- Home
- Standard 11
- Mathematics
13.Statistics
hard
Find the mean and variance for the following frequency distribution.
| Classes | $0-10$ | $10-20$ | $20-30$ | $30-40$ | $40-50$ |
| Frequencies | $5$ | $8$ | $15$ | $16$ | $6$ |
A
$132$
B
$132$
C
$132$
D
$132$
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, $\left( {{\sigma ^2}} \right) = \frac{{{h^2}}}{{{N^2}}}\left[ {N\sum\limits_{i = 1}^5 {{f_i}{y_i}^2 – {{\left( {\sum\limits_{i = 1}^5 {{f_i}{y_i}} } \right)}^2}} } \right]$
$=\frac{(10)^{2}}{(50)^{2}}\left[50 \times 68-(10)^{2}\right]$
$=\frac{1}{25}[3400-100]=\frac{3300}{25}$
$=132$
Standard 11
Mathematics
Similar Questions
Find the mean, variance and standard deviation using short-cut method
| Height in cms | $70-75$ | $75-80$ | $80-85$ | $85-90$ | $90-95$ | $95-100$ | $100-105$ | $105-110$ | $110-115$ |
| No. of children | $3$ | $4$ | $7$ | $7$ | $15$ | $9$ | $6$ | $6$ | $3$ |
hard
