Calculate mean, variance and standard deviation for the following distribution.
Classes
$30-40$
$40-50$
$50-60$
$60-70$
$70-80$
$80-90$
$90-100$
${f_i}$
$3$
$7$
$12$
$15$
$8$
$3$
$2$
Calculate mean, variance and standard deviation for the following distribution.
| Classes | $30-40$ | $40-50$ | $50-60$ | $60-70$ | $70-80$ | $80-90$ | $90-100$ |
| ${f_i}$ | $3$ | $7$ | $12$ | $15$ | $8$ | $3$ | $2$ |
Let the assumed mean $A =65 .$ Here $h=10$
We obtain the following Table from the given data :
| Class |
Frequency ${f_i}$ |
Mid-point ${x_i}$ |
${y_i} = \frac{{{x_i} - 65}}{{10}}$ | ${y_i}^2$ | ${f_i}{y_i}$ | ${f_i}{y_i}^2$ |
| $30-40$ | $3$ | $35$ | $-3$ | $9$ | $-9$ | $27$ |
| $40-50$ | $7$ | $45$ | $-2$ | $4$ | $-14$ | $28$ |
| $50-60$ | $12$ | $55$ | $-1$ | $1$ | $-12$ | $12$ |
| $60-70$ | $15$ | $65$ | $0$ | $0$ | $0$ | $0$ |
| $70-80$ | $8$ | $75$ | $1$ | $1$ | $8$ | $8$ |
| $80-90$ | $3$ | $85$ | $2$ | $4$ | $6$ | $12$ |
| $90-100$ | $2$ | $95$ | $3$ | $9$ | $6$ | $18$ |
| $N=50$ | $-15$ | $105$ |
Therefore $\bar x = A + \frac{{\sum {{f_i}{y_i}} }}{{50}} \times h = 65 - \frac{{15}}{{50}} \times 10 = 62$
Variance ${\sigma ^2} = \frac{{{h^2}}}{{{N^2}}}\left[ {N{{\sum {{f_i}{y_i}} }^2} - {{\left( {\sum {{f_i}{y_i}} } \right)}^2}} \right]$
$=\frac{(10)^{2}}{(50)^{2}}\left[50 \times 105-(-15)^{2}\right]$
$=\frac{1}{25}[5250-225]=201$
and standard deviation $(\sigma)=\sqrt{201}=14.18$
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${x_i}$
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$109$
${f_i}$
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