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 $B$ $10$ $20$ $30$ $25$ $43$ $15$ $7$

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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.

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