Question

In an experiment with 10 observations on $'x'$, $\sum{x}=60$ and $\sum{{{x}^{2}}=1000}$ one observation that was 20 was found to be wrong and was replaced by the correct value 30 then the correct variance is
(a) 64
(b) 111
(c) 52
(d) 101

Answer 1

We solve this problem first by converting the given data of $\sum{x},\sum{{{x}^{2}}}$ to the correct values after replacing. Then we find the variance of given data by using the formula as
${{\sigma }^{2}}=\dfrac{\sum{{{x}^{2}}}}{N}-{{\left( \dfrac{\sum{x}}{N} \right)}^{2}}$
Where, $N$ is total number of observations given as 10

Complete step-by-step solution:
We are given that for the experiment of 10 observations
$\sum{x}=60$
We are given that the number 20 was replaced by 30
Now, let us assume that the summation of numbers of correct observations as $\sum{X}$ then we say the subtracting 20 from old summation and subtracting 30 from new observations have the same value that is
$\Rightarrow \sum{X}-30=\sum{x}-20$
Now by substituting $\sum{x}=60$ in above equation we get

& \Rightarrow \sum{X}=60-20+30 \\\ & \Rightarrow \sum{X}=70 \\\ \end{aligned}$$ Now, we are given that for an experiment of 10 observations we have $$\sum{{{x}^{2}}}=1000$$ We are given that the number 20 was replaced by 30 Now let us assume the correct summation as $$\sum{{{X}^{2}}}$$ then we can say that subtracting square of 30 from new summation and subtracting square of 20 from old summation gives the same value that is $$\Rightarrow \sum{{{X}^{2}}}-{{30}^{2}}=\sum{{{x}^{2}}}-{{20}^{2}}$$ Now, by substituting $$\sum{{{x}^{2}}}=1000$$ in above equation we get $$\begin{aligned} & \Rightarrow \sum{{{X}^{2}}}=1000-400+900 \\\ & \Rightarrow \sum{{{X}^{2}}}=1500 \\\ \end{aligned}$$ Let us assume the number of observations as $$\Rightarrow N=10$$ We know that the formula of variance is given as $${{\sigma }^{2}}=\dfrac{\sum{{{x}^{2}}}}{N}-{{\left( \dfrac{\sum{x}}{N} \right)}^{2}}$$ Now by substituting the required values in the above equation for the correct observations we get $$\begin{aligned} & \Rightarrow {{\sigma }^{2}}=\dfrac{\sum{{{X}^{2}}}}{N}-{{\left( \dfrac{\sum{X}}{N} \right)}^{2}} \\\ & \Rightarrow {{\sigma }^{2}}=\dfrac{1500}{10}-{{\left( \dfrac{70}{10} \right)}^{2}} \\\ & \Rightarrow {{\sigma }^{2}}=150-49 \\\ & \Rightarrow {{\sigma }^{2}}=101 \\\ \end{aligned}$$ Therefore, we can see that the variance of correct data is 101 **So, option (d) is the correct answer.** **Note:** Students may do mistake in the formula of variance that is the formula of variance is given as $${{\sigma }^{2}}=\dfrac{\sum{{{x}^{2}}}}{N}-{{\left( \dfrac{\sum{x}}{N} \right)}^{2}}$$ But the students may take the formula as $$\sigma =\sqrt{\dfrac{\sum{{{x}^{2}}}}{N}-{{\left( \dfrac{\sum{x}}{N} \right)}^{2}}}$$ Here, $$'\sigma '$$ is the standard deviation from mean and $$'{{\sigma }^{2}}'$$ is the variance. Both are different. This part needs to be taken care of because both the mean deviation and the variance are denoted by the same symbol but there is a square which is important to differentiate the both values.