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.]

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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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