Q.Let $a, b, c, d, e$ be the observations with mean $m$ and standard deviation $s$. The standard deviation of the observations $a + k, b + k, c + k, d + k, e + k$ is
(A) $s$
(B) $ks$
(C) $s + k$
(D) $\dfrac{s}{k}$

Concept: Effect of Translation on Standard Deviation

Standard deviation measures the spread of data around the mean. When we add a constant $k$ to every observation, we shift the entire dataset by $k$ units, but the spread remains unchanged.

Step 1: The original mean is $m = \frac{a+b+c+d+e}{5}$.

Step 2: The new mean after adding $k$ to each observation is:

Step 3: Standard deviation is defined as $s = \sqrt{\frac{\sum(x_i - m)^2}{n}}$. For the transformed data:

$s' = \sqrt{\frac{\sum[(x_i + k) - (m + k)]^2}{5}} = \sqrt{\frac{\sum(x_i - m)^2}{5}} = s$ …

Why Adding a Constant Preserves Spread

Standard deviation measures how far observations scatter around their mean. When you add the same constant $k$ to every single observation, you're shifting the entire dataset along the number line by $k$ units. The mean shifts by $k$ too, so the distance of each observation from the new mean stays exactly what it was before.

Think of it this way: if five people are standing at distances $a, b, c, d, e$ meters from a lamppost (the mean), and you move the lamppost $k$ meters down the road, everyone moves with it. Their distances from the lamppost don't change.

Step-by-Step Verification

1. Original setup

   The observations $a, b, c, d, e$ have mean $m = \frac{a+b+c+d+e}{5}$ and standard deviation $s = \sqrt{\frac{(a-m)^2 + (b-m)^2 + (c-m)^2 + (d-m)^2 + (e-m)^2}{5}}$.

2. New mean after adding $k$

   The transformed observations are $a+k, b+k, c+k, d+k, e+k$. Their mean is

The mean shifts by exactly $k$.

3. Deviations from the new mean For any observation, say $a+k$, its deviation from the new mean is

The $k$ cancels out. Every observation's deviation from its mean is unchanged.

4. New variance and standard deviation The variance of the transformed data is …