Question

The standard deviation of 10, 16, 10, 16, 10, 10, 16, 16 is?
A) 4
B) 6
C) 3
D) 0

Answer 1

Here we have been asked to find the standard deviation of an ungrouped data. The standard deviation is a summary measure of the differences of each observation from the mean.

& \sigma =\sqrt{\dfrac{\sum{{{\left( x-\overline{x} \right)}^{2}}}}{n}} \\\ & \\\ \end{aligned}$$ Here , $$\begin{aligned} & \sigma =\text{Standard Deviation} \\\ & \overline{x}=\text{mean} \\\ & \text{n = number of data values} \\\ & \text{x = the data values} \\\ & \sum{x}=\text{ Summation of all the data values} \\\ & \sum{{{\left( x-\overline{x} \right)}^{2}}=\text{ Subtract the mean from each data value,square and finally add up the resulting values}} \\\ \end{aligned}$$ An alternative, yet equivalent formula, which is often easier to use is $${{\sigma }^{2}}=\dfrac{\sum{{{x}^{2}}}}{n}-{{\left( \overline{x} \right)}^{2}}$$ **Complete step by step answer:** Firstly, to find the value of Standard Deviation let us find the mean of the given data The data values given to us are 10, 16, 10, 16, 10, 10, 16, 16 Number of data values= n= 8 It would also be easy if solved using a tabular form Data values (x)| $${{x}^{2}}$$ ---|--- 10| 100 16| 256 10| 100 16| 256 10| 100 10| 100 16| 256 16| 256 ∑x=10+16+10+16+10+10+16+16=104| $$\sum{{{x}^{2}}}$$ =1424 $$\begin{aligned} & Mean=\dfrac{\sum{x}}{n}=\dfrac{10+16+10+16+10+10+16+16}{8}=\dfrac{104}{8}=13 \\\ & \overline{x}=13 \\\ \end{aligned}$$ Now let us find the value of summation of squares of the given data values $$\begin{aligned} & \sum{{{\left( x \right)}^{2}}={{\left( 10 \right)}^{2}}+{{\left( 16 \right)}^{2}}+{{\left( 10 \right)}^{2}}+{{\left( 16 \right)}^{2}}+{{\left( 10 \right)}^{2}}+{{\left( 10 \right)}^{2}}+{{\left( 16 \right)}^{2}}+{{\left( 16 \right)}^{2}}} \\\ &\Rightarrow \sum{{{\left( x \right)}^{2}}=4\left( 100+256 \right)=1424} \\\ \end{aligned}$$ Substituting the values in the given formula $${{\sigma }^{2}}=\dfrac{\sum{{{x}^{2}}}}{n}-{{\left( \overline{x} \right)}^{2}}$$ $$\begin{aligned} &\Rightarrow \sum{{{x}^{2}}=1424} \\\ &\Rightarrow n=8 \\\ &\Rightarrow \overline{x}=13 \\\ &\Rightarrow {{\sigma }^{2}}=\dfrac{1424}{8}-{{\left( 13 \right)}^{2}} \\\ &\Rightarrow {{\sigma }^{2}}=\dfrac{1424}{8}-169 \\\ &\Rightarrow {{\sigma }^{2}}=178-169=9 \\\ &\Rightarrow {{\sigma }^{2}}=9 \\\ &\Rightarrow \sigma =\sqrt{9}=3 \\\ &\Rightarrow \sigma =3 \\\ \end{aligned}$$ **Note:** This problem can also be solved using the formula$$\begin{aligned} & \sigma =\sqrt{\dfrac{\sum{{{\left( x-\overline{x} \right)}^{2}}}}{n}} \\\ & \\\ \end{aligned}$$, but here first we have to find the deviation of all the data values from the mean and then add their squares. It would be easier if we write down all the values and calculate using a tabular form Data values (x)| $$x-\overline{x}$$ | $${{\left( x-\overline{x} \right)}^{2}}$$ ---|---|--- 10| $$10-13=-3$$ | 9 16| $$16-13=3$$| 9 10| $$10-13=-3$$| 9 16| $$16-13=3$$| 9 10| $$10-13=-3$$| 9 10| $$10-13=-3$$| 9 16| $$16-13=3$$| 9 16| $$16-13=3$$| 9 | | $$\sum{{{\left( x-\overline{x} \right)}^{2}}=9\times 72}$$ Substituting the values in the formula $$\sigma =\sqrt{\dfrac{72}{8}}=\sqrt{9}=3$$