Question

Let $X = \left\{ {x \in N:1 \le x \le 17} \right\}$, and $Y = \left\{ {ax + b: x \in X and a,b \in R, a > 0} \right\}$. If mean and variance of elements of $Y$ are $17$ and $216$ respectively. Then what is the value of $a + b$?
A. $27$
B. 7
C. $ - 7$
D. 9

Answer 1

Hint: First, use the definitions of the set and calculate the elements of the set. Then use formulas of the mean and variance to calculate the mean and variance of the set $Y$. In the end, solve both equations to reach the required answer.

Formula Used:
For a data set $\left\{ {{x_1},{x_2},{x_3},{x_4},.....{x_n}} \right\}$:
Mean: $\mu = \dfrac{{\sum\limits_{i = 1}^n {{x_i}} }}{n}$
Variance: ${\sigma ^2} = \dfrac{{\sum\limits_{i = 1}^n {{{\left( {{x_i} - \mu } \right)}^2}} }}{n}$
The sum of natural numbers: $\sum\limits_{i = 1}^n i = \dfrac{{n\left( {n + 1} \right)}}{2}$
The sum of squares of natural numbers: $\sum\limits_{i = 1}^n {{i^2}} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}$

Complete step by step solution:
The given definitions of sets are $X = \left\{ {x \in N:1 \le x \le 17} \right\}$, and $Y = \left\{ {ax + b: x \in X and a,b \in R, a > 0} \right\}$.
The mean and variance of a set $Y$ are $17$ and $216$ respectively
Let’s get the elements of the given sets.
$X = \left\{ {1,2,3,...,17} \right\}$
$Y = \left\{ {a + b, 2a + b,3a + b,...,17a + b} \right\}$
Now apply the formula of mean for the set $Y$.
$\mu = \dfrac{{\sum\limits_{x = 1}^{17} {\left( {ax + b} \right)} }}{{17}}$
$ \Rightarrow 17 = \dfrac{{\sum\limits_{x = 1}^{17} {\left( {ax} \right) + \sum\limits_{x = 1}^{17} {\left( b \right)} } }}{{17}}$
$ \Rightarrow 17 = \dfrac{{a\sum\limits_{x = 1}^{17} {\left( x \right) + b\sum\limits_{x = 1}^{17} {\left( 1 \right)} } }}{{17}}$
Now use the formula of the sum of natural numbers
$17 = \dfrac{{a\left( {\dfrac{{17\left( {18} \right)}}{2}} \right) + 17b}}{{17}}$ [ Since $\sum\limits_{n = 1}^n 1 = n$]
$ \Rightarrow 9a + b = 17$
$ \Rightarrow b - 17 = - 9a$ $.....\left( 1 \right)$

Now apply the formula of mean for the set $Y$.
${\sigma ^2} = \dfrac{{\sum\limits_{x = 1}^{17} {{{\left( {ax + b - 17} \right)}^2}} }}{{17}}$
Substitute the values in the above equation.
$216 = \dfrac{{\sum\limits_{x = 1}^{17} {{{\left( {ax - 9a} \right)}^2}} }}{{17}}$ [ Since $b - 17 = - 9a$]
$ \Rightarrow 216 = \dfrac{{\sum\limits_{x = 1}^{17} {{a^2}{{\left( {x - 9} \right)}^2}} }}{{17}}$
$ \Rightarrow 216 = \dfrac{{\sum\limits_{x = 1}^{17} {{a^2}\left( {{x^2} - 18x + 81} \right)} }}{{17}}$
$ \Rightarrow 216 = \dfrac{{{a^2}\sum\limits_{x = 1}^{17} {{x^2} - 18{a^2}\sum\limits_{x = 1}^{17} x + 81{a^2}\sum\limits_{x = 1}^{17} 1 } }}{{17}}$
Now use the formulas of the sum of natural numbers and sum of squares of natural numbers.
$ \Rightarrow 216 = \dfrac{{{a^2}\left( {\dfrac{{17\left( {18} \right)\left( {35} \right)}}{6}} \right) - 18{a^2}\left( {\dfrac{{17\left( {18} \right)}}{2}} \right) + 81{a^2}\left( {17} \right)}}{{17}}$
$ \Rightarrow 216 = {a^2}\left( {3\left( {35} \right)} \right) - 18{a^2}\left( 9 \right) + 81{a^2}$
$ \Rightarrow 216 = 105{a^2} - 162{a^2} + 81{a^2}$
$ \Rightarrow 216 = 24{a^2}$
Divide both sides by $24$.
${a^2} = 9$
Take square roots on both sides.
$a = \pm 3$
Since $a > 0$.
So, the only possible value is $a = 3$.

Now substitute $a = 3$ in the equation $\left( 1 \right)$.
$b - 17 = - 9\left( 3 \right)$
$ \Rightarrow b = - 27 + 17$
$ \Rightarrow b = - 10$

Therefore,
$a + b = 3 + \left( { - 10} \right)$
$ \Rightarrow a + b = - 7$

Option ‘C’ is correct

Note: Students often get confused with the formulas of the standard deviation and variance. The formula of standard deviation is $\sigma = \sqrt {\dfrac{{\sum\limits_{i = 1}^n {{{\left( {{x_i} - \mu } \right)}^2}} }}{n}} $ and the formula of the variance is ${\sigma ^2} = \dfrac{{\sum\limits_{i = 1}^n {{{\left( {{x_i} - \mu } \right)}^2}} }}{n}$.