Question

Find the standard deviation of 40,42, and 48. If each value is multiplied by 3, find the standard deviation of the new data.

Answer 1

First of all, calculate the mean of 40, 42, and 48 using the formula, $Mean\left( \mu \right)=\dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{n}$ . Here, the value of n is 3. Now, calculate $\sum{\left( x-\mu \right)}$ and ${{\sum{\left( x-\mu \right)}}^{2}}$ . Use the formula, $\sigma =\sqrt{\dfrac{\sum{{{\left( x-\mu \right)}^{2}}}}{n}-{{\left( \dfrac{\sum{\left( x-\mu \right)}}{n} \right)}^{2}}}$ and calculate the standard deviation. Now, multiply by 3 in 40, 42, and 48. Similarly, calculate the mean of 120, 126, and 144 using the formula, $Mean\left( \mu \right)=\dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{n}$ . Here, the value of n is 3. Now, calculate $\sum{\left( x-\mu \right)}$ and ${{\sum{\left( x-\mu \right)}}^{2}}$ . Use the formula, $\sigma =\sqrt{\dfrac{\sum{{{\left( x-\mu \right)}^{2}}}}{n}-{{\left( \dfrac{\sum{\left( x-\mu \right)}}{n} \right)}^{2}}}$ and calculate the standard deviation.

Complete step by step answer:
According to the question,
In the ${{1}^{st}}$ case, we have,
The number of data, $n$ = 3 ………………………………………..(1)
The first numeric data = 40 ………………………………………….(2)
The second numeric data = 42 ………………………………………….(3)
The third numeric data = 48 ………………………………………….(4)
First of all, let us calculate the mean of the above data.
We know the formula for the mean of ${{x}_{1}},{{x}_{2}},$ and ${{x}_{3}}$ , $Mean\left( \mu \right)=\dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{n}$ ………………………………………(5)
Now, from equation (1), equation (2), equation (3), equation (4), and equation (5), we get
$Mean\left( \mu \right)=\dfrac{40+42+48}{3}=\dfrac{130}{3}=43.34$ ……………………………………………..(6)
Here, on arranging the data in the table below, we get

$x$	$Mean\left( \mu \right)$	$\left( x-\mu \right)$	${{\left( x-\mu \right)}^{2}}$
40	43.34	-3.34	11.1556
42	43.34	-1.34	1.7956
48	43.34	4.66	21.7156

$\sum{\left( x-\mu \right)=\left( -3.34 \right)+\left( -1.34 \right)+4.66=}-0.02$ ………………………………………………(7)
$\sum{{{\left( x-\mu \right)}^{2}}=11.1556+1.7956+21.7156=}34.6668$ ………………………………………………..(8)
We know the formula for the standard deviation, $\sigma =\sqrt{\dfrac{\sum{{{\left( x-\mu \right)}^{2}}}}{n}-{{\left( \dfrac{\sum{\left( x-\mu \right)}}{n} \right)}^{2}}}$ ……………………………………….(9)
Now, from equation (7), equation (8), and equation (9), we get

& \Rightarrow \sigma =\sqrt{\dfrac{\sum{{{\left( x-\mu \right)}^{2}}}}{n}-{{\left( \dfrac{\sum{\left( x-\mu \right)}}{n} \right)}^{2}}} \\\ & \Rightarrow \sigma =\sqrt{\dfrac{34.668}{3}-{{\left( \dfrac{-0.02}{3} \right)}^{2}}} \\\ & \Rightarrow \sigma =\sqrt{11.556-0.000045} \\\ & \Rightarrow \sigma =11.555955 \\\ \end{aligned}$$ $$\Rightarrow \sigma =3.399$$ ………………………………………………(10) In the $${{2}^{nd}}$$ case, On multiplying by 3, we get The number of data, $$n$$ = 3 ………………………………………..(11) The first numeric data = 120 ………………………………………….(12) The second numeric data = 126 ………………………………………….(13) The third numeric data = 144 ………………………………………….(14) Now, from equation (5), equation (11), equation (12), equation (13), and equation (14), we get $$Mean\left( \mu \right)=\dfrac{120+126+144}{3}=130$$ …………………………………………..(15) Here, on arranging the data in the table below, we get $$x$$ | $$Mean\left( \mu \right)$$| $$\left( x-\mu \right)$$ | $${{\left( x-\mu \right)}^{2}}$$ ---|---|---|--- 120| 130| -10| 100 126| 130| -4| 16 144| 130| 14| 196 $$\sum{\left( x-\mu \right)=-10+\left( -4 \right)+\left( 14 \right)=}0$$ ………………………………………….(16) $$\sum{{{\left( x-\mu \right)}^{2}}=100+16+196=312}$$ ……………………………………………(17) Now, from equation (9), equation (16), and equation (17), we get $$\sigma =\sqrt{\dfrac{\sum{{{\left( x-\mu \right)}^{2}}}}{n}-{{\left( \dfrac{\sum{\left( x-\mu \right)}}{n} \right)}^{2}}}$$ $$\begin{aligned} & \Rightarrow \sigma =\sqrt{\dfrac{312}{3}-{{\left( \dfrac{0}{3} \right)}^{2}}} \\\ & \Rightarrow \sigma =\sqrt{104} \\\ \end{aligned}$$ $$\Rightarrow \sigma =10.198$$ ……………………………………(18) From equation (10) and equation (18), we have the standard deviation for the first case and second case respectively. **Hence, the standard deviation of 40, 42, and 48 is 3.399, and after multiplying by 3 the standard deviation is 10.198.** **Note:** We can also solve this question by using a shortcut method. We know the property that if the standard deviation of the numbers $$a,b,\,\,\text{and}\,c$$ is $$d$$ . Then, on multiplying the numbers $$a,b,\,\,\text{and}\,c$$ by $$p$$, the standard deviation will be $$pd$$ . The standard deviation of 40, 42, and 48 is 3.399. Since the numbers 120, 126, and 144 are three times 40, 42, and 48. Therefore, the standard deviation of 40,42, and 48 is $$3\times 3.399=10.198\left( approx \right)$$ .