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