Question

Following are the runs scored by two batsmen in 5 cricket matches. Who is more consistent in scoring runs?

Batsman A	38	47	34	18	33
Batsman B	37	35	41	27	35

Answer 1

Hint: In this question, we are given scores of two batsmen and we have to find who is more consistent. For this, we have to find coefficient of variation for both data and then compare it to find who have less coefficient of variation and that batsman will be more consistent. For calculating coefficient of variation we will use formula given by $$CV={\displaystyle \frac{\sigma }{\bar{X}}}\times 100$$ where, $\sigma$ is standard deviation and $\bar{X}$ is the mean of data. For mean $\bar{X}$ we will divide the sum of the terms by the number of terms. For standard deviation we will use formula $$\sigma =\sqrt{{\displaystyle \frac{\sum x^2}{N}}}$$ where, x is deviation from mean and N is the number of terms.

Complete step-by-step answer:
Let us calculate coefficient of variation for both data one by one:
For batsman A, data is as follows:
38, 47, 34, 18, 33.
Let us first find the mean of the data for which we can see the number of terms are 5 and sum of terms is $38+47+34+18+33=170$. Hence,
$$\text{Mean }{\bar{X}}_A={\displaystyle \frac{170}{5}}=34$$.
For calculating standard deviation let us draw table with columns: Runs (X), deviation from mean $(x=X-\bar{X})$ and $x^2$ is square of deviation.

Runs (X)	$(x=X-\bar{X})$$\bar{X}=34$	$x^2={(X-\bar{X})}^2$
38	4	16
47	13	169
34	0	0
18	-16	256
33	-1	1

Now, let us calculate sum of $x^2$ terms which becomes
$$\sum x^2=16+169+256+1=442$$.
Standard deviation for batsman A becomes
$${\sigma }_A=\sqrt{{\displaystyle \frac{\sum x^2}{N}}}=\sqrt{{\displaystyle \frac{442}{5}}}=\sqrt{88.4}\simeq 9.4$$.
Coefficient of variation for batsman A becomes
$$C{V}_A={\displaystyle \frac{{\sigma }_A}{\overline{X_A}}}\times 100={\displaystyle \frac{9.4}{34}}\times 100=27.65$$.
Hence, coefficient of variation for batsman A is equal to 27.65.
For batman B, data is as follows:
37, 35, 41, 27, 35
Let us find the mean of the data for which we can see the number of terms are 5 and sum of terms is $37+35+41+27+35=175$. Hence,
$$\text{Mean }{\bar{X}}_B={\displaystyle \frac{175}{5}}=35$$.
For calculating standard deviation let us draw table with columns: Runs (X), deviation from mean $(x=X-\bar{X})$ and $x^2$ is square of deviation

Runs (X)	$(x=X-\bar{X})$$\bar{X}=35$	$x^2={(X-\bar{X})}^2$
37	2	4
35	0	0
41	6	36
27	-8	64
35	0	0

Now, let us calculate sum of $x^2$ terms which becomes
$$\sum x^2=4+36+64=104$$.
Standard deviation for batsman B becomes
$${\sigma }_B=\sqrt{{\displaystyle \frac{\sum x^2}{N}}}=\sqrt{{\displaystyle \frac{104}{5}}}=\sqrt{20.8}\simeq 4.6$$.
Coefficient of variation for batsman B becomes
$$C{V}_B={\displaystyle \frac{{\sigma }_B}{\overline{X_B}}}\times 100={\displaystyle \frac{4.6}{35}}\times 100=13.14$$.
Hence, coefficient of variation for batsman B is equal to 13.14.
As we can see $C{V}_A=27.65\text{ and }C{V}_B=13.14$ therefore, coefficient of variation for batsman B is less than coefficient of variation for batsman A.
Therefore, batsman B is more consistent than batsman A in scoring the runs.

Note: Students should note that, sum of deviation from mean is always zero. Students should carefully apply the formula for calculating standard deviation and coefficient of variation. Squares and square roots must be calculated carefully. We have used coefficient of variation for comparing because coefficient of variation measures how consistent the different values of the set are from the mean of the data set. The lesser the variation the more is the consistency.