From the data given below state which group is more variable, $A$ or $B$ ?

Marks $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ $60-70$ $70-80$
Group $A$ $9$ $17$ $32$ $33$ $40$ $10$ $9$
Group $B$ $10$ $20$ $30$ $25$ $43$ $15$ $7$

Vedclass pdf generator app on play store
Vedclass iOS app on app store

Firstly, the standard deviation of group $A$ is calculated as follows.

Marks

Group $A$

${f_i}$

Mid-point

${x_i}$

${y_i} = \frac{{{x_i} - 45}}{{10}}$ ${y_i}^2$ ${f_i}{y_i}$ ${f_i}{y_i}^2$
$10-20$ $9$ $15$ $-3$ $9$ $-27$ $81$
$20-30$ $17$ $25$ $-2$ $4$ $-34$ $68$
$30-40$ $32$ $35$ $-1$ $1$ $-32$ $32$
$40-50$ $33$ $45$ $0$ $0$ $0$ $0$
$50-60$ $40$ $55$ $1$ $1$ $40$ $40$
$60-70$ $10$ $65$ $2$ $4$ $20$ $40$
$70-80$ $9$ $75$ $3$ $9$ $27$ $81$
  $150$       $-6$ $342$

Here, $h =10, N =150, A =45$

Mean $ = A + \frac{{\sum\limits_{i = 1}^7 {{x_i}} }}{N} \times h$

$ = 45 + \frac{{\left( { - 6} \right) \times 10}}{{150}} \times 45 - 0.4 = 44.6$

$\sigma _1^2 = \frac{{{h^2}}}{{{N^2}}}\left[ {N\sum\limits_{i = 1}^7 {{f_i}{y_i}^2 - {{\left( {\sum\limits_{i = 1}^7 {{f_i}{y_i}} } \right)}^2}} } \right]$

$=\frac{100}{22500}\left[150 \times 342-(-6)^{2}\right]$

$=\frac{1}{225}(51264)$

$=227.84$

$\therefore$ Standard deviation $\left(\sigma_{1}\right)=\sqrt{227.84}=15.09$

The standard deviation of group $B$ is calculated as follows.

Marks

Group $A$

${f_i}$

Mid-point

${x_i}$

${y_i} = \frac{{{x_i} - 45}}{{10}}$ ${y_i}^2$ ${f_i}{y_i}$ ${f_i}{y_i}^2$
$10-20$ $9$ $15$ $-3$ $9$ $9$ $-30$
$20-30$ $17$ $25$ $-2$ $4$ $4$ $-40$
$30-40$ $32$ $35$ $-1$ $1$ $1$ $-30$
$40-50$ $33$ $45$ $0$ $0$ $0$ $0$
$50-60$ $40$ $55$ $1$ $1$ $1$ $43$
$60-70$ $10$ $65$ $2$ $4$ $4$ $30$
$70-80$ $9$ $75$ $3$ $9$ $9$ $21$
  $150$         $-6$

Mean  $ = A + \frac{{\sum\limits_{i = 1}^7 {{f_i}{y_i}} }}{N} \times h$

$ = 45 + \frac{{\left( { - 6} \right) \times 10}}{{150}} \times 45 - 0.4 = 44.6$

$\sigma _2^2 = \frac{{{h^2}}}{{{N^2}}}\left[ {N\sum\limits_{i = 1}^7 {{f_i}{y_i}^2 - {{\left( {\sum\limits_{i = 1}^7 {{f_i}{y_i}} } \right)}^2}} } \right]$

$=\frac{100}{22500}\left[150 \times 366-(-6)^{2}\right]$

$=\frac{1}{225}(54864)=243.84$

$\therefore$ Standard deviation $\left(\sigma_{1}\right)=\sqrt{243.84}=15.61$

Since the mean of both the groups is same, the group with greater standard deviation will be more variable.

Thus, group $B$ has more variability in the marks.

Similar Questions

The following values are calculated in respect of heights and weights of the students of a section of Class $\mathrm{XI}:$

  Height Weight
Mean $162.6\,cm$ $52.36\,kg$
Variance $127.69\,c{m^2}$ $23.1361\,k{g^2}$

Can we say that the weights show greater variation than the heights?

If $x_1, x_2,.....x_n$ are $n$ observations such that $\sum\limits_{i = 1}^n {x_i^2}  = 400$ and $\sum\limits_{i = 1}^n {{x_i}}  = 100$ , then possible value of $n$ among the following is 

If the mean deviation about the mean of the numbers $1,2,3, \ldots ., n$, where $n$ is odd, is $\frac{5(n+1)}{n}$, then $n$ is equal to

  • [JEE MAIN 2022]

The data is obtained in tabular form as follows.

${x_i}$ $60$ $61$ $62$ $63$ $64$ $65$ $66$ $67$ $68$
${f_i}$ $2$ $1$ $12$ $29$ $25$ $12$ $10$ $4$ $5$

The variance of $20$ observations is $5 .$ If each observation is multiplied by $2,$ find the new variance of the resulting observations.