Let A={x∈N:1≤x≤17} and B={ax+b:x∈X,a,b∈R,a>0}.If mean and variance of elements of B are 17 and 216 respectively, than a+ b is equal to

A
7
B
9
C
-7
D
-27

Solution:

We know that the mean and variance of first $n$

natural numbers are $\frac{n + 1}{2}$ and $\frac{n^{2} - 1}{12}$ respectively.

$∴ \bar{X} = \frac{17 + 1}{2} = 9$ and Var $(X) = \frac{17^{2} - 1}{12} = 24$

Now, $y_{i} = a x_{i} + b, i = 1, 2, \dots, 17$

$\begin{array}{ll} \Rightarrow \bar{Y} = a \bar{X} + b and Var (Y) = a^{2} Var (X) \\ \Rightarrow 17 = 9 a + b and 216 = a^{2} (24) \\ \Rightarrow a = 3, b = - 10 [∵∣ z > 0] \\ ∴ a + b = - 7 \end{array}$

Let $A = {x \in N : 1 \leq x \leq 17}$ and $B = {a x + b : x \in X, a, b \in R, a > 0} .$
If mean and variance of elements
of $B$ are 17 and 216 respectively, than $a + b$ is equal to

Solution:

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