Higher Secondary Mathematics: mean

Monday, July 17, 2023

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:
$\text{M.D.(M)}~=~\frac{\sum{(f_i |x_i - \text{M}|)}}{\sum{f_i}}$

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:
$\text{M.D.(M)}~=~\frac{58}{11}~=~5.27$

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

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Sunday, July 16, 2023

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

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.

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.

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.

In the next section, we will see mean deviation.