Question

The mean of 6 numbers is 6.5 and its variance is 10.25. If 4 numbers are 2, 4, 5 and 7, then find the other two

Answer 1

Hint: We know that mean is the ratio of sum of all the observations to the total number of observations and variance is given by \[{{\sigma }^{2}}=\dfrac{\sum{{{\left( x
\right)}^{2}}}}{n}-{{\left( \overline{x} \right)}^{2}}\]. Using these two definition we can find the two unknown numbers.

Formula used: ${\sigma ^2} = \dfrac{{\sum {{{(x)}^2}} }}{n} - {(\bar x)^2}$

Complete Step by step solution:
Given, that the mean of 6 numbers is \[6.5\].
That is \[\overline{x}=6.5\].
The four numbers are 2, 4, 5 and 7.
Let $a$ and $b$be the two unknown numbers.
Then by definition, we have,
\[\Rightarrow \dfrac{a+b+2+4+5+7}{6}=6.5\]
\[\Rightarrow \dfrac{a+b+18}{6}=6.5\]
\[\Rightarrow a+b+18=6.5\times 6\]
\[\Rightarrow a+b+18=39\]
\[\Rightarrow a+b=39-18\]
\[\Rightarrow a+b=21\,\,\,\,--(1)\]
Also, given the variance is 10.25.
That is \[{{\sigma }^{2}}=10.25\]
Using the formula \[{{\sigma }^{2}}=\dfrac{\sum{{{\left( x
\right)}^{2}}}}{n}-{{\left( \overline{x} \right)}^{2}}\] we have
\[\Rightarrow \dfrac{\left(
{{a}^{2}}+{{b}^{2}}+{{2}^{2}}+{{4}^{2}}+{{5}^{2}}+{{7}^{2}}
\right)}{6}-{{\left( \overline{x} \right)}^{2}}=10.25\]
Substituting the value of mean,
\[\Rightarrow \dfrac{\left( {{a}^{2}}+{{b}^{2}}+4+16+25+49
\right)}{6}-{{\left( 6.5 \right)}^{2}}=10.25\]
\[\Rightarrow \dfrac{\left( {{a}^{2}}+{{b}^{2}}+94
\right)}{6}-42.25=10.25\]
\[\Rightarrow \dfrac{\left( {{a}^{2}}+{{b}^{2}}+94
\right)}{6}=10.25+42.25\]
\[\Rightarrow {{a}^{2}}+{{b}^{2}}+94=52.5\times 6\]
\[\Rightarrow {{a}^{2}}+{{b}^{2}}+94=315\]
\[\Rightarrow {{a}^{2}}+{{b}^{2}}=315-94\]
\[\Rightarrow {{a}^{2}}+{{b}^{2}}=221\,\,\,--(2)\]
Now from equation (1) we have,
\[\Rightarrow a+b=21\]
Squaring on both side we have
\[\Rightarrow {{\left( a+b \right)}^{2}}={{\left( 21
\right)}^{2}}\]
\[\Rightarrow {{a}^{2}}+{{b}^{2}}+2ab=441\]
\[\Rightarrow 221+2ab=441\]
\[\Rightarrow 2ab=441-221\]
\[\Rightarrow 2ab=220\]
\[\Rightarrow ab=110\]
\[\Rightarrow a=\dfrac{110}{b}\,\,\,--(3)\]
Substituting equation (3) in equation (1) we have
\[\Rightarrow \dfrac{110}{b}+b=21\]
\[\Rightarrow \dfrac{110+{{b}^{2}}}{b}=21\]
\[\Rightarrow 110+{{b}^{2}}=21b\]
\[\Rightarrow {{b}^{2}}-21b+110=0\]
Solving the quadratic equation by factorization method,
\[\Rightarrow {{b}^{2}}-10b-11b+110=0\]
\[\Rightarrow b\left( b-10 \right)-11\left( b-10 \right)=0\]
\[\Rightarrow \left( b-10 \right)\left( b-11 \right)=0\]
\[\Rightarrow b=10\] or \[b=11\]
Now from equation (1) we have
\[\Rightarrow a+b=21\]
If \[b=10\], then equation (1) becomes
\[\Rightarrow a+10=21\]
\[\Rightarrow a=11\]
If \[b=11\], then equation (1) becomes
\[\Rightarrow a+11=21\]
\[\Rightarrow a=10\]

Hence, the two numbers are \[\Rightarrow a=11,\,b=10\] or \[\Rightarrow a=10,\,b=11\]

Note: We can also solve the above quadratic equation by quadratic formula or by completing the square method. In all the methods we will get the same answer. It is impossible to solve a problem like this where there are two unknowns by using a single equation. Try to create distinct relationships to solve the values of two unknowns.