Q.The variance of 20 observations is 5. If each observation is multiplied by 2, find the new variance of the resulting observations.

Concept and Intuition

Variance measures spread around the mean, and shifting every value by the same constant $k$ shifts the mean by $k$ too — so every deviation from the mean is completely unchanged. This is why $\text{Var}(X+k)=\text{Var}(X)$ always, regardless of $k$.

Step-by-Step Solution

1. First dataset (sorted): $1,15,35,53,64,72$.
2. Second dataset (sorted): $-1,13,33,51,62,70$.
3. Check the difference at each position: $1-(-1)=2,\ 15-13=2,\ 35-33=2,\ 53-51=2,\ 64-62=2,\ 72-70=2$.
4. So the second dataset is exactly the first dataset with every value reduced by $2$.
5. Since variance is translation-invariant, $\text{Var}(\text{second set}) = \text{Var}(\text{first set})$.
6. The standard deviation of the first set is given as $x$, so its variance is $x^2$.
7. Therefore the variance of the second set is also $x^2$.

Common Mistakes

• Not noticing (or not checking) that the two datasets are related by a simple shift, and instead trying to compute the variance of the second dataset from scratch.
• Confusing "standard deviation is $x$" with "variance is $x$" — the question specifically asks for variance, which is $x^2$.

ANSWER: D

Concept and Intuition

Expanding both given identities as perfect squares produces two linear equations in $\sum x_i$ and $\sum x_i^2$ (with $n$ as a common factor); adding and subtracting isolates each unknown cleanly, after which the standard variance formula finishes the job.

Step-by-Step Solution

1. $\sum(x_i+2)^2=\sum x_i^2+4\sum x_i+4n=28n$ ... (i)
2. $\sum(x_i-2)^2=\sum x_i^2-4\sum x_i+4n=12n$ ... (ii)
3. (i) $-$ (ii): $8\sum x_i=16n \Rightarrow \sum x_i=2n \Rightarrow \bar x=2$.
4. (i) $+$ (ii): $2\sum x_i^2+8n=40n \Rightarrow 2\sum x_i^2=32n \Rightarrow \sum x_i^2=16n$.
5. Variance $=\dfrac{\sum x_i^2}{n}-\bar x^2=16-2^2=16-4=12$.

Common Mistakes

• Forgetting the $4n$ constant term when expanding the squares, which throws off both equations.
• Confusing $\sum x_i^2/n$ with $\left(\sum x_i/n\right)^2$.

ANSWER: A

Concept and Intuition

Variance measures squared spread from the mean. If $Y_i=kX_i$, then $\bar Y=k\bar X$ and $Y_i-\bar Y=k(X_i-\bar X)$, so $(Y_i-\bar Y)^2=k^2(X_i-\bar X)^2$. Averaging gives $\mathrm{Var}(Y)=k^2\mathrm{Var}(X)$ — the mean's shift doesn't matter (variance is a "spread" measure, translation-invariant), only the scaling factor matters, and it enters squared.

Step-by-Step Solution

1. Given $\mathrm{Var}(X)=5$, and $Y_i=2X_i$ for each of the 20 observations.
2. $\mathrm{Var}(Y)=2^2\cdot\mathrm{Var}(X)=4\times5=20$.
3. The count of observations (20) is irrelevant to this scaling law.

Common Mistakes

• Scaling variance by $k$ instead of $k^2$ (confusing it with how the mean/SD would scale — SD scales by $k$, not variance).
• Getting distracted by the "20 observations" figure, which plays no role here.

ANSWER: C

Concept and Intuition

Variance $=\dfrac{1}{n}\sum(x_i-\bar x)^2$ depends only on how far each point is from the mean of the data, not on the absolute values. Shifting every point by $k$ shifts the mean by $k$ too, so every deviation $(x_i-\bar x)$ is exactly unchanged.

Step-by-Step Solution

1. Let $y_i = x_i+k$. Then $\bar y = \bar x+k$.
2. $y_i-\bar y = (x_i+k)-(\bar x+k) = x_i-\bar x$ — the deviation is unchanged.
3. $\text{Var}(y)=\dfrac1n\sum(y_i-\bar y)^2=\dfrac1n\sum(x_i-\bar x)^2=\text{Var}(x)$.

Common Mistakes

• Confusing this with scaling (multiplying by $k$), which DOES change variance by a factor of $k^2$ — that rule only applies to multiplication, not addition/subtraction.

ANSWER: B

Concept and Intuition

Variance measures squared deviation from the mean. If every data point is multiplied by a constant $k$, both the mean and every deviation scale by $k$, so the squared deviations (and hence the variance) scale by $k^2$ — a well-known linear-transformation property of variance.

Step-by-Step Solution

1. Property: $\text{Var}(kw,kx,ky,kz) = k^2\,\text{Var}(w,x,y,z)$.
2. Here $k=5$ and $\text{Var}(w,x,y,z)=9$.
3. New variance $=5^2\times9=25\times9=225$.

Common Mistakes

• Multiplying the variance by $k$ instead of $k^2$ (confusing it with the scaling of standard deviation, or of the mean).
• Confusing this with adding a constant to each number, which leaves variance unchanged (only scaling changes it).

ANSWER: A

Concept and Intuition

For an AP with common difference $d$ and $n$ terms, there is a ready-made formula for the standard deviation: $\sigma=d\sqrt{\dfrac{n^2-1}{12}}$. This avoids computing the mean and every squared deviation by hand.

Step-by-Step Solution

1. The numbers are $4,8,12,\dots,40$: an AP with $d=4$, $n=10$.
2. Apply $\sigma=d\sqrt{\dfrac{n^2-1}{12}}=4\sqrt{\dfrac{100-1}{12}}=4\sqrt{\dfrac{99}{12}}=4\sqrt{8.25}$.
3. $\sqrt{8.25}\approx2.8723$, so $\sigma\approx4\times2.8723\approx11.49\approx11.5$.

Common Mistakes

• Using $n$ instead of $n^2-1$ in the numerator.
• Forgetting the common difference multiplies the whole square root, not just the $99/12$ part.

ANSWER: C

Concept and Intuition

If a dataset $Y = kX$ (each value scaled by constant $k$), then $\text{SD}(Y) = |k|\cdot\text{SD}(X)$. All four given sets are exact multiples of $\{1,2,3,4\}$: by $10$, by $2$, by $3$, and by $1$ respectively. So whichever has the smallest multiplier has the smallest SD.

Step-by-Step Solution

1. Recognize: $\{10,20,30,40\}=10\times\{1,2,3,4\}$; $\{2,4,6,8\}=2\times\{1,2,3,4\}$; $\{3,6,9,12\}=3\times\{1,2,3,4\}$; $\{1,2,3,4\}=1\times\{1,2,3,4\}$.
2. Compute the base SD of $\{1,2,3,4\}$: mean $=2.5$; deviations $-1.5,-0.5,0.5,1.5$; squared sum $=2.25+0.25+0.25+2.25=5$; variance $=5/4=1.25$; SD $=\sqrt{1.25}\approx1.118$.
3. Scaled SDs: $10\times1.118\approx11.18$ (A), $2\times1.118\approx2.236$ (B), $3\times1.118\approx3.354$ (C), $1\times1.118\approx1.118$ (D).
4. The smallest is clearly option (D), the unscaled base set.

Common Mistakes

• Confusing standard deviation (absolute spread) with coefficient of variation (relative spread) — CV would actually be equal across all four sets since they're proportional, but the question asks for the (absolute) SD itself.
• Assuming the set with the smallest numbers must have the smallest range and stopping there without checking SD formally (though here it happens to agree).

ANSWER: D

Concept and Intuition

The arithmetic mean is a linear operator: $\text{mean}(kx_i) = k\cdot\text{mean}(x_i)$, because $\dfrac{1}{n}\sum(kx_i) = k\cdot\dfrac1n\sum x_i$.

Step-by-Step Solution

1. Original mean: $\bar x = \dfrac1n\sum x_i = 10$.
2. New data: $y_i = 2x_i$ for each observation.
3. New mean: $\bar y = \dfrac1n\sum y_i = \dfrac1n\sum 2x_i = 2\cdot\dfrac1n\sum x_i = 2\bar x = 2(10)=20$.

Common Mistakes

• Confusing how variance/standard deviation scale (by $k^2$ and $k$ respectively) with how the mean scales (simply by $k$).

ANSWER: D

Effect of Scaling Variance

The mean is a linear statistic: if you scale every data point by a factor $k$, the mean scales by exactly the same factor. Here, dividing each number by 5 is scaling by $1/5$. So if the original mean is $\bar{X}$, the new mean is $\bar{X}/5$. The problem gives the new mean as $X/5$, so we can equate and solve.

1. Define the original mean Let the original set of $n$ numbers be $a_1, a_2, \dots, a_n$. Their mean is

2. Apply the division by 5 Each number is divided by 5, so the new set is $\frac{a_1}{5}, \frac{a_2}{5}, \dots, \frac{a_n}{5}$. The new mean is

3. Equate to the given value The problem states this new mean equals $\frac{X}{5}$. Therefore

Multiply both sides by 5:

4. Interpret the result The original mean $\bar{X}$ is exactly $X$. So the mean of the original $n$ numbers is $X$.

Concept and Intuition

If every data value is scaled by a constant $k$ ($x_i=ky_i$), then $\text{Var}(x) = k^2\,\text{Var}(y)$ (variance scales with the square of the multiplier, since it's built from squared deviations). Spotting a common factor before crunching the numbers keeps the numbers small and the arithmetic error-free.

Step-by-Step Solution

1. Data: $180,198,90,126,72,144,18,81,27,54$. Each is $9\times$ one of $20,22,10,14,8,16,2,9,3,6$.
2. Let $y_i$ be this smaller set. Sum of $y_i = 20+22+10+14+8+16+2+9+3+6=110$, so mean $\bar y = 11$.
3. Sum of $y_i^2 = 400+484+100+196+64+256+4+81+9+36 = 1630$.
4. $\text{Var}(y) = \dfrac{\sum y_i^2}{n} - \bar y^2 = \dfrac{1630}{10}-11^2 = 163-121 = 42$.
5. Since $x_i=9y_i$: $\text{Var}(x)=9^2\times\text{Var}(y) = 81\times42 = 3402$.

Common Mistakes

• Computing variance on the large numbers directly, inviting arithmetic slips.
• Forgetting variance scales by $k^2$ (not $k$) under a linear scaling of the data.

ANSWER: B