Question

The batting scores of two cricket players $A$ and $B$ in $10$ innings are as follow: -
Batsman $A - 15,17,19,27,30,36,40,90,95,110$
Batsman $B - 10,16,21,28,37,41,36,80,82,65$
Find which of the players is more consistent.

Answer 1

Hint: In the given question we need to find which player is more consistent so finding this we need to find the value of the standard deviation of both players then we will compare both the standard deviation and player which has a minimum value of standard deviation is the more consistent.

Complete step by step Solution:
Given that we have two players $A$ and $B$ in $10$ innings are as follow: -
Batsman $A - 15,17,19,27,30,36,40,90,95,110$
Batsman $B - 10,16,21,28,37,41,36,80,82,65$
Now first we will calculate the standard deviation for $A$ and then similarly find for $B$.
We know that the formula for standard deviation
If the given data is the $x$and the total number of data is $N$ and the mean of given data is $\overline x $ then the standard deviation of given data is the denoted by the $\sigma $ then
$\sigma = \sqrt {\sum\limits_n^{i = 1} {\dfrac{{{{({x_i} - \overline x )}^2}}}{N}} } $
Where the $\sigma $ is the standard deviation of given data and $x$is value of given data and $\overline x $ is the mean or average value of the given data ,$N$ is the value of the number of data which is given for $A$where $i = 1,2,3,4,5,6,.................n$
Now we will calculate standard deviation for $A$
Let - $x = 15,17,19,27,30,36,40,90,95,110$
First, we will find the mean value of given data for $A$ then we will get
$ = \dfrac{{15 + 17 + 19 + 27 + 30 + 36 + 40 + 90 + 95 + 110}}{{10}}$
Now after completely solving the above addition we will get the value of the mean for $A$
$ = 47.9$
The mean value or we can say it is the average value of the given data for $A$so we can denote it by the $\overline x $ which is the sign for the mean and average of data
So, we can write it as $\overline x = 47.9$
Now we will find the value of the $(x - \overline x )$ which is we will get then
$(x - \overline x ) = (x - 47.9)$
Now after calculating we will get the value of $(x - \overline x )$
$ \Rightarrow (x - \overline x ) = - 32.9, - 30.9, - 28.9, - 20.9, - 17.9, - 11.9, - 7.9,42.1,47.1,62.1$
Now we will calculate the square of all values of $(x - \overline x )$ that is we will calculate the value of the ${(x - \overline x )^2}$ then after calculating the value of the we will get the value which are
${(x - \overline x )^2}$= $1082.41,954.81,8635.21,436.81,320.41,141.61,62.41,1772.41,2218.41,3856.41$
Now we will calculate the sum of the all value of the ${(x - \overline x )^2}$ then we will get the value of the $\sum\limits_{}^{} {} {(x - \overline x )^2}$ then
cv$ \Rightarrow \sum\limits_{}^{} {} {(x - \overline x )^2} = 11680.9$
And the given number of data is the $N = 10$
Now we will put all values in the formula of standard deviation which is $\sigma = \sqrt {\sum\limits_n^{i = 1} {\dfrac{{{{({x_i} - \overline x )}^2}}}{N}} } $ then we will get the value of the standard deviation $\sigma $
$ \Rightarrow {\sigma _A} = \sqrt {\sum\limits_n^{i = 1} {\dfrac{{{{({x_i} - \overline {{x_A}} )}^2}}}{N}} } $
Now after putting values we will get
$ \Rightarrow {\sigma _A} = \sqrt {\dfrac{{11680.9}}{{10}}} $
Now after calculating we will get the value of the standard deviation ${\sigma _A}$
$ \Rightarrow {\sigma _A} = \sqrt {\dfrac{{116809}}{{100}}} $
$ \Rightarrow {\sigma _A} = \dfrac{{341.77}}{{10}}$
Now after calculating we will get the value of the standard deviation ${\sigma _A}$
$ \Rightarrow {\sigma _A} = \dfrac{{341.77}}{{10}} = 34.177$
Therefore, the standard deviation of $A$ is the ${\sigma _A} = 34.177$
Similarly, we will find standard deviation for the Batsman $B - 10,16,21,28,37,41,36,80,82,65$
Then we will get the standard deviation for the $B$ then we will get the value of the ${\sigma _B}$
Then the value of the ${\sigma _B} = 7.727$
Now we will compare both standard deviation ${\sigma _A}$ and the ${\sigma _B}$ then
we have ${\sigma _A} = 34.177$ and ${\sigma _B} = 7.727$ we can see clearly that the value of the ${\sigma _A} > {\sigma _B}$

so, we can say that the player $B$ is more consistent than the player $A$.

Note:
We can solve this question by directly putting the value of the $(x - \overline x )$ but this type of question is more calculative so always calculate the approximation value and don't try to find the exact value.