Q.Calculate the mean deviation about the mean of the set of first $n$ natural numbers when $n$ is an odd number.

Concept and Intuition

Although the sample space has 12 outcomes (coin $\times$ die), the random variable $X(a,b)=b$ ignores the coin entirely. Each value $b=1,\dots,6$ occurs with exactly two of the twelve equally likely outcomes (once with H, once with T), so $X$ is a discrete uniform variable on $\{1,\dots,6\}$ — exactly like a single fair die.

Step-by-Step Solution

1. Each outcome $(a,b)$ has probability $\frac{1}{12}$; for a fixed $b$, both $(H,b)$ and $(T,b)$ give $X=b$, so $P(X=b) = \frac{2}{12}=\frac16$ for each $b=1,\dots,6$.
2. Mean: $E(X) = \dfrac{1+2+3+4+5+6}{6} = \dfrac{21}{6} = \dfrac{7}{2}$.
3. $E(X^2) = \dfrac{1^2+2^2+3^2+4^2+5^2+6^2}{6} = \dfrac{91}{6}$.
4. Variance $= E(X^2) - [E(X)]^2 = \dfrac{91}{6} - \dfrac{49}{4} = \dfrac{182}{12} - \dfrac{147}{12} = \dfrac{35}{12}$.

Common Mistakes

• Treating the sample space as needing a joint variance calculation over 12 outcomes as if $X$ depended on both coordinates — it only depends on $b$.
• Using the wrong standard variance formula for a die, e.g. $(n^2-1)/12$ with $n$ mistaken for something other than 6.

ANSWER: D

Concept and Intuition

Variance measures the average squared spread of data around its mean:

Here we first need to pin down the unknown value $n$ using the given mean, and only then compute the variance over the completed data set.

Step-by-Step Solution

1. Mean condition: $\dfrac{3+6+9+10+n}{5}=9 \implies 3+6+9+10+n=45 \implies 28+n=45 \implies n=17$.
2. The data set is $\{3,6,9,10,17\}$ with mean $9$.
3. Deviations from the mean: $3-9=-6$, $6-9=-3$, $9-9=0$, $10-9=1$, $17-9=8$.
4. Squares of deviations: $36, 9, 0, 1, 64$.
5. Sum of squared deviations $=36+9+0+1+64=110$.
6. Variance $=\dfrac{110}{5}=22$.

Common Mistakes

• Computing variance using $n-1$ (sample variance) instead of $n$ (population variance) — for this kind of "variance of a given set of values" question, the population formula (divide by $N$) is standard.
• Forgetting to first solve for $n$ and instead treating $n$ symbolically throughout, which needlessly complicates the arithmetic.

ANSWER: B

Concept and Intuition

Variance measures the average squared spread of data around its mean: $\sigma^2=\dfrac1n\sum(x_i-\bar x)^2$. The steps are always the same — find the mean, subtract it from each value, square, average.

Step-by-Step Solution

1. Sum the data: $4+7+8+10+13+16+19=77$. There are $n=7$ values, so the mean $\bar x=\dfrac{77}{7}=11$.
2. Compute deviations from the mean: $4-11=-7$, $7-11=-4$, $8-11=-3$, $10-11=-1$, $13-11=2$, $16-11=5$, $19-11=8$.
3. Square each: $49, 16, 9, 1, 4, 25, 64$.
4. Sum the squares: $49+16+9+1+4+25+64=168$.
5. Variance $=\dfrac{168}{7}=24$.

Common Mistakes

• Dividing by $n-1$ (sample variance) instead of $n$ (population variance) when the question intends the simple population variance — here dividing by $7$ gives the listed answer.
• Arithmetic slips in computing the mean or the sum of squared deviations.

ANSWER: B

Concept and Intuition

The first $n$ even natural numbers $2,4,\ldots,2n$ are just the first $n$ natural numbers scaled by 2. Variance scales by the square of a scaling factor, so if $\mathrm{Var}(1,\ldots,n)=\dfrac{n^2-1}{12}$, then $\mathrm{Var}(2,4,\ldots,2n)=2^2\cdot\dfrac{n^2-1}{12}=\dfrac{n^2-1}{3}$.

Step-by-Step Solution

1. Recall $\mathrm{Var}(1,2,\ldots,n)=\dfrac{n^2-1}{12}$.
2. Scaling every value by 2 multiplies variance by $2^2=4$: $\mathrm{Var}(2,4,\ldots,2n)=4\cdot\dfrac{n^2-1}{12}=\dfrac{n^2-1}{3}$.
3. Statement I claims this equals $\dfrac{n^2-1}{4}$ — this is wrong (the correct denominator is 3), so Statement I is false.
4. For $n=20$: variance $=\dfrac{20^2-1}{3}=\dfrac{399}{3}=133$.
5. Mean of $2,4,\ldots,40$ is $2\cdot\dfrac{1+2+\cdots+20}{20}=2\cdot\dfrac{210}{20}=2\cdot10.5=21$.
6. Difference $=133-21=112$, exactly matching Statement II, so Statement II is true.

Common Mistakes

• Misremembering the variance formula for $1,\ldots,n$ as having denominator other than 12.
• Forgetting to square the scale factor (2) when scaling variance.

ANSWER: C

Concept and Intuition

The variance of the first $n$ natural numbers $1,2,\ldots,n$ has the closed form $\dfrac{n^2-1}{12}$. If every value in a data set is scaled by a constant $k$ (as with even numbers $2,4,\ldots,2m$, which are $2\times(1,2,\ldots,m)$), the variance scales by $k^2$. This lets us reuse the same base formula for both parts of the problem.

Step-by-Step Solution

1. Variance of first $n$ natural numbers: $\mathrm{Var}=\dfrac{n^2-1}{12}$.
2. Given this equals 10: $\dfrac{n^2-1}{12}=10 \Rightarrow n^2-1=120 \Rightarrow n^2=121 \Rightarrow n=11$.
3. The first $m$ even natural numbers are $2,4,6,\ldots,2m = 2\times(1,2,\ldots,m)$.
4. Scaling every data point by 2 multiplies the variance by $2^2=4$: $\mathrm{Var}(2,4,\ldots,2m)=4\cdot\dfrac{m^2-1}{12}=\dfrac{m^2-1}{3}$.
5. Given this equals 16: $\dfrac{m^2-1}{3}=16 \Rightarrow m^2-1=48 \Rightarrow m^2=49 \Rightarrow m=7$.
6. Therefore $n:m=11:7$.

Common Mistakes

• Forgetting the scaling factor is squared (variance is a squared-deviation measure) — using a factor of 2 instead of 4 for the even numbers would give a wrong $m$.
• Taking the negative root of $n^2=121$ or $m^2=49$ — since $n,m$ count natural numbers, only the positive root is meaningful.

ANSWER: C

Concept and Intuition

$\sum_{k=1}^n k^2$ grows like $\dfrac{n^3}{3}$ for large $n$ (its exact value is $\dfrac{n(n+1)(2n+1)}{6}$, a cubic in $n$ with leading coefficient $\dfrac13$). Dividing by $n^3$ and letting $n\to\infty$ isolates exactly that leading coefficient.

Step-by-Step Solution

1. Recall $\displaystyle\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$.
2. So

3. As $n\to\infty$, $\dfrac1n\to0$, so this tends to

Common Mistakes

• Misremembering the sum-of-squares formula or its degree (it's cubic in $n$, matching the $n^3$ in the denominator).
• Forgetting to pull the constant $x$ out of the sum.
• Arithmetic slip in simplifying $\dfrac{2}{6}$ to $\dfrac13$.

ANSWER: C

Concept and Intuition

The discrete uniform distribution on $\{1,\dots,n\}$ has well-known closed forms for its mean (the average of $1$ to $n$) and its second moment (from the sum-of-squares formula). Variance is then just $E[X^2]-(E[X])^2$ — a standard identity that avoids summing $(X-\bar X)^2$ directly.

Step-by-Step Solution

1. $E[X]=\displaystyle\sum_{k=1}^n k\cdot\tfrac1n=\tfrac1n\cdot\tfrac{n(n+1)}2=\tfrac{n+1}2$.
2. $E[X^2]=\displaystyle\sum_{k=1}^n k^2\cdot\tfrac1n=\tfrac1n\cdot\tfrac{n(n+1)(2n+1)}6=\tfrac{(n+1)(2n+1)}6$.
3. $\text{Var}(X)=E[X^2]-(E[X])^2=\tfrac{(n+1)(2n+1)}6-\tfrac{(n+1)^2}4$.
4. Factor $(n+1)$: $=(n+1)\left[\tfrac{2n+1}6-\tfrac{n+1}4\right]=(n+1)\cdot\dfrac{2(2n+1)-3(n+1)}{12}=(n+1)\cdot\dfrac{4n+2-3n-3}{12}=(n+1)\cdot\dfrac{n-1}{12}$.
5. So $\text{Var}(X)=\dfrac{(n+1)(n-1)}{12}=\dfrac{n^2-1}{12}$.

Common Mistakes

• Misremembering the sum-of-squares formula $\sum k^2=\tfrac{n(n+1)(2n+1)}6$.
• Arithmetic slip combining the fractions over a common denominator of 12.

ANSWER: A

Concept and Intuition

The first $n$ even natural numbers are $2,4,\dots,2n = 2\times(1,2,\dots,n)$. Scaling a data set by $k$ scales its variance by $k^2$. Since $\text{Var}(1,\dots,n)=\dfrac{n^2-1}{12}$, we get $\text{Var}(2,4,\dots,2n)=4\times\dfrac{n^2-1}{12}=\dfrac{n^2-1}{3}$.

Step-by-Step Solution

1. True variance formula: $\dfrac{n^2-1}{3}$. Statement-I asserts $\dfrac{n^2-1}{4}$ — this is incorrect in general, so Statement-I is false.
2. For $n=20$: variance $=\dfrac{20^2-1}{3}=\dfrac{399}{3}=133$.
3. Arithmetic mean of $2,4,\dots,40$ is $\dfrac{2+40}{2}=21$.
4. Difference $=133-21=112$, exactly matching Statement-II's claim, so Statement-II is true.
5. Hence Statement-I is false, Statement-II is true.

Common Mistakes

• Misremembering the variance formula as $\frac{n^2-1}{4}$ instead of $\frac{n^2-1}{3}$ (a common trap this question is testing).
• Forgetting the scaling-by-$k$ squares the variance.

ANSWER: D

Concept and Intuition

Given the shape of a probability distribution ($P(X=k)\propto k^2$), the actual constant of proportionality is fixed by requiring all probabilities to sum to 1. The mean is then just $E[X]=\sum kP(X=k)$, using the known sums $\sum k^2$ and $\sum k^3$ for $k=1$ to $6$.

Step-by-Step Solution

1. Normalization: $\sum_{k=1}^{6}P(X=k)=P\sum_{k=1}^{6}k^2=1$.
2. $\sum_{k=1}^{6}k^2=1+4+9+16+25+36=91$, so $91P=1 \Rightarrow P=\dfrac{1}{91}$.
3. Mean $=E[X]=\sum_{k=1}^{6}k\cdot P(X=k)=\sum_{k=1}^{6}k\cdot k^2P=P\sum_{k=1}^{6}k^3$.
4. $\sum_{k=1}^{6}k^3=1+8+27+64+125+216=441$ (equivalently $\left(\dfrac{6\cdot7}{2}\right)^2=21^2=441$).
5. Mean $=441P=\dfrac{441}{91}$.

Common Mistakes

• Forgetting to first solve for $P$ using the normalization condition before computing the mean.
• Using $\sum k^2$ in the mean calculation instead of $\sum k^3$.

ANSWER: D

Concept and Intuition

For a discrete uniform random variable on $\{1,\dots,6\}$ (each with probability $1/6$), mean and variance follow directly from $E[X]=\sum xP(x)$ and $\text{Var}(X)=E[X^2]-(E[X])^2$ — a standard result worth memorizing but easy to re-derive.

Step-by-Step Solution

1. $E[X]=\dfrac{1}{6}(1+2+3+4+5+6)=\dfrac{21}{6}=\dfrac{7}{2}$.
2. $E[X^2]=\dfrac16(1+4+9+16+25+36)=\dfrac{91}{6}$.
3. $\text{Var}(X)=E[X^2]-(E[X])^2=\dfrac{91}{6}-\left(\dfrac72\right)^2=\dfrac{91}{6}-\dfrac{49}{4}$.
4. Common denominator 12: $\dfrac{182}{12}-\dfrac{147}{12}=\dfrac{35}{12}$.

Common Mistakes

• Using $n(n+1)(2n+1)/6$ formulas incorrectly for $E[X^2]$ instead of direct summation.
• Forgetting to square the mean before subtracting.

ANSWER: D

Concept and Intuition

For a binomial distribution $B(n,p)$, mean $=np$ and variance $=npq$ where $q=1-p$. Dividing variance by mean eliminates $n$, giving $q$ directly, which then unlocks $p$ and $n$.

Step-by-Step Solution

1. Mean: $np=15$.
2. Variance: $npq=10$.
3. Divide: $\frac{npq}{np}=q=\frac{10}{15}=\frac23$.
4. So $p = 1-q = \frac13$.
5. From $np=15$: $n = \frac{15}{p} = \frac{15}{1/3} = 45$.

Common Mistakes

• Confusing which ratio gives $q$ vs $p$ (variance/mean gives $q$, not $p$).
• Arithmetic slip converting $\frac{10}{15}$ to $\frac23$.

ANSWER: C

Concept and Intuition

The mean of any list is (sum) $\div$ (count). Here the "list" is $1^2,2^2,\dots,n^2$, and the sum of squares of the first $n$ naturals has a well-known closed form.

Step-by-Step Solution

1. Sum of squares: $\sum_{k=1}^n k^2 = \dfrac{n(n+1)(2n+1)}{6}$.
2. Mean $= \dfrac{1}{n}\cdot\dfrac{n(n+1)(2n+1)}{6} = \dfrac{(n+1)(2n+1)}{6}$.
3. Expand the numerator: $(n+1)(2n+1) = 2n^2+n+2n+1 = 2n^2+3n+1$.
4. So the mean is $\dfrac{2n^2+3n+1}{6}$.

Common Mistakes

• Confusing this with $\left[\dfrac{n(n+1)}{2}\right]^2$, which is the square of the sum, not the sum/mean of squares.
• Arithmetic slip expanding $(n+1)(2n+1)$.

ANSWER: C