Question

Find the mean, variance and standard deviation using a short-cut method.

Height (in cm)	No. of children
$70 - 75$	$3$
$75 - 80$	$4$
$80 - 85$	$7$
$85 - 90$	$7$
$90 - 95$	$15$
$95 - 100$	$9$
$100 - 105$	$6$
$105 - 110$	$6$
$110 - 115$	$3$

Answer 1

Hint: In this question, we have been asked to find mean, standard deviation and variance using a short-cut method. This method is also known as the assumed mean method. First, we will start by calculating mean. To find mean, calculate the mid points of the interval, then assume one of the mid-points as mean. Using this assumed mean, calculate ${y_i}$ and then, ${f_i}{y_i}$. Put all the values in the formula and simplify to find mean.
Next, we have to calculate variance. As we have already calculated ${y_i}$, use this to find ${y_i}^2$ and ${\left( {{f_i}{y_i}} \right)^2}$. Put the values in the formula and find variance.
Using variance, we will find standard deviation. Find the square root of the variance and that will be our required answer.

Formula used: 1) ${y_i} = \dfrac{{{x_i} - A}}{h}$
2) Mean = $A + \dfrac{{\sum {{f_i}{y_i}} }}{{\sum f }} \times h$
3) Variance, ${\sigma ^2} = \dfrac{{{h^2}}}{{{N^2}}}\left[ {N\sum {{f_i}{y_i}^2 - {{\left( {\sum {{f_i}{y_i}} } \right)}^2}} } \right]$
4) Standard deviation $\left( \sigma \right) = \sqrt {{\sigma ^2}} = \sqrt {{\text{Variance}}} $

Complete step-by-step solution:
We are given a table with observations and frequency and we have been asked to find mean, variance and standard deviation, but using a short-cut method.
Let’s start by calculating mean. For this, we have to first calculate the mid-points. Then, we have to assume a mean out of these mid-points. Let us take our assumed mean ($A$) to be $92.5$. After this, we have to calculate ${y_i}$. $\left( {{y_i} = \dfrac{{{x_i} - A}}{h}} \right)$, where $h$ is the class interval. $h = 75 - 70 = 5$. Using ${y_i}$, calculate ${f_i}{y_i}$.

Height (in cm)	No. of children (${f_i}$)	Mid-point $({x_i})$	${y_i} = \dfrac{{{x_i} - A}}{h}$	${f_i}{y_i}$	${y_i}^2$	${f_i}{y_i}^2$
$70 - 75$	$3$	$72.5$	$ - 4$	$ - 12$	$16$	$48$
$75 - 80$	$4$	$77.5$	$ - 3$	$ - 12$	$9$	$36$
$80 - 85$	$7$	$82.5$	$ - 2$	$ - 14$	$4$	$28$
$85 - 90$	$7$	$87.5$	$ - 1$	$ - 7$	$1$	$7$
$90 - 95$	$15$	$92.5$$ = A$	$0$	$0$	$0$	$0$
$95 - 100$	$9$	$97.5$	$1$	$9$	$1$	$9$
$100 - 105$	$6$	$102.5$	$2$	$12$	$4$	$24$
$105 - 110$	$6$	$107.5$	$3$	$18$	$9$	$54$
$110 - 115$	$3$	$112.5$	$4$	$12$	$16$	$48$
	$\sum {{f_i} = 60} $			$\sum {{f_i}{y_i} = 6} $		$\sum {{f_i}{y_i}^2 = 254} $

After we have calculated all that we need, let’s calculate arithmetic mean using the short cut method.
Mean = $A + \dfrac{{\sum {{f_i}{y_i}} }}{{\sum f }} \times h$.
Putting all the values in the formula,
Mean = $92.5 + \dfrac{6}{{60}} \times 5$
Simplifying we get,
$ \Rightarrow 92.5 + \left( {\dfrac{1}{2}} \right)$
$ \Rightarrow 93$
$\therefore $ The mean = $93$
Now, let us calculate variance using a short-cut method. $\left( {{\sigma ^2} = \dfrac{{{h^2}}}{{{N^2}}}\left[ {N\sum {{f_i}{y_i}^2 - {{\left( {\sum {{f_i}{y_i}} } \right)}^2}} } \right]} \right)$
In this formula, $h$ = class interval and $N = \sum {{f_i}} $. First, we have to calculate ${f_i}{y_i}^2$. Refer to the table shown above. Putting all the values,
$ \Rightarrow {\sigma ^2} = \dfrac{{{5^2}}}{{{{60}^2}}}\left[ {60 \times 254 - {6^2}} \right]$
Simplifying,
$ \Rightarrow {\sigma ^2} = \dfrac{{25}}{{3600}}\left( {15240 - 36} \right)$
Subtracting the terms,
$ \Rightarrow {\sigma ^2} = \dfrac{{25}}{{3600}} \times 15204$
Solving we get,
$ \Rightarrow {\sigma ^2} = 105.58$
$\therefore $ The variance = $105.58$
Now, let us calculate standard deviation.
Standard deviation $\left( \sigma \right) = \sqrt {{\sigma ^2}} = \sqrt {{\text{Variance}}} $
$ \Rightarrow \sigma = \sqrt {105.58} = 10.27$

$\therefore $ The standard deviation $\left( \sigma \right) = 10.27$

Note: Standard deviation looks at how spread out a group of numbers is from the mean, by looking at the square root of the variance. The variance measures the average degree to which each point differs from the mean-the average of all data points. The two concepts are useful and significant for traders, who use them to measure market volatility.