• So we can think of a method to validate such statements:
In this method, we need to show two items:
(i) If p is true, then q is true.
(ii) If q is true, then p is true.

Rule 5
This rule is used to prove that, a statement is true. In this rule, we make use of contradiction. The steps are:
(i) We assume that p is false.
This is same as assuming ~p to be true.
(ii) Then we arrive at some result which contradicts the assumption.
(iii) Due to the contradiction, we conclude that, p is true.

Rule 6
This rule is used to prove that, a statement is false. In this rule, we simply present an example where the statement is not valid. Such an example is called counter example. When the counter example is presented, the statement is immediately written off as false.
◼ We may be able to present numerous examples where a statement is valid. Whatever be the number of such examples we present, we cannot conclude that the statement is true.
◼ We may be able to present numerous counter examples where a statement is not valid.  Just one counter example is sufficient to conclude that the statement is false.

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Let us see some solved examples:

Solved example 14.10
Check whether the following statement is true or not.
If x, y ∈ Z are such that x and y are odd, then xy is odd.
Solution:
1. Consider the statement:
r: If x, y ∈ Z are such that x and y are odd, then xy is odd.
2. The component statements are:
p: x, y ∈ Z are such that x and y are odd.
q: xy is odd.
3. This is a statement with "if-then". So we will apply Rule 3. We will use the direct method.
    ♦ First, we assume that p is true.
    ♦ Based on this assumption, we check q.
4. Assuming p to be true:
• Let us assume that, both x and y are indeed elements of the set Z and they are indeed odd.
• We know that, Z is the set of integers. So we can write:
We assume that both x and y are indeed odd integers.
5. Since they are odd integers, we can write:
    ♦ x = 2m + 1
    ♦ y = 2n + 1
        ✰ Where m and n are some integers.
6. Now we calculate the product xy. We get:
$\begin{array}{ll} {}&{xy} & {~=~}& {(2m+1)(2n+1)} &{} \\ {}&{} & {~=~}& {4mn + 2m + 2n + 1} &{} \\ {}&{} & {~=~}& {2(2mn + m + n) + 1} &{} \\ \end{array}$

7. Consider the result in (6):
• m and n are integers
⇒ (2mn + m + n) can be odd or even.
⇒ 2(2mn + m + n) will be even.
⇒ 2(2mn + m + n) + 1 will be odd.

8. Based on (6) and (7), we can write:
The product xy is odd.
• So the statement q is true.
9. We obtained the truth value of q as T.
• We obtained this by assuming that the truth value of p is T.
• So the truth value of “if p then q” is T.
• We can write:
The truth value of the given compound statement r is T.

Alternatively, using the contrapositive method:
1. We have: if ~q then ~p
• The contrapositive statements are:
~q: xy is even.
~p: One among x, y is even.
2. For applying the contrapositive method,
    ♦ First, we assume that ~q is true.
    ♦ Based on this assumption, we check ~p.
3. Assuming ~q to be true:
~q: xy is even.
4. When xy is even, x or y has to be even.
(supposing x = 2n, we get xy = 2ny, which is even)
• xy will never become even if both x and y are odd.
• So ~p is true.
5. We obtained the truth value of ~p as T.
• We obtained this by assuming that the truth value of ~q is T.
• So the truth value of “if ~q then ~p” is T.
• We can write:
The truth value of the given compound statement r is T.

Solved example 14.11
Check whether the following statement is true or false by proving its contrapositive.
If x, y ∈ Z such that xy is odd, then both x and y are odd.
Solution:
1. Consider the statement:
r: If x, y ∈ Z such that xy is odd, then both x and y are odd.
2. The component statements are:
p: x, y ∈ Z such that xy is odd.
q: both x and y are odd.
3. This is a statement with "if-then". So we will apply the contrapositive method for validation.
• We have: if ~q then ~p
• The contrapositive statements are:
~q: One among x, y is even.
~p: xy is even.
2. For applying the contrapositive method,
    ♦ First, we assume that ~q is true.
    ♦ Based on this assumption, we check ~p.
3. Assuming ~q to be true:
~q: One among x, y is even.
4. When one among x, y is even, xy will be even.
(supposing x = 2n, we get xy = 2ny, which is even)
• xy will never become even if both x and y are odd.
• So ~p is true.
5. We obtained the truth value of ~p as T.
• We obtained this by assuming that the truth value of ~q is T.
• So the truth value of “if ~q then ~p” is T.
• We can write:
The truth value of the given compound statement r is T.

Solved example 14.12
Prove that the following statement is true.
p: √7 is irrational
Solution:
1. Assume that √7 is rational.
2. If √7 is rational, we can write: $\sqrt{7} = \frac{a}{b}$
Where the ratio $\frac{a}{b}$ is in the simplest form.
• That is, a and b do not have any common factors.
3. Squaring both sides, we get: $7 = \frac{a^2}{b^2}$
4. Based on this we can write: $b^2 = \frac{a^2}{7}$
• b is an integer. So b² will be an integer.   
5. Let us analyze the result in (4):
$\text{An integer}(b^2)~=~\frac{\text{An integer}(a) \times \text{An integer}(a)}{7}$
• So $\frac{a}{7}$ must be an integer. Let it be c.
6. We can write: $\frac{a}{7}= c$
• Thus we get a = 7c
• Squaring both sides, we get: a² = 49c².
7. But from (3) we have a² = 7b².
8. Equating the results in (6) and (7), we get:
49c² = 7b².
• From this we get: 7c² = b².
9. Based on this we can write: $c^2 = \frac{b^2}{7}$
• c is an integer. So c² will be an integer.
10. Let us analyze the result in (9):
$\text{An integer}(c^2)~=~\frac{\text{An integer}(b) \times \text{An integer}(b)}{7}$
• So $\frac{b}{7}$ must be an integer. Let it be d.   
11. Let us compare two results:
   ♦ In (5), we saw that, a is divided by 7 to give an integer.
   ♦ In (10), we saw that, b is divided by 7 to give an integer.
• That means, a and b has a common factor 7. This contradicts our assumption in (2) that, a and b have no common factors.
12. So it is clear that the assumption that √7 is rational is wrong.
• We can write:
p: √7 is irrational
is true.

Solved example 14.13
By giving a counter example, prove that the following statement is false.
If n is an odd integer, then n is prime.
Solution:
1. Consider the statement:
r: If n is an odd integer, then n is prime.
2. The component statements are:
p: n is an odd integer.
q: n is prime.
3. This is a statement with "if-then".
• We have to prove "if p then q" is false.
4. For statement p, we take n = 9 as an example. This is acceptable because 9 is an odd number.
5. Now we consider statement q.
• We know that 9 is not a prime number. So when n = 9, the statement q becomes false.
6. So we presented a counter example which proved "if p then q" to be false.
• That is., r: If n is an odd integer, then n is prime.
is false.

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The link to some solved examples is given below:

Exercise 14.5

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In the next section, we will see some miscellaneous examples.

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