Question

If the standard deviation of the numbers 2, 3, a and 11 is 3.5, then which of the following is true?
(a) $3{{a}^{2}}-26a+55=0$
(b) $3{{a}^{2}}-32a+84=0$
(c) $3{{a}^{2}}-34a+91=0$
(d) $3{{a}^{2}}-23a+44=0$

Answer 1

To solve this problem we first need to know the relation between standard deviation and the variance and also how to calculate the variance. Variance (v) is the square of the Standard deviation (s) and variance of given n terms when mean is $\overline{x}$ is given by, v = $\dfrac{\sum\limits_{i=0}^{i=n}{{{x}_{i}}^{2}}}{n}-{{\left( \overline{x} \right)}^{2}}$, so using this formula and the relation mentioned we will find the value of a.

Complete step by step answer:
We are given that there are 4 terms as 2, 3, a, 11 and their standard deviation is 3.5,
To solve this we need to know the relation between variance and standard deviation first,
Variance (v) is given as the square of the standard deviation (s),
And also variance of n terms is given as,
$\dfrac{\sum\limits_{i=0}^{i=n}{{{x}_{i}}^{2}}}{n}-{{\left( \overline{x} \right)}^{2}}$,
Where $\overline{x}$ represents the mean and ${{x}_{i}}$ is the ${{i}^{th}}$ term.
So first we will have to calculate the mean,
We will do this as follows,
$\begin{aligned} & \overline{x}=\dfrac{sum\,of\,all\,terms}{total\,number\,of\,terms} \\\ & \overline{x}=\dfrac{2+3+a+11}{4} \\\ & \overline{x}=\dfrac{16+a}{4} \\\ \end{aligned}$
Now we will find variance as,
Variance = ${{\left( standard\,deviation \right)}^{2}}$
Variance = ${{\left( 3.5 \right)}^{2}}$
Variance = ${{\left( \dfrac{7}{2} \right)}^{2}}$= $\dfrac{49}{4}\,\,....\left( 1 \right)$
And also
Variance = $\dfrac{\sum\limits_{i=0}^{i=n}{{{x}_{i}}^{2}}}{n}-{{\left( \overline{x} \right)}^{2}}$
Variance = $\dfrac{{{2}^{2}}+{{3}^{2}}+{{a}^{2}}+{{11}^{2}}}{4}-{{\left( \dfrac{16+a}{4} \right)}^{2}}$
Variance = $\dfrac{4+9+{{a}^{2}}+121}{4}-\left( \dfrac{256+{{a}^{2}}+32a}{16} \right)$
Variance = $\dfrac{134+{{a}^{2}}}{4}-\dfrac{\left( 256+{{a}^{2}}+32a \right)}{16}\,\,\,....\left( 2 \right)$
Equating equations 1 and 2, we get
$\dfrac{49}{4}=\dfrac{134+{{a}^{2}}}{4}-\dfrac{\left( 256+{{a}^{2}}+32a \right)}{16}$
Multiplying both sides with 16, we get
$196=536+4{{a}^{2}}-256-{{a}^{2}}-32a$
Further simplifying this we get,
$3{{a}^{2}}-32a+84=0$
Hence we get option (b) as the correct answer.

Note:
Take care of the mistake here of taking square of denominator in the first part of the variance i.e. $\dfrac{\sum\limits_{i=0}^{i=n}{{{x}_{i}}^{2}}}{n}$ but it is simply total number of terms and not square so kindly remember that. Try to show all the steps mentioned in the solution if this problem appears in subjective type otherwise you may lose marks.