Question

Standard deviation for first $10$ natural numbers is
A $5.5$
B $3.87$
C $2.97$
D $2.87$

Answer 1

Hint: We use standard deviation to solve the statistics problems. In statistics, the standard deviation is a number used to tell how measurements for a group are spread out from average or mean value. The low standard deviation means that most of the numbers are close to the average. The high standard deviation means that the numbers are more spread out. The formula to find the standard deviation is $\sigma =\sqrt{\dfrac{\sum{{{\left( {{x}_{i}}-\mu \right)}^{2}}}}{N}}$ where, σ is the population standard deviation, ${{x}_{i}}$ is the each value from population, $N$ is the size of the population and μ is the population mean.

Complete step-by-step solution:
To find the standard deviation of the natural number first we will find the mean of the $10$ natural number.
The formula of mean is $m=\sum{\dfrac{{{x}_{i}}}{N}}$ .
The first $10$ natural number are $1,2,3,4,5,6,7,8,9,10$. Now the mean is:
$\begin{align}
  & \Rightarrow m=\sum{\dfrac{{{x}_{i}}}{N}} \\
 & \Rightarrow m=\dfrac{1+2+3+4+5+6+7+8+9+10}{10} \\
 & \Rightarrow m=5.5 \\
\end{align}$
Now after we will calculate the variance of the $10$ natural numbers. The formula of variance is:
$\Rightarrow \operatorname{var}={{\sigma }^{2}}=\dfrac{\sum{{{\left( {{x}_{i}}-\mu \right)}^{2}}}}{N}$
Now by putting values and subtracting each natural number with $5.5$ then we get,$\begin{align}
  & \Rightarrow \operatorname{var}=\dfrac{{{\left( -4.5 \right)}^{2}}+{{\left( -3.5 \right)}^{2}}++{{\left( -2.5 \right)}^{2}}+{{\left( -1.5 \right)}^{2}}+{{\left( -.5 \right)}^{2}}+{{\left( .5 \right)}^{2}}+{{\left( 1.5 \right)}^{2}}+{{\left( 2.5 \right)}^{2}}+{{\left( 3.5 \right)}^{2}}+{{\left( 4.5 \right)}^{2}}}{10} \\
 & \Rightarrow \operatorname{var}=\dfrac{20.25+12.25+6.252.25+0.25+0.25+2.25+6.2512.25+20.25}{10} \\
 & \\
\end{align}$ Now by adding them we get
$\begin{align}
  & \Rightarrow \operatorname{var}=\dfrac{82.5}{10} \\
 & \Rightarrow \operatorname{var}=8.25 \\
\end{align}$
Now the standard deviation is $\sqrt{\operatorname{var}iance}$
$\Rightarrow \sigma =\sqrt{\operatorname{var}}=\sqrt{8.25}=2.87$
Hence we get the standard deviation of $10$ natural numbers is $2.87$. Option D is correct.

Note: We can also find out the standard deviation of the $10$ natural number by direct applying the formula of standard deviation of $10$ natural numbers which is as,
$\Rightarrow \sigma =\sqrt{\dfrac{{{n}^{2}}-1}{12}}$
Where, $n$ is the size of the population.
Now putting values in the above formula, we get
$\Rightarrow \sigma =\sqrt{\dfrac{{{10}^{2}}-1}{12}}=\sqrt{\dfrac{100-1}{12}}=\sqrt{\dfrac{99}{12}}=2.87$
Hence we get the same answer as we calculated above, which is $\sigma =2.87$ . so option D is correct.