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.]
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 $(\sigma)=5.55$
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