Let $x_1, x_2, \ldots x_n$ be $n$ observations. Let $w_i = lx_i + k$ for $i = 1, 2, \ldots n$, where $l$ and $k$ are constants. If the mean of $x_i$'s is 48 and their standard deviation is 12, the mean of $w_i$'s is 55 and standard deviation of $w_i$'s is 15, the values of $l$ and $k$ should be (A) $l = 1.25, k = -5$ (B) $l = -1.25, k = 5$ (C) $l = 2.5, k = -5$ (D) $l = 2.5, k = 5$

> [!TLDR] > When observations are linearly transformed as $w_i = lx_i + k$, their mean transforms as $\bar{w} = l\bar{x} + k$ and their standard deviation transforms as $\sigma_w = |l|\sigma_x$. Using the given values, we find $l = 1.25$ and $k = -5$. The problem asks us to find the constants $l$ and $k$ given how the mean and standard deviation change after a linear transformation of the observations. This requires understanding how these statistical measures behave under such transformations. ### Concept: Effect of Scaling and Shifting on Mean and Standard Deviation Imagine you have a set of numbers. * **Shifting (adding a constant $k$):** If you add the same constant $k$ to every number, the entire distribution shifts. The mean will also shift by $k$. However, the spread of the numbers (how far they are from each other or from the mean) does not change. So, the standard deviation remains the same. * **Scaling (multiplying by a constant $l$):** If you multiply every number by a constant $l$, the entire distribution stretches or shrinks. The mean will also be multiplied by $l$. The spread of the numbers will also be scaled. If $l$ is positive, the standard deviation will be multiplied by $l$. If $l$ is negative, the numbers flip their order and the spread is still scaled by the *magnitude* of $l$. Therefore, the standard deviation is multiplied by $|l|$. Combining these, for a transformation $w_i = lx_i + k$: * The new mean, $\bar{w}$, will be $l\bar{x} + k$. * The new standard deviation, $\sigma_w$, will be $|l|\sigma_x$. Let's derive these formally to solidify the understanding. > [!PROOF] > **Derivation for Mean:** > Given $w_i = lx_i + k$. > The mean of $w_i$'s is $\bar{w} = \frac{1}{n}\sum_{i=1}^n w_i$. > Substitute $w_i$: > $\bar{w} = \frac{1}{n}\sum_{i=1}^n (lx_i + k)$ > $\bar{w} = \frac{1}{n}\left( \sum_{i=1}^n lx_i + \sum_{i=1}^n k \right)$ > $\bar{w} = \frac{1}{n}\left( l\sum_{i=1}^n x_i + nk \right)$ > $\bar{w} = l\left(\frac{1}{n}\sum_{i=1

Let x₁, x₂, … xₙ be n observations. Let wᵢ = lxᵢ + k for i = 1, 2, … n, where l and k are constants. If the me

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Concept understanding — Effect of Scaling Variance

Effect of Scaling Variance – First Encounter

Let’s start with a simple question: what happens to the spread of a dataset if you multiply every number by 2? Or by 0.5? Or by –3?

If you have a set of marks: 40, 50, 60, 70, 80, the variance is some number. Now imagine the teacher doubles every mark: 80, 100, 120, 140, 160. The marks are now twice as far apart from each other. The spread has clearly changed — but by how much?

That’s exactly what the effect of scaling tells you: a precise rule for how variance changes when you multiply (or divide) every observation by a constant.

Intuition first

Variance measures average squared distance from the mean. If you scale all values by a factor $k$ , two things happen:

The mean also gets scaled by $k$ (because mean is linear).
Each individual distance from the mean also gets scaled by $k$ .

But variance squares those distances. So if each distance is multiplied by $k$ , each squared distance is multiplied by $k^2$ . The average of those squared distances — the variance — therefore gets multiplied by $k^2$ .

Note

Scaling by $k$ multiplies the variance by $k^2$ , not by $k$ . This is because variance is in squared units of the original data.

The precise statement

Let $X$ be a random variable (or a dataset) with variance $\text{Var}(X)$ . Let $k$ be any real constant. Then:

\text{Var}(kX) = k^2 \cdot \text{Var}(X)

That’s it. No extra terms, no dependence on the mean. Just $k^2$ times the original variance.

What about adding a constant?

This is a separate but related idea: if you add a constant $c$ to every value, the spread doesn’t change — the whole distribution just shifts. So:

\text{Var}(X + c) = \text{Var}(X)

Combining both: for a linear transformation $Y = aX + b$ ,

\text{Var}(Y) = a^2 \cdot \text{Var}(X)

The constant $b$ has no effect on variance.

A quick example

Suppose the variance of heights (in cm) is 25. Convert to metres: divide by 100, i.e., multiply by $0.01$ .

\text{Var}(\text{height in m}) = (0.01)^2 \times 25 = 0.0001 \times 25 = 0.0025

That’s a tiny number — but it’s correct, because metres are a larger unit, so the spread in metres is much smaller numerically.

Common mistake to avoid

Watch out

Do not say “variance gets multiplied by $k$ ”. It’s $k^2$ . …

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When observations are linearly transformed as $w_i = lx_i + k$ , their mean transforms as $\bar{w} = l\bar{x} + k$ and their standard deviation transforms as $\sigma_w = |l|\sigma_x$ . Using the given values, we find $l = 1.25$ and $k = -5$ .

The problem asks us to find the constants $l$ and $k$ given how the mean and standard deviation change after a linear transformation of the observations. This requires understanding how these statistical measures behave under such transformations.

Concept: Effect of Scaling and Shifting on Mean and Standard Deviation

Imagine you have a set of numbers.

Shifting (adding a constant $k$ ): If you add the same constant $k$ to every number, the entire distribution shifts. The mean will also shift by $k$ . However, the spread of the numbers (how far they are from each other or from the mean) does not change. So, the standard deviation remains the same.
Scaling (multiplying by a constant $l$ ): If you multiply every number by a constant $l$ , the entire distribution stretches or shrinks. The mean will also be multiplied by $l$ . The spread of the numbers will also be scaled. If $l$ is positive, the standard deviation will be multiplied by $l$ . If $l$ is negative, the numbers flip their order and the spread is still scaled by the magnitude of $l$ . Therefore, the standard deviation is multiplied by $|l|$ .

Combining these, for a transformation $w_i = lx_i + k$ :

The new mean, $\bar{w}$ , will be $l\bar{x} + k$ .
The new standard deviation, $\sigma_w$ , will be $|l|\sigma_x$ .

Let's derive these formally to solidify the understanding.

›Proof

Derivation for Mean:

Given $w_i = lx_i + k$ .

The mean of $w_i$ 's is $\bar{w} = \frac{1}{n}\sum_{i=1}^n w_i$ .

Substitute $w_i$ :

$\bar{w} = \frac{1}{n}\sum_{i=1}^n (lx_i + k)$

$\bar{w} = \frac{1}{n}\left( \sum_{i=1}^n lx_i + \sum_{i=1}^n k \right)$

$\bar{w} = \frac{1}{n}\left( l\sum_{i=1}^n x_i + nk \right)$

$\bar{w} = l\left(\frac{1}{n}\sum_{i=1}^n x_i\right) + k$

Since $\bar{x} = \frac{1}{n}\sum_{i=1}^n x_i$ , we have:

$\bar{w} = l\bar{x} + k$

Derivation for Standard Deviation:

The variance of $w_i$ 's is $\sigma_w^2 = \frac{1}{n}\sum_{i=1}^n (w_i - \bar{w})^2$ .

Substitute $w_i = lx_i + k$ and $\bar{w} = l\bar{x} + k$ :

$\sigma_w^2 = \frac{1}{n}\sum_{i=1}^n ((lx_i + k) - (l\bar{x} + k))^2$

$\sigma_w^2 = \frac{1}{n}\sum_{i=1}^n (lx_i + k - l\bar{x} - k)^2$

$\sigma_w^2 = \frac{1}{n}\sum_{i=1}^n (l(x_i - \bar{x}))^2$

$\sigma_w^2 = \frac{1}{n}\sum_{i=1}^n l^2(x_i - \bar{x})^2$

$\sigma_w^2 = l^2 \left(\frac{1}{n}\sum_{i=1}^n (x_i - \bar{x})^2\right)$

Since $\sigma_x^2 = \frac{1}{n}\sum_{i=1}^n (x_i - \bar{x})^2$ , we have:

$\sigma_w^2 = l^2 \sigma_x^2$

Taking the square root to find the standard deviation:

$\sigma_w = \sqrt{l^2 \sigma_x^2} = \sqrt{l^2}\sqrt{\sigma_x^2} = |l|\sigma_x$

Now, let's apply these formulas to the given problem.

Identify the given information.

We are given:
- Mean of $x_i$ 's: $\bar{x} = 48$
- Standard deviation of $x_i$ 's: $\sigma_x = 12$
- Mean of $w_i$ 's: $\bar{w} = 55$
- Standard deviation of $w_i$ 's: $\sigma_w = 15$
- The transformation: $w_i = lx_i + k$
Formulate an equation using the mean transformation.

Using the formula $\bar{w} = l\bar{x} + k$ , we substitute the given values:

$55 = l(48) + k$

This gives us our first equation:

$48l + k = 55 \quad \text{(Equation 1)}$
Formulate an equation using the standard deviation transformation.

Using the formula $\sigma_w = |l|\sigma_x$ , we substitute the given values: …

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Concept: Effect of Scaling and Shifting on Mean and Standard Deviation

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