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.
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| Table 15.1 |
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.
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| 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.
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| 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.
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| Fig.15.1 |
♦ Green circles denote the scores of batsman A.
♦ Red diamonds denote scores of batsman B.
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We see that:
♦ The ten red diamonds are close together.
♦ The ten green circles are scattered.
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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.
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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.
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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.
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For batsman A,
♦ Maximum value is 117.
♦ Minimum value is 0.
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Maximum value – Minimum value = (117 – 0) = 117
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This number 117 is the range of the data of A.
2. For batsman B,
♦ Maximum value is 60
♦ Minimum value is 46.
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Maximum value – Minimum value = (60 – 46) = 14
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This number 14 is the range of the data of B.
3. We see that:
Range of A > Range of B
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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.
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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.
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While calculating the range, mean and median are not taken into account.
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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.
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