Question

If the mean and standard deviation of 5 observations \[{{x}_{1}},{{x}_{2}},{{x}_{3}},{{x}_{4}},{{x}_{5}}\] are 10 and 3, respectively, then the variance of 6 observations \[{{x}_{1}},{{x}_{2}},.....,{{x}_{5}}\] and -50 is equal to: -
(a) 582.5
(b) 507.5
(c) 586.5
(d) 509.5

Answer 1

Hint: To calculate the variance of 6 observations \[{{x}_{1}},{{x}_{2}},.....,{{x}_{5}}\], use the formula: - \[{{\sigma }^{2}}=\dfrac{1}{n}\sum\limits_{i=1}^{n}{{{\left( {{x}_{i}}-\overline{x} \right)}^{2}}}\], where ‘n’ is the number of terms, ‘\[{{\sigma }^{2}}\]’ is the variance and ‘\[\overline{x}\]’ is the mean of ‘n’ observations. Substitute the value of \[\dfrac{\sum\limits_{i=1}^{5}{{{x}_{i}}}}{5}\] and \[\sqrt{\dfrac{\sum\limits_{i=1}^{5}{{{\left( x-\overline{x} \right)}^{2}}}}{5}}\] using the given information about 5 observations, to get the answer. Here, \[\dfrac{\sum\limits_{i=1}^{5}{{{x}_{i}}}}{5}\] is the mean and \[\dfrac{\sum\limits_{i=1}^{5}{{{\left( x-\overline{x} \right)}^{2}}}}{5}\] is the standard deviation of \[{{x}_{1}},{{x}_{2}},.....,{{x}_{5}}\].

Complete step by step answer:
Here, we have been provided that mean of 5 observations \[{{x}_{1}},{{x}_{2}},.....,{{x}_{5}}\] is 10. Therefore, applying the formula for mean given by \[\overline{x}=\dfrac{\sum\limits_{i=1}^{n}{{{x}_{i}}}}{n}\], where ‘n’ is the number of terms and ‘\[\overline{x}\]’ is the mean, we get,
\[\Rightarrow \overline{x}=\dfrac{1}{5}\times \sum\limits_{i=1}^{5}{{{x}_{i}}=10}\] - (1)
Now, we have also been provided that standard deviation of \[{{x}_{1}},{{x}_{2}},.....,{{x}_{5}}\] is 3. Therefore, applying the formula for standard deviation, given by \[\sqrt{\dfrac{1}{n}\times \sum\limits_{i=1}^{n}{{{\left( {{x}_{1}}-\overline{x} \right)}^{2}}}}\], we get,
Standard deviation = \[\sqrt{\dfrac{1}{5}\times \sum\limits_{i=1}^{5}{{{\left( {{x}_{i}}-\overline{x} \right)}^{2}}}}\]
Substituting the value of \[\overline{x}\] from equation (1), we get,
Standard deviation = \[\sqrt{\dfrac{1}{5}\sum\limits_{i=1}^{5}{{{\left( {{x}_{i}}-10 \right)}^{2}}}}=3\]
\[\Rightarrow \sqrt{\dfrac{1}{5}\sum\limits_{i=1}^{5}{{{\left( {{x}_{i}}-10 \right)}^{2}}}}=3\]
On squaring both sides, we get,
\[\Rightarrow \dfrac{1}{5}\sum\limits_{i=1}^{5}{{{\left( {{x}_{i}}-10 \right)}^{2}}}=9\]
\[\Rightarrow \sum\limits_{i=1}^{5}{{{\left( {{x}_{i}}-10 \right)}^{2}}}=45\] - (2)
Now, we have been asked to find the variance of 6 observations \[{{x}_{1}},{{x}_{2}},.....,{{x}_{5}}\] and -50.
We know that variance of ‘n’ observation is given by the formula: -
\[{{\sigma }^{2}}=\dfrac{1}{n}\sum\limits_{i=1}^{n}{{{\left( {{x}_{i}}-\overline{x} \right)}^{2}}}\], where ‘\[\overline{x}\]’ is the mean of ‘n’ observations and ‘\[{{\sigma }^{2}}\]’ is the notation of variance.
So, we have 6 observations \[{{x}_{1}},{{x}_{2}},.....,{{x}_{5}}\] and -50. Let us calculate its mean first.
\[\Rightarrow \] Mean = \[\dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}+{{x}_{5}}+\left( -50 \right)}{6}\]
This can be written as,
\[\Rightarrow \] Mean = \[\dfrac{\left( \sum\limits_{i=1}^{5}{{{x}_{i}}} \right)-50}{6}\]
Now, substituting the value of \[\sum\limits_{i=1}^{5}{{{x}_{i}}}\] from equation (1), we get,
\[\Rightarrow \] Mean = \[\dfrac{5\times 10-50}{6}=\dfrac{50-50}{6}=0\]
Therefore, the mean of the given 6 observations is 0. So, the expression of variance for these 6 observation becomes,
\[\begin{align}
  & \Rightarrow {{\sigma }^{2}}=\dfrac{1}{6}\sum\limits_{i=1}^{6}{x_{i}^{2}} \\
 & \Rightarrow {{\sigma }^{2}}=\dfrac{1}{6}\left[ x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+x_{5}^{2}+{{\left( -50 \right)}^{2}} \right] \\
 & \Rightarrow {{\sigma }^{2}}=\dfrac{1}{6}\left[ \left( \sum\limits_{i=1}^{5}{x_{i}^{2}} \right)+{{\left( -50 \right)}^{2}} \right] \\
\end{align}\]
\[\Rightarrow {{\sigma }^{2}}=\dfrac{1}{6}\left[ \sum\limits_{i=1}^{5}{x_{i}^{2}}+2500 \right]\] - (3)
Now, let us come back to equation (2), we have,
\[\Rightarrow \sum\limits_{i=1}^{5}{{{\left( {{x}_{i}}-10 \right)}^{2}}}=45\]
Applying the identity, \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\], we get,
\[\Rightarrow \sum\limits_{i=1}^{5}{\left( x_{i}^{2}+{{10}^{2}}-20{{x}_{i}} \right)}=45\]
\[\begin{align}
  & \Rightarrow \sum\limits_{i=1}^{5}{x_{i}^{2}}+\sum\limits_{i=1}^{5}{{{10}^{2}}}-20\sum\limits_{i=1}^{5}{{{x}_{i}}}=45 \\
 & \Rightarrow \sum\limits_{i=1}^{5}{x_{i}^{2}}+\sum\limits_{i=1}^{5}{100}-20\sum\limits_{i=1}^{5}{{{x}_{i}}}=45 \\
\end{align}\]
Using the value of \[\sum\limits_{i=1}^{5}{{{x}_{i}}}\] from equation (1), we get,
\[\begin{align}
  & \Rightarrow \sum\limits_{i=1}^{5}{x_{i}^{2}}+100\times 5-20\times 50=45 \\
 & \Rightarrow \sum\limits_{i=1}^{5}{x_{i}^{2}}+500-1000=45 \\
\end{align}\]
\[\Rightarrow \sum\limits_{i=1}^{5}{x_{i}^{2}}=545\] - (4)
Therefore, substituting the value of \[\sum\limits_{i=1}^{5}{x_{i}^{2}}\] from equation (4) in equation (3), we get,
\[\begin{align}
  & \Rightarrow {{\sigma }^{2}}=\dfrac{1}{6}\left[ 545+2500 \right] \\
 & \Rightarrow {{\sigma }^{2}}=\dfrac{1}{6}\times 3045 \\
 & \Rightarrow {{\sigma }^{2}}=507.5 \\
\end{align}\]

Hence, option (b) is the correct answer.

Note:
One may note that, actually variance is nothing but the square of standard deviation. Standard deviation is denoted by (\[\sigma \]) and variance by (\[{{\sigma }^{2}}\]). In the above question the main problem was to use the given information to find the variance. We were to find the variance of 6 observations in which 5 were the previous observations. So, we formed different equations about these 5 observations and used them for the calculation of variance.