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What are the mean, median, mode, variance and standard deviation of $\left\{ 4,6,7,5,9,4,3,4 \right\}$?

Answer
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Hint: In this problem we need to calculate the mean, median, mode, variance and standard deviation of given data. We know that mean is the average of the given data which is calculated by taking the ratio of the sum of the observations to the number of observations. So, we will calculate the sum of observations and the number of observations after that we can find the mean. Median is the middle most term of the given data. For this we are going to write the given data in ascending order or descending order and calculate the middle term. To find the mode we need to observe the given data and how many times that each variable is repeated. The most repeated variable is taken as mode. We have the formula for the variance as ${{s}^{2}}=\dfrac{\sum{{{\left( x-\bar{x} \right)}^{2}}}}{n}$. From the above formula we can calculate the variance and to find the standard deviation we will calculate the square root of the variance.

Complete step by step answer:
Given data, $\left\{ 4,6,7,5,9,4,3,4 \right\}$.
Sum of the observations in the given data is given by
$sum=4+6+7+5+9+4+3+4$
Simplifying the above equation, then we will get
$sum=42$
Number of observations we have in the given data is $8$.
Hence the mean is calculated as
$\begin{align}
  & \bar{x}=\dfrac{42}{8} \\
 & \Rightarrow \bar{x}=5.25 \\
\end{align}$
The ascending order of the given data will be $\text{3,4,4,4,5,6,7,9}$.
The number of observations in the given data is $8$ which is an even number, so the median of the given data is given by
$median=\dfrac{{{\left( \dfrac{n}{2} \right)}^{th}}\text{ term}+{{\left( \dfrac{n}{2}+1 \right)}^{th}}\text{ term}}{2}$
Substituting $n=8$ in the above equation, then we will get
$\begin{align}
  & median=\dfrac{{{\left( \dfrac{8}{2} \right)}^{th}}\text{ term}+{{\left( \dfrac{8}{2}+1 \right)}^{th}}\text{ term}}{2} \\
 & \Rightarrow median=\dfrac{{{4}^{th}}\text{ term}+{{\text{5}}^{th}}\text{ term}}{2} \\
\end{align}$
In the given data we have $4,5$ as ${{4}^{th}}$ and ${{5}^{th}}$ terms respectively. Substituting these values in the above equation, then we will get
$\begin{align}
  & median=\dfrac{4+5}{2} \\
 & \Rightarrow median=\dfrac{9}{2} \\
 & \Rightarrow median=4.5 \\
\end{align}$
On observing the given data which is $\text{3,4,4,4,5,6,7,9}$. We can say that the term $4$ is repeated three times which is highest in the data. So, the Mode of the given data is $4$.
We have the formula for the variance as ${{s}^{2}}=\dfrac{\sum{{{\left( x-\bar{x} \right)}^{2}}}}{n}$. Substituting the all the variables in the above equation, then we will get
${{s}^{2}}=\dfrac{{{\left( 3-5.25 \right)}^{2}}+{{\left( 4-5.25 \right)}^{2}}+{{\left( 4-5.25 \right)}^{2}}+{{\left( 4-5.25 \right)}^{2}}++{{\left( 5-5.25 \right)}^{2}}+{{\left( 6-5.25 \right)}^{2}}+{{\left( 7-5.25 \right)}^{2}}+{{\left( 9-5.25 \right)}^{2}}}{8}$
Simplifying the values in the above equation, then we will get
$\begin{align}
  & {{s}^{2}}=\dfrac{{{\left( -2.25 \right)}^{2}}+{{\left( -1.25 \right)}^{2}}+{{\left( -1.25 \right)}^{2}}+{{\left( -1.25 \right)}^{2}}+{{\left( -0.25 \right)}^{2}}+{{\left( 0.75 \right)}^{2}}+{{\left( 1.75 \right)}^{2}}+{{\left( 3.75 \right)}^{2}}}{8} \\
 & \Rightarrow {{s}^{2}}=\dfrac{5.0625+1.5625+1.5625+1.5625+0.0625+0.5625+3.0625+14.0625}{8} \\
 & \Rightarrow {{s}^{2}}=\dfrac{27.5}{8} \\
 & \Rightarrow {{s}^{2}}=3.4375 \\
\end{align}$
Now the standard deviation is calculated by taking square root for the above calculated variance value, then we will have
$\begin{align}
  & s=\sqrt{3.4375} \\
 & \Rightarrow s=1.854 \\
\end{align}$

Hence the mean, median, mode, variance and standard deviation of the given data are $5.25$, $4.5$, $4$, $3.4375$ and $1.854$ respectively.

Note: This type of problems consists of a scientific calculation like dealing with decimals and squares of the decimal values. So, it is better to have a calculator or good presence of mind to deal with decimals. Since the parameters like variance, standard deviations are depending on the value of mean. If you get the mean as wrong then the whole problem will be wrong.