Q103 / 180 · COMEDK 2023 (Evening Shift) · Mathematics
Consider the first 10 natural numbers. If we multiply each number by $-$1 and add 1 to each number, the variance of the numbers so obtained is
- Correct answerA8.25
- B6.5
- C3.87
- D2.87
Solution
Let the original set of numbers be $X = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$.
The variance of the first $n$ natural numbers is given by the formula $Var(X) = \dfrac{n^2 - 1}{12}$.
For $n = 10$, $Var(X) = \dfrac{10^2 - 1}{12} = \dfrac{99}{12} = 8.25$.
The new set of numbers is obtained by the transformation $Y = -X + 1$.
The property of variance states that $Var(aX + b) = a^2 Var(X)$.
Here, $a = -1$ and $b = 1$.
Therefore, $Var(Y) = (-1)^2 Var(X) = 1 \times 8.25 = 8.25$.
Answer: 8.25
The variance of the first $n$ natural numbers is given by the formula $Var(X) = \dfrac{n^2 - 1}{12}$.
For $n = 10$, $Var(X) = \dfrac{10^2 - 1}{12} = \dfrac{99}{12} = 8.25$.
The new set of numbers is obtained by the transformation $Y = -X + 1$.
The property of variance states that $Var(aX + b) = a^2 Var(X)$.
Here, $a = -1$ and $b = 1$.
Therefore, $Var(Y) = (-1)^2 Var(X) = 1 \times 8.25 = 8.25$.
Answer: 8.25
Hence, the correct answer is option: 'A'.