Question

The given distribution shows the number of runs scored by some by some top batsmen of the world in one-day international cricket matches. Find the mode of the data.

Runs scored	Number of batsmen
$3000 - 4000$	$4$
$4000 - 5000$	$18$
$5000 - 6000$	$9$
$6000 - 7000$	$7$
$7000 - 8000$	$6$
$8000 - 9000$	$3$
$9000 - 10000$	$1$
$10000 - 11000$	$1$

Answer 1

Hint: For finding the mode of the group data we use the formula Estimated mode = $L + \dfrac{{{f_m} - {f_{m - 1}}}}{{({f_m} - {f_{m - 1}}) + ({f_m} - {f_{m + 1}})}} \times w$ ,Modal group is the group which have maximum frequency Modal group is $4000 - 5000$ , L = $4000$ is the lower class boundary of the modal group ,${f_{m - 1}}$ = $4$ is the frequency of the group before the modal group,${f_m} = 18$ is the frequency of the modal group, ${f_{m + 1}}$ = $9$ is the frequency of the group after the modal group ,w = $1000$ is the group width

Complete step by step answer:
In this case we have to find the mode of the given data ,
Mode is the number that appears most frequently in a data set . In the grouped data we use the formula for mode that is ,
Estimated mode = $L + \dfrac{{{f_m} - {f_{m - 1}}}}{{({f_m} - {f_{m - 1}}) + ({f_m} - {f_{m + 1}})}} \times w$
Modal group is the group which have maximum frequency.
where $L$ is the lower class boundary of the modal group
${f_{m - 1}}$ is the frequency of the group before the modal group
${f_m}$ is the frequency of the modal group
${f_{m + 1}}$ is the frequency of the group after the modal group
$w$ is the group width.
For the given question
The modal group is $4000 - 5000$ because its frequency is maximum that is $18$.
$L= 4000$ is the lower class boundary of the modal group
${f_{m - 1}}$ = $4$ is the frequency of the group before the modal group
${f_m} = 18$ is the frequency of the modal group
${f_{m + 1}}$ = $9$ is the frequency of the group after the modal group
w = $1000$ is the group width .
By putting these values in the mode equation we get ,
Estimated mode = $4000 + \dfrac{{18 - 4}}{{(18 - 4) + (18 - 9)}} \times 1000$
= $4000 + \dfrac{{14}}{{14 + 9}} \times 1000$
= $4000 + \dfrac{{14}}{{23}} \times 1000$
After dividing $14$ by $23$ we get $0.608695$ hence ,
$4000 + 0.608695 \times 1000$
$4000 + 608.695$
$4608.695$

$\therefore $ The mode of the given question is $4608.695$.

Note:
For the mean of grouped data Mean = $\dfrac{{\sum {f \times } X}}{{\sum f }}$ where $X$ is the midpoint of group and f is frequency of data .
For the Median of the group data, we use the formula Median = $ = L + \dfrac{{\dfrac{n}{2} - B}}{G} \times w$ where: $L$ is the lower class boundary of the group containing the median, $n$ is the total number of values, $B$ is the cumulative frequency of the groups before the median group, $G$ is the frequency of the median group,w is the group width