The diameters of circles (in mm) drawn in a design are given below:
Diameters | $33-36$ | $37-40$ | $41-44$ | $45-48$ | $49-52$ |
No. of circles | $15$ | $17$ | $21$ | $22$ | $25$ |
Calculate the standard deviation and mean diameter of the circles.
[ Hint : First make the data continuous by making the classes as $32.5-36.5,36.5-40.5,$ $40.5-44.5,44.5-48.5,48.5-52.5 $ and then proceed.]
Class Interval |
Frequency ${f_i}$ |
Mid=point ${x_i}$ |
${y_i} = \frac{{{x_i} - 42.5}}{4}$ | ${f_i}^2$ | ${f_i}{y_i}$ | ${f_i}{y_i}^2$ |
$33-36$ | $15$ | $34.5$ | $-2$ | $4$ | $-30$ | $60$ |
$37-40$ | $17$ | $38.5$ | $-1$ | $1$ | $-17$ | $17$ |
$41-44$ | $21$ | $42.5$ | $0$ | $0$ | $0$ | $0$ |
$45-48$ | $22$ | $46.5$ | $1$ | $1$ | $22$ | $22$ |
$49-52$ | $25$ | $50.5$ | $2$ | $4$ | $50$ | $100$ |
$100$ | $25$ | $199$ |
here, $N=100,$ $h=4$
Let the assumed mean, $A,$ be $42.5$
Mean, $\bar x = A + \frac{{\sum\limits_{i = 1}^5 {{f_i}{y_i}} }}{N} \times h$
$ = 42.5 + \frac{{25}}{{100}} \times 4 = 43.5$
Variance, $\left( {{\sigma ^2}} \right) = \frac{{{h^2}}}{{{N^2}}}\left[ {N\sum\limits_{i = 1}^5 {{f_i}{y_i}^2 - {{\left( {\sum\limits_{i = 1}^5 {{f_i}{y_i}} } \right)}^2}} } \right]$
$=\frac{16}{10000}\left[100 \times 199-(25)^{2}\right]$
$=\frac{16}{10000}[19900-625]$
$=\frac{16}{10000} \times 19275$
$=30.84$
$\therefore$ Standard deviation $(\sigma)=5.55$
Find the mean and variance of the frequency distribution given below:
$\begin{array}{|l|l|l|l|l|} \hline x & 1 \leq x<3 & 3 \leq x<5 & 5 \leq x<7 & 7 \leq x<10 \\ \hline f & 6 & 4 & 5 & 1 \\ \hline \end{array}$
The variance $\sigma^2$ of the data is $ . . . . . .$
$x_i$ | $0$ | $1$ | $5$ | $6$ | $10$ | $12$ | $17$ |
$f_i$ | $3$ | $2$ | $3$ | $2$ | $6$ | $3$ | $3$ |
If the mean of the frequency distribution
Class: | $0-10$ | $10-20$ | $20-30$ | $30-40$ | $40-50$ |
Frequency | $2$ | $3$ | $x$ | $5$ | $4$ |
is $28$ , then its variance 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