If the difference between the standard deviation and variance of a data is 12, then find the sum of the variance and standard deviation of that data.
A. 20
B. 15
C. 18
D. 22
Answer
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Hint: We first find the general form of the standard deviation and variance of n observations. We find the difference and form the equation. We then solve the quadratic equation to find the solution for the standard deviation. At the end we find the sum of the variance and standard deviation.
Complete step by step answer:
For given observations of ${{x}_{i}},i=1\left( 1 \right)n$, sample variance can be expressed as ${{\sigma }^{2}}=\dfrac{1}{n}\sum{{{x}_{i}}^{2}}-{{\left( \overline{x} \right)}^{2}}$ and sample standard deviation can be expressed as \[\sigma =\sqrt{{{\sigma }^{2}}}=\sqrt{\dfrac{1}{n}\sum{{{x}_{i}}^{2}}-{{\left( \overline{x} \right)}^{2}}}\].
So, standard deviation is the square root value of variance.
Let’s take standard deviation as $p$ and variance becomes ${{p}^{2}}$.
It’s given that the difference between the standard deviation and variance of a data is 12.
So, ${{p}^{2}}-p=12$ which gives ${{p}^{2}}-p-12=0$.
We know for a general equation of quadratic $a{{x}^{2}}+bx+c=0$, the value of the roots of $x$ will be $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
In the given equation we have ${{p}^{2}}-p-12=0$. The values of a, b, c is $1,-1,-12$ respectively.
We put the values and get $p$ as \[p=\dfrac{-\left( -1 \right)\pm \sqrt{{{\left( -1 \right)}^{2}}-4\times \left( -12 \right)\times 1}}{2\times 1}=\dfrac{1\pm \sqrt{49}}{2}=\dfrac{1\pm 7}{2}=-3,4\].
So, the value of the standard deviation is $p=4$.
We need to find the sum of the variance and standard deviation of that data which is ${{p}^{2}}+p$.
Putting the value, we get ${{p}^{2}}+p={{4}^{2}}+4=20$.
So, the correct answer is “Option A”.
Note: These all-required attributes are the measurement of dispersion. The sample standard deviation and variance are the parts of dispersion. It is obvious that measures of central tendency and measures of dispersion are both important and complementary. We can have two datasets with the same median or mode, but their spread may be different.
Complete step by step answer:
For given observations of ${{x}_{i}},i=1\left( 1 \right)n$, sample variance can be expressed as ${{\sigma }^{2}}=\dfrac{1}{n}\sum{{{x}_{i}}^{2}}-{{\left( \overline{x} \right)}^{2}}$ and sample standard deviation can be expressed as \[\sigma =\sqrt{{{\sigma }^{2}}}=\sqrt{\dfrac{1}{n}\sum{{{x}_{i}}^{2}}-{{\left( \overline{x} \right)}^{2}}}\].
So, standard deviation is the square root value of variance.
Let’s take standard deviation as $p$ and variance becomes ${{p}^{2}}$.
It’s given that the difference between the standard deviation and variance of a data is 12.
So, ${{p}^{2}}-p=12$ which gives ${{p}^{2}}-p-12=0$.
We know for a general equation of quadratic $a{{x}^{2}}+bx+c=0$, the value of the roots of $x$ will be $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
In the given equation we have ${{p}^{2}}-p-12=0$. The values of a, b, c is $1,-1,-12$ respectively.
We put the values and get $p$ as \[p=\dfrac{-\left( -1 \right)\pm \sqrt{{{\left( -1 \right)}^{2}}-4\times \left( -12 \right)\times 1}}{2\times 1}=\dfrac{1\pm \sqrt{49}}{2}=\dfrac{1\pm 7}{2}=-3,4\].
So, the value of the standard deviation is $p=4$.
We need to find the sum of the variance and standard deviation of that data which is ${{p}^{2}}+p$.
Putting the value, we get ${{p}^{2}}+p={{4}^{2}}+4=20$.
So, the correct answer is “Option A”.
Note: These all-required attributes are the measurement of dispersion. The sample standard deviation and variance are the parts of dispersion. It is obvious that measures of central tendency and measures of dispersion are both important and complementary. We can have two datasets with the same median or mode, but their spread may be different.
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