Question

The sum of 100 observations and the sum of their squares are 400 and 2475, respectively. Later on, three observations, 3,4 and 5, were found to be incorrect. If the incorrect observations are omitted, then the variance of the remaining observations is.
(a) 8.00
(b) 8.50
(c) 8.25
(d) 9.00

Answer 1

It is given to us that there are 100 observations and the sum of the values and sum of the squares of their values are given to us. Now, if we remove three observations, the number of observations also decreases by three, and the sum of the values of the observations and sum of the squares of the values of the observations also decrease. We will find decreased sums. From this information we will find the mean of the observations and after finding mean, we can find the variation as $\dfrac{\sum{{{X}^{2}}}}{N}-{{\left( \overline{X} \right)}^{2}}$, where $\overline{X}$ is the mean of the data set.

Complete step-by-step solution:
It is given to us that there are 100 observations and the sum of the values of the observation is 400. We are also given the sum of the squares of the values of the observations as 2475.
Let ${{x}_{1}},{{x}_{2}},....,{{x}_{100}}$ be the observations.
$\begin{aligned} & \Rightarrow \sum\limits_{i=1}^{100}{{{x}_{i}}}=400 \\\ & \Rightarrow \sum\limits_{i=1}^{100}{{{x}_{i}}^{2}=2475} \\\ \end{aligned}$
We know that the mean is of a data set is defined as the quotient of the sum of the values of all the observations and number of observations.
Therefore, if there are 100 observations and the sum of the values of the observations is 400, the mean of the data set will be the quotient of 400 and 100. That is 4.
But we are given that the 3 observations are not correct and their values are 3, 4 and 5. These observations are removed from the data set.
The number of observations in the modified data set will thus be 100 – 3 = 97.
As these observations are removed, the modified sum of the observations will be
$\Rightarrow \sum\limits_{i=1}^{97}{{{x}_{i}}}=$ 400 – 3 – 4 – 5 = 388.
Therefore, the mean of the modified data set will be given as follows:
$\begin{aligned} & \overline{X}=\dfrac{\sum\limits_{i=1}^{97}{{{x}_{i}}}}{97} \\\ & =\dfrac{388}{97} \\\ & =4 \\\ \end{aligned}$.
The sum of the squares of the values of the observations will also decrease. It will be as follows:
$\begin{aligned} & \Rightarrow \sum{{{x}^{2}}=2475-{{\left( 3 \right)}^{2}}-{{\left( 4 \right)}^{2}}-{{\left( 5 \right)}^{2}}} \\\ & \Rightarrow \sum{{{x}^{2}}=2425} \\\ \end{aligned}$
We know that the variance in a data set is given as $\dfrac{\sum{{{X}^{2}}}}{N}-{{\left( \overline{X} \right)}^{2}}$
$\begin{aligned} & \Rightarrow \text{Var}\left( X \right)=\dfrac{2425}{97}-{{\left( 4 \right)}^{2}} \\\ & \Rightarrow \text{Var}\left( X \right)=25-16 \\\ & \Rightarrow \text{Var}\left( X \right)=9 \\\ \end{aligned}$
Hence, option (d) is the correct option.

Note: The variance of a data set is defined as $\text{Var}\left( X \right)=\dfrac{{{\sum{\left( {{x}_{i}}-\overline{X} \right)}}^{2}}}{N}$. We will solve the parenthesis. $\dfrac{\sum{\left( {{x}_{i}}^{2}-2{{x}_{i}}\overline{X}+{{\overline{X}}^{2}} \right)}}{N}$ . Then we split the denominator. $\dfrac{\sum{\left( {{x}_{i}}^{2} \right)}}{N}-\dfrac{2\sum{{{x}_{i}}\overline{X}}}{N}+\dfrac{\sum{{{\overline{X}}^{2}}}}{N}=\dfrac{\sum{\left( {{x}_{i}}^{2} \right)}}{N}-2{{\overline{X}}^{2}}+\dfrac{N{{\overline{X}}^{2}}}{N}$. Thus, we get the modified formula for variance as $\text{Var}\left( X \right)=\dfrac{\sum{\left( {{x}_{i}}^{2} \right)}}{N}-{{\overline{X}}^{2}}$ .