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Find the mean and variance for the following frequency distribution.
Classes | $0-10$ | $10-20$ | $20-30$ | $30-40$ | $40-50$ |
Frequencies | $5$ | $8$ | $15$ | $16$ | $6$ |
$132$
$132$
$132$
$132$
Solution
Class | Frequency ${f_i}$ | Mid-point ${x_i}$ | ${y_i} = \frac{{{x_i} – 25}}{{10}}$ | ${y_i}^2$ | ${f_i}{y_i}$ | ${f_i}{y_i}^2$ |
$0-10$ | $5$ | $5$ | $-2$ | $4$ | $-10$ | $20$ |
$10-20$ | $8$ | $15$ | $-1$ | $1$ | $-8$ | $8$ |
$20-30$ | $15$ | $25$ | $0$ | $0$ | $0$ | $0$ |
$30-40$ | $16$ | $35$ | $1$ | $1$ | $16$ | $16$ |
$40-50$ | $6$ | $45$ | $2$ | $4$ | $12$ | $24$ |
$50$ | $10$ | $68$ |
Mean, $\bar x = A + \frac{{\sum\limits_{i = 1}^5 {{f_i}{y_i}} }}{N} \times h$
$ = 25 + \frac{{10}}{{50}} \times 10 = 25 + 2 = 27$
Variance, $\left( {{\sigma ^2}} \right) = \frac{{{h^2}}}{{{N^2}}}\left[ {N\sum\limits_{i = 1}^5 {{f_i}{y_i}^2 – {{\left( {\sum\limits_{i = 1}^5 {{f_i}{y_i}} } \right)}^2}} } \right]$
$=\frac{(10)^{2}}{(50)^{2}}\left[50 \times 68-(10)^{2}\right]$
$=\frac{1}{25}[3400-100]=\frac{3300}{25}$
$=132$
Similar Questions
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 mean and variance of the frequency distribution
$\mathrm{x}$ | $\mathrm{x}_{1}=2$ | $\mathrm{x}_{2}=6$ | $\mathrm{x}_{3}=8$ | $\mathrm{x}_{4}=9$ |
$\mathrm{f}$ | $4$ | $4$ | $\alpha$ | $\beta$ |
be $6$ and $6.8$ respectively. If $x_{3}$ is changed from $8$ to $7 ,$ then the mean for the new data will be: