Question

The diameters of circles $\left( {in{\text{ mm}}} \right)$ drawn in a design are given below:

Diameters	No. of circles
$33 - 36$	$15$
$37 - 40$	$17$
$41 - 44$	$21$
$45 - 48$	$22$
$49 - 52$	$25$

Find the standard deviation.

Answer 1

Hint: Here diameters of circles are given in millimeters and number of circles also given. In this problem we are going to consider the diameter of circles as the class intervals and the number of circles as frequencies and from the class interval we fix a mid-point, from this way students need to precede the problem. Mainly here we are going to find the mean value, variance and standard deviation.

Formula used: Mean $\left( {\overline X } \right) = A + \dfrac{{\sum {{f_i}{y_i}} }}{{\sum {{f_i}} }} \times h$
Variance ${\left( \sigma \right)^2} = \dfrac{{{h^2}}}{{\sum {{f_i}} }}\left[ {\sum {{f_i} \times \sum {{f_i}{y_i}^2} - {{\left( {\sum {{f_i}{y_i}} } \right)}^2}} } \right]$
Where
$A$ = assumed mean,
${f_i}$ = frequency,
${y_i} = \dfrac{{{x_i} - A}}{h}$ and $h$ = class size. Here
$A = 42.5$

Complete step-by-step solution:
Now for the given diameters and frequency we are going to find class interval, mid-point $\left( {{x_i}} \right)$, class size $\left( h \right)$, ${y_i}$, ${y_i}^2$, ${f_i}{y_i}$, ${f_i}{y_i}^2$
Let us consider for the given diameters $33 - 36$ and frequency $15$:
Class interval = $32.5 - 36.5$
Mid-point $\left( {{x_1}} \right) = 34.5$
Class size $\left( h \right) = 36.5 - 32.5 = 4$
${y_1} = \dfrac{{{x_i} - A}}{h}$
Substituting the given values we get,
$ \Rightarrow \dfrac{{34.5 - 42.5}}{4}$
\[{y_1} = - 2\]
${y_1}^2 = 4$, ${f_1}{y_1} = - 30$, ${f_1}{y_1}^2 = 60$
Let us consider for the diameter $37 - 40$ and frequency $17$:
Class interval $ = 36.5 - 40.5$
Mid-point $\left( {{x_2}} \right) = 38.5$
Class size $\left( h \right) = 4$
${y_1} = \dfrac{{{x_i} - A}}{h}$
Substituting the given values we get,
$\Rightarrow$${y_2} = \dfrac{{38.5 - 42.5}}{4}$
Simplifying we get,
$\Rightarrow$${y_2} = - 1$
${y_2}^2 = 1$, ${f_2}{y_2} = - 17$, ${f_2}{y_2}^2 = 17$
For the diameter $41 - 44$ and frequency $21$
Class interval = $40.5 - 44.5$
Mid-point $\left( {{x_3}} \right)$= $42.5$
Class size $\left( h \right)$= $4$
${y_1} = \dfrac{{{x_i} - A}}{h}$
Substituting the given values we get,
$\Rightarrow$${y_3} = \dfrac{{42.5 - 42.5}}{4}$
$\Rightarrow$${y_3} = 0$
${y_3}^2 = 0$, ${f_3}{y_3} = 0$, ${f_3}{y_3}^2 = 0$
For the diameter $45 - 48$ and frequency $22$:
Class interval $ = 44.5 - 48.5$
Mid-point $\left( {{x_4}} \right) = 46.5$
${y_1} = \dfrac{{{x_i} - A}}{h}$
Substituting the given values we get,
${y_1} = \dfrac{{{x_i} - A}}{h}$
Substituting the given values we get,
$\Rightarrow$${y_4} = \dfrac{{46.5 - 42.5}}{4}$
$\Rightarrow$${y_4} = 1$
${y_4}^2 = 1$, ${f_4}{y_4} = 22$, \[{f_4}{y_4}^2 = 22\]
For the diameter $49 - 52$ and frequency $25$:
Class interval $ = 48.5 - 52.5$
Mid-point $\left( {{x_5}} \right) = 50.5$
${y_1} = \dfrac{{{x_i} - A}}{h}$
Substituting the given values we get,
$\Rightarrow$${y_5} = \dfrac{{50.5 - 42.5}}{4}$
$\Rightarrow$${y_5} = 2$
${y_5}^2 = 4$, ${f_5}{y_5} = 50$, ${f_5}{y_5}^2 = 100$
Now $\sum\limits_{i = 1}^5 {{f_i}} = 100$ , $\sum\limits_{i = 1}^5 {{f_i}{y_i}} = 25$, $\sum\limits_{i = 1}^5 {{f_i}{y_i}^2} = 199$.
We know that, Mean $\left( {\overline X } \right)$=$A + \dfrac{{\sum {{f_i}{y_i}} }}{{\sum {{f_i}} }} \times h$
Substituting the values we get,
\[\overline X = 42.5 + \dfrac{{25}}{{100}} \times 4\]
Simplifying we get,
$ \Rightarrow 42.5 + \dfrac{1}{4} \times 4$
Multiplying the terms,
$ \Rightarrow 42.5 + 1$
Hence we get,
$ \Rightarrow 43.5$
$\therefore \overline X = 43.5$
We have to find variance:
We know that, ${\left( \sigma \right)^2} = \dfrac{{{h^2}}}{{\sum {{f_i}} }}\left[ {\sum {{f_i} \times \sum {{f_i}{y_i}^2} - {{\left( {\sum {{f_i}{y_i}} } \right)}^2}} } \right]$
Substitute the values we get,
\[\left( {{\sigma ^2}} \right) = \dfrac{{{4^2}}}{{{{100}^2}}}\left[ {100 \times \left( {199} \right) - {{\left( {25} \right)}^2}} \right]\]
Simplifying we get,
$ \Rightarrow \dfrac{{16}}{{10000}}\left[ {19900 - 625} \right]$
Subtracting we get,
$ \Rightarrow \dfrac{{16}}{{10000}} \times 19275$
Hence we get,
$ \Rightarrow 30.84$
${\left( \sigma \right)^2} = 30.84$
 $\therefore $Standard deviation $\left( \sigma \right) = \sqrt {{\text{variance}}} $
$ \Rightarrow \sigma = \sqrt {30.84} $
Taking square root we get,
$ \Rightarrow 5.55$
$\therefore \sigma = 5.55$

Standard deviation is equal to 5.55.

Note: Standard deviation is a statistic that measures the dispersion of a dataset relative to its mean and is calculated as the square root of the variance. The standard deviation is calculated as the square root of variance by determining each data point’s deviation relative to the mean. If the data points are further from the mean, there is a higher deviation within the data set. Thus, the more spread out the data, the higher the standard deviation.