Question

The mean of the numbers $a,b,8,5,10$ is $6$ and the variance is $6.80$ . Then which one of the following gives possible values of ‘a’ and ‘b’?
(A) $a = 0,b = 7$
(B) $a = 5,b = 2$
(C) $a = 1,b = 6$
(D) $a = 3,b = 4$

Answer 1

Hint: Use the definition of mean to establish a linear relationship between ‘a’ and ‘b’. And then use the formula for a variance to form a quadratic expression. Use the linear relation to eliminate one unknown from this equation. You’ll get a quadratic equation in a single variable. Now find roots to find one unknown. Then use the linear relation to figure out the other. This will give you both the values and then match the options.

Complete step-by-step answer:
Here in this problem, a group of numbers is given as $a,b,8,5,10$ , with a given mean of $6$ and the variance is given as $6.80$. With this given information we need to find the unknown values of the term ‘a’ and ‘b’
Before starting with the solution, we should first understand the concepts of mean and variance. Mean, in mathematics, is the average of all the numbers present in the set or calculated “central” value of a set of numbers. The mean is calculated by dividing the sum of the numbers by the total number of quantities.
$ \Rightarrow Mean{\text{ = }}\dfrac{{{\text{Sum of all the numbers }}}}{{{\text{Number of terms}}}}$
$ = \dfrac{{\sum {{x_i}} }}{N}{\text{ ; where }}\sum {{x_i}} {\text{ represents sum of all numbers, i}}{\text{.e}}{\text{. }}{x_1},{x_2},{x_3}......{\text{ }}$and N is the number of terms.
In probability theory and statistics, the variance is the expectation of the squared deviation of a random variable from its mean. Informally, it measures how far a set of numbers is spread out from their average value.
$ \Rightarrow Variance = \dfrac{{{\text{Sum of the square of difference of each term and mean}}}}{{{\text{Number of terms in the group}}}} = \dfrac{{\sum {{{\left( {{x_i} - \bar x} \right)}^2}} }}{N}$
Where $\bar x$ represents the mean of the set and $\left( {{x_i} - \bar x} \right)$ represents the difference of each term with the mean.
Let’s use the mean formula to establish a relationship by substituting the known values:
$ \Rightarrow Mean{\text{ = }}\dfrac{{{\text{Sum of all the numbers }}}}{{{\text{Number of terms}}}} \Rightarrow 6 = \dfrac{{a + b + 8 + 5 + 10}}{5}$
Now we can transpose the denominator from RHS to LHS, and further solve the above relation:
$ \Rightarrow 6 = \dfrac{{a + b + 8 + 5 + 10}}{5} \Rightarrow 6 \times 5 = a + b + 23$
On transforming the equation, we can write the value of ‘b’ in terms of ‘a’:
$ \Rightarrow 6 \times 5 = a + b + 23 \Rightarrow b = 30 - 23 - a \Rightarrow b = 7 - a$ ............(i)
Now let’s use the formula for a variance to find another relationship by substituting the known values.
$ \Rightarrow Variance = \dfrac{{\sum {{{\left( {{x_i} - \bar x} \right)}^2}} }}{N} \Rightarrow 6.8 = \dfrac{{{{\left( {a - 6} \right)}^2} + {{\left( {b - 6} \right)}^2} + {{\left( {8 - 6} \right)}^2} + {{\left( {5 - 6} \right)}^2} + {{\left( {10 - 6} \right)}^2}}}{5}$
We can further simplify this by transposing and squaring the terms of the difference:
$ \Rightarrow 6.8 = \dfrac{{{{\left( {a - 6} \right)}^2} + {{\left( {b - 6} \right)}^2} + {{\left( 2 \right)}^2} + {{\left( { - 1} \right)}^2} + {{\left( 4 \right)}^2}}}{5} \Rightarrow 6.8 \times 5 = {\left( {a - 6} \right)^2} + {\left( {b - 6} \right)^2} + 4 + 1 + 16$
This can be further solved as:
$ \Rightarrow 6.8 \times 5 = {\left( {a - 6} \right)^2} + {\left( {b - 6} \right)^2} + 4 + 1 + 16 \Rightarrow {\left( {a - 6} \right)^2} + {\left( {b - 6} \right)^2} = 34 - 21 = 13$ ..........(ii)
Now we can use relations (i) and (ii) to eliminate one of the two unknowns:
$ \Rightarrow {\left( {a - 6} \right)^2} + {\left( {7 - a - 6} \right)^2} = 13 \Rightarrow {\left( {a - 6} \right)^2} + {\left( {1 - a} \right)^2} = 13$
Using the identity ${\left( {a - b} \right)^2} = {a^2} + 2ab + {b^2}$ in the above expression, we get:
$ \Rightarrow {a^2} + 36 - 2 \times 6 \times a + {a^2} + 1 - 2 \times 1 \times a = 13 \Rightarrow 2{a^2} + 37 - 14a = 13 \Rightarrow 2{a^2} + 24 - 14a = 0$
So we get a quadratic equation as: ${a^2} - 7a + 12 = 0$
The roots of this quadratic can found as:
$ \Rightarrow {a^2} - 3a - 4a + 12 = 0 \Rightarrow a\left( {a - 3} \right) - 4\left( {a - 3} \right) = 0 \Rightarrow \left( {a - 3} \right)\left( {a - 4} \right) = 0$
According to this, we get $a = 3$
Again by using (i), we can say: $b = 7 - a = 7 - 3 = 4$
Therefore, we get the values as: $a = 3{\text{ and }}b = 4$
Hence, the option (D) is the correct answer.

Note: In questions like this, it is not possible to find two unknowns from using just one equation. Try to establish separate relations for solving values for two unknowns. An alternative approach for the same problem can be to use the Quadratic formula, according to which the roots of a quadratic equation of the form $a{x^2} + bx + c = 0$ , can be defined by $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ .