Q.The mean of 5 observations is 4.4 and their variance is 8.24. If three of the observations are 1, 2 and 6, find the other two observations.

Showing the 12 most recent of 17 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.If the median and mode of 4 observations are 4, 6 respectively and the sum of the squares of the observations is 48, then their coefficient of variation is (A) $100\sqrt{3}$ (B) $100/\sqrt{3}$ (C) $100\sqrt{2}$ (D) $100/\sqrt{5}$

   ›Reveal solutionSolution

   Apply the empirical relation Mode $=3$ Median $-2$ Mean to pin down the mean, then use $\Sigma x_i^2$ to get the variance and hence the coefficient of variation.

   Concept and Intuition

   For four raw numbers, the exact identity of "the mode" and "the median" don't leave enough information to reconstruct every individual observation — but statistics has a well-known empirical (approximate) relationship for moderately skewed distributions: $\text{Mean}-\text{Mode}=3(\text{Mean}-\text{Median})$, equivalently $\text{Mode}=3\,\text{Median}-2\,\text{Mean}$. This is exactly the tool such problems are testing: once you have the mean and the sum of squares, the variance and CV follow from the standard formulas without needing the raw data at all.

   Step-by-Step Solution

   1. Empirical relation: $\text{Mode}=3\,\text{Median}-2\,\text{Mean}$.
   2. Substitute Median $=4$, Mode $=6$: $6=3(4)-2\,\text{Mean}=12-2\,\text{Mean}\Rightarrow 2\,\text{Mean}=6\Rightarrow\text{Mean}=3$.
   3. With $n=4$ observations, Mean $=3\Rightarrow \Sigma x_i=12$.
   4. Variance $=\dfrac{1}{n}\Sigma x_i^2-(\text{Mean})^2=\dfrac{48}{4}-3^2=12-9=3$.
   5. Standard deviation $=\sqrt3$.
   6. Coefficient of variation $=\dfrac{\text{SD}}{\text{Mean}}\times100=\dfrac{\sqrt3}{3}\times100=\dfrac{100}{\sqrt3}$.

   Common Mistakes

   • Trying to solve for the four raw observations directly — the given data (median, mode, sum of squares alone) is insufficient for that and even contradictory if you assume two observations literally equal 6.
   • Mixing up which of mean/median/mode goes where in the empirical formula.

   ✓Final answer

   The correct option is (B) — $\dfrac{100}{\sqrt3}$.

   ANSWER: B

2. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.The mean and variance of a discrete data $x_i (i=1,2,...15)$ are 35 and 10 respectively. The mean and variance of another discrete data $y_j (j=1,2,...10)$ are 15 and 5 respectively. If these two data are combined, then the variance of the obtained data of 25 items is (A) $20$ (B) $75$ (C) $104$ (D) $82$

   ›Reveal solutionSolution

   This tests the combined-variance formula for two pooled datasets, which requires first finding the combined mean and then correcting each group's variance for how far its own mean sits from the combined mean.

   Concept and Intuition

   When two datasets are merged, you cannot simply average their variances — each group's spread must be measured about the new combined mean, not its own original mean. The correction term $(\bar x_i-\bar X)^2$ accounts for the extra spread introduced because each subgroup's mean itself differs from the overall mean.

   Step-by-Step Solution

   1. Combined mean: $\bar X=\dfrac{n_1\bar x_1+n_2\bar x_2}{n_1+n_2}=\dfrac{15(35)+10(15)}{25}=\dfrac{525+150}{25}=\dfrac{675}{25}=27$.
   2. Deviation of group 1's mean from combined mean: $35-27=8$, squared $=64$.
   3. Deviation of group 2's mean from combined mean: $15-27=-12$, squared $=144$.
   4. Combined variance $=\dfrac{n_1(\text{var}_1+64)+n_2(\text{var}_2+144)}{n_1+n_2}=\dfrac{15(10+64)+10(5+144)}{25}=\dfrac{15(74)+10(149)}{25}$.
   5. $=\dfrac{1110+1490}{25}=\dfrac{2600}{25}=104$.

   Common Mistakes

   • Simply averaging the two variances (weighted or not) without adding the mean-correction terms — this is the single most common error in combined-variance problems.
   • Using the wrong combined mean (e.g. a plain average of $35$ and $15$ instead of the sample-size-weighted mean).

   ✓Final answer

   The correct option is (C) — $104$.

   ANSWER: C

3. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.The variance of the frequency distribution given below is Class interval: 0-10, 10-20, 20-30, 30-40, 40-50, 50-60 Frequency: 2, 2, 3, 4, 1, 3 (A) 264 (B) 88 (C) 84 (D) 90

   ›Reveal solutionSolution

   Apply the step-deviation method for grouped data: compute $d_i=(x_i-A)/h$, get the variance in $d$-units, then scale by $h^2$; the variance is $264$.

   Concept and Intuition

   For grouped/frequency data, working directly with large midpoint values is error-prone, so we shift to a convenient assumed mean $A$ and scale by the class width $h$ using $d_i=(x_i-A)/h$. Variance transforms cleanly under this linear substitution: $\text{Var}(x)=h^2\,\text{Var}(d)$, where $\text{Var}(d)=\dfrac{\sum fd^2}{N}-\left(\dfrac{\sum fd}{N}\right)^2$.

   Step-by-Step Solution

   1. Midpoints: $5,15,25,35,45,55$ for the six classes, with frequencies $2,2,3,4,1,3$; $N=2+2+3+4+1+3=15$.
   2. Take $A=35,\ h=10$: $d=\dfrac{x-35}{10}$ gives $d=-3,-2,-1,0,1,2$ respectively.
   3. $fd$: $2(-3)=-6,\ 2(-2)=-4,\ 3(-1)=-3,\ 4(0)=0,\ 1(1)=1,\ 3(2)=6$. Sum $=-6-4-3+0+1+6=-6$.
   4. $fd^2$: $2(9)=18,\ 2(4)=8,\ 3(1)=3,\ 4(0)=0,\ 1(1)=1,\ 3(4)=12$. Sum $=18+8+3+0+1+12=42$.
   5. Mean of $d$: $\bar d=\dfrac{-6}{15}=-0.4$.
   6. Variance of $d$: $\dfrac{42}{15}-(-0.4)^2=2.8-0.16=2.64$.
   7. Variance of $x$: $h^2\times2.64=100\times2.64=264$.

   Common Mistakes

   • Forgetting to square the mean when subtracting (variance $=E[d^2]-(E[d])^2$, not $E[d^2]-E[d]$).
   • Forgetting to scale the final variance by $h^2$ (not just $h$) to return to original units.

   ✓Final answer

   The correct option is (A) 264.

   ANSWER: A

4. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.The variance of the ungrouped data 2, 12, 3, 11, 5, 10, 6, 7 is (A) $11.875$ (B) $11$ (C) $12$ (D) $10.765$

   ›Reveal solutionSolution

   Variance of ungrouped data is the mean of the squared deviations from the mean. Here the mean is $7$ and the variance works out to $12$.

   Concept and Intuition

   For ungrouped data $x_1,\dots,x_n$, variance $=\dfrac{1}{n}\sum (x_i-\bar x)^2$ where $\bar x$ is the mean. It measures the average squared spread of the data around its centre.

   Step-by-Step Solution

   1. Data: $2,12,3,11,5,10,6,7$ ($n=8$).
   2. Mean: $\bar x=\dfrac{2+12+3+11+5+10+6+7}{8}=\dfrac{56}{8}=7$.
   3. Deviations from mean: $-5,5,-4,4,-2,3,-1,0$.
   4. Squared deviations: $25,25,16,16,4,9,1,0$.
   5. Sum of squared deviations $=25+25+16+16+4+9+1+0=96$.
   6. Variance $=\dfrac{96}{8}=12$.

   Common Mistakes

   • Using $n-1$ (sample variance) instead of $n$ when the question asks for the plain variance of ungrouped data — CBSE/EAPCET convention here divides by $n$.
   • Arithmetic slip in computing the mean (miscounting the sum as something other than 56).

   ✓Final answer

   The correct option is (C) — $12$.

   ANSWER: C

5. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.Variance of the following discrete frequency distribution is Class Interval: 0-2, 2-4, 4-6, 6-8, 8-10 Frequency: 2, 3, 5, 3, 2 (A) $\dfrac{463}{15}$ (B) $\dfrac{838}{15}$ (C) $\dfrac{44}{5}$ (D) $\dfrac{88}{15}$

   ›Reveal solutionSolution

   Using midpoints and the shortcut formula $\text{Var}=\frac{\sum fx^2}{N}-\left(\frac{\sum fx}{N}\right)^2$ gives $\dfrac{88}{15}$.

   Concept and Intuition

   For a grouped/discrete frequency distribution, replace each class interval by its midpoint and treat it as a discrete random variable, then apply the standard variance formula.

   Step-by-Step Solution

   1. Midpoints: $1,3,5,7,9$ for the classes $0$–$2,\dots,8$–$10$; frequencies $2,3,5,3,2$; $N=2+3+5+3+2=15$.
   2. $\sum fx = 2(1)+3(3)+5(5)+3(7)+2(9) = 2+9+25+21+18=75$, so mean $\bar x = \dfrac{75}{15}=5$.
   3. $\sum fx^2 = 2(1)+3(9)+5(25)+3(49)+2(81) = 2+27+125+147+162=463$.
   4. Variance $=\dfrac{\sum fx^2}{N}-\bar x^2 = \dfrac{463}{15}-25 = \dfrac{463-375}{15}=\dfrac{88}{15}$.

   Common Mistakes

   • Using the class boundaries (e.g. 0 and 2) directly instead of the midpoint 1.
   • Forgetting to subtract $\bar x^2$ (computing only $\sum fx^2/N$ and stopping there).

   ✓Final answer

   The correct option is (D) — $\dfrac{88}{15}$.

   ANSWER: D

6. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.The mean and variance of the observations $x_1, x_2, x_3, \ldots, x_{15}$ are respectively 2 and 4. If the mean and variance of the observations $y_1, y_2, \ldots, y_{10}$ are respectively 2 and 5, then the variance of the observations $x_1, x_2, \ldots x_{15}, y_1, y_2, \ldots, y_{10}$ is (A) $6.5$ (B) $5.3$ (C) $3.4$ (D) $4.4$

   ›Reveal solutionSolution

   When two groups share the exact same mean, the variance of their pooled data is simply the sample-size-weighted average of the two individual variances — no cross-term correction is needed.

   Concept and Intuition

   In general, combining two data sets requires accounting for how far each group's own mean is from the overall combined mean (a "between-group spread" correction). But when both groups already share the same mean as each other (and hence as the combined data), that correction term is exactly zero, and the combined variance collapses to a simple weighted average of the two variances by sample size.

   Step-by-Step Solution

   1. Group 1: $n_1=15$, mean $\bar x_1=2$, variance $\sigma_1^2=4$.
   2. Group 2: $n_2=10$, mean $\bar x_2=2$, variance $\sigma_2^2=5$.
   3. Since $\bar x_1=\bar x_2=2$, the combined mean is also $2$ — no shift between groups' means and the pooled mean.
   4. General combined-variance formula: $\sigma^2=\dfrac{n_1\sigma_1^2+n_2(\bar x_1-\bar X)^2+n_2\sigma_2^2+n_2(\bar x_2-\bar X)^2}{n_1+n_2}$ — but since $\bar x_1=\bar x_2=\bar X=2$, both shift terms are zero.
   5. So $\sigma^2=\dfrac{n_1\sigma_1^2+n_2\sigma_2^2}{n_1+n_2}=\dfrac{15(4)+10(5)}{25}=\dfrac{60+50}{25}=\dfrac{110}{25}=4.4$.

   Common Mistakes

   • Applying the full combined-variance formula with nonzero shift terms even though the means are identical (unnecessary, but not wrong if done correctly — the shift terms would just evaluate to 0 anyway).
   • Averaging the variances unweighted (i.e. $(4+5)/2=4.5$) instead of weighting by sample size.

   ✓Final answer

   The correct option is (D) — $4.4$.

   ANSWER: D

7. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.Let $x_1,x_2,\ldots,x_{11}$ be the observations satisfying $\sum_{i=1}^{11}(x_i-4)=22$ and $\sum_{i=1}^{11}(x_i-4)^2=154$. If the mean and variance of the observations are $\alpha$ and $\beta$, then the quadratic equation having the roots $\dfrac{\alpha}{\beta}$ and $\dfrac{\beta}{\alpha}$ is (A) $15x^2-16x+15=0$ (B) $15x^2-34x+15=0$ (C) $x^2-16x+60=0$ (D) $12x^2-25x+20=0$

   ›Reveal solutionSolution

   Shift-invariance of variance lets us compute the mean and variance directly from the given sums; the resulting quadratic in $\alpha/\beta,\beta/\alpha$ is $15x^2-34x+15=0$.

   Concept and Intuition

   Variance is unaffected by shifting all observations by a constant, so working with $d_i=x_i-4$ is exactly as good as working with $x_i$ directly — it just makes the arithmetic simpler.

   Step-by-Step Solution

   1. Let $d_i=x_i-4$. Then $\sum d_i=22$ and $\sum d_i^2=154$, over $n=11$ observations.
   2. Mean of $x$: $\bar x=4+\dfrac{\sum d_i}{11}=4+2=6=\alpha$.
   3. Variance is shift-invariant, so $\beta=\text{Var}(x)=\text{Var}(d)=\dfrac{\sum d_i^2}{11}-\left(\dfrac{\sum d_i}{11}\right)^2=14-4=10$.
   4. Roots required: $\dfrac{\alpha}{\beta}=\dfrac{6}{10}=\dfrac{3}{5}$ and $\dfrac{\beta}{\alpha}=\dfrac{10}{6}=\dfrac{5}{3}$.
   5. Sum of roots $=\dfrac{3}{5}+\dfrac{5}{3}=\dfrac{9+25}{15}=\dfrac{34}{15}$; product of roots $=1$.
   6. Quadratic: $x^2-\dfrac{34}{15}x+1=0$. Multiplying through by 15: $15x^2-34x+15=0$.

   Common Mistakes

   • Forgetting to shift the mean back by $+4$ (using $\sum d_i/11$ alone as the mean of $x$).
   • Applying the variance formula to $x_i$ directly without recognizing $d_i=x_i-4$ already centers the data conveniently.

   ✓Final answer

   The correct option is (B) — $15x^2-34x+15=0$.

   ANSWER: B

8. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
   Q.Variance of the following continuous frequency distribution is Class Interval: 0-4, 4-8, 8-12, 12-16 Frequency: 1, 2, 2, 1 (A) 16 (B) $\dfrac{44}{3}$ (C) 23 (D) $\dfrac{22}{3}$

   ›Reveal solutionSolution

   This tests computing variance of a grouped/continuous frequency distribution using class midpoints. Answer: $\dfrac{44}{3}$.

   Concept and Intuition

   For a continuous frequency distribution we treat each class by its midpoint (the representative value) and compute mean and variance exactly as for discrete data, weighting by frequency. Variance measures the average squared deviation from the mean.

   Step-by-Step Solution

   1. Midpoints of classes $0$–$4,4$–$8,8$–$12,12$–$16$ are $x_i = 2,6,10,14$ with frequencies $f_i=1,2,2,1$. Total $N=\sum f_i=6$.
   2. Mean: $\bar x=\dfrac{\sum f_ix_i}{N}=\dfrac{1(2)+2(6)+2(10)+1(14)}{6}=\dfrac{2+12+20+14}{6}=\dfrac{48}{6}=8$.
   3. Deviations squared times frequency:

   • $x=2$: $(2-8)^2=36$, times $f=1$ → $36$
   • $x=6$: $(6-8)^2=4$, times $f=2$ → $8$
   • $x=10$: $(10-8)^2=4$, times $f=2$ → $8$
   • $x=14$: $(14-8)^2=36$, times $f=1$ → $36$

   4. Sum $=36+8+8+36=88$.
   5. Variance $=\dfrac{\sum f_i(x_i-\bar x)^2}{N}=\dfrac{88}{6}=\dfrac{44}{3}$.

   Common Mistakes

   • Using $N-1$ (sample variance) instead of $N$ (population variance) — CBSE/EAPCET convention here uses $N$.
   • Forgetting to weight the squared deviations by frequency.

   ✓Final answer

   The correct option is (B) — $\dfrac{44}{3}$.

   ANSWER: B

9. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
   Q.If $\displaystyle\sum_{i=1}^{9}(x_i-5)=9$ and $\displaystyle\sum_{i=1}^{9}(x_i-5)^2=45$, then the standard deviation of the nine observations $x_1,x_2,\ldots,x_9$ is (A) $2$ (B) $4$ (C) $3$ (D) $9$

   ›Reveal solutionSolution

   This tests computing standard deviation from a shifted variable using $\sum(x_i-a)$ and $\sum(x_i-a)^2$. The standard deviation is $2$.

   Concept and Intuition

   Standard deviation is shift-invariant: subtracting a constant from every observation doesn't change the spread. So working with $d_i=x_i-5$ instead of $x_i$ directly is valid, and the variance formula $\text{Var}=\dfrac{\sum d_i^2}{n}-\left(\dfrac{\sum d_i}{n}\right)^2$ applies exactly as it would to the $x_i$ themselves.

   Step-by-Step Solution

   1. Let $d_i=x_i-5$ for $i=1,\dots,9$. Given $\sum d_i=9$ and $\sum d_i^2=45$.
   2. Mean of $d$: $\bar d=\dfrac{\sum d_i}{9}=\dfrac99=1$.
   3. Variance of $d$ (equals variance of $x$, since shifting doesn't change variance): $\sigma^2=\dfrac{\sum d_i^2}{9}-\bar d^2=\dfrac{45}{9}-1^2=5-1=4$.
   4. Standard deviation $\sigma=\sqrt4=2$.

   Common Mistakes

   • Forgetting to subtract the square of the mean (just reporting $\sum d_i^2/n=5$ as the variance).
   • Confusing variance with standard deviation and skipping the final square root.

   ✓Final answer

   The correct option is (A) — $2$.

   ANSWER: A

10. AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
    Q.If the mean of the data $7, 8, 9, 7, 8, 7, \lambda, 8$ is $8$ then variance of the data (A) $2$ (B) $\dfrac{7}{8}$ (C) $\dfrac{9}{8}$ (D) $1$

    ›Reveal solutionSolution

    First find $\lambda=10$ from the mean condition, then compute variance directly; the answer is $1$.

    Concept and Intuition

    With the mean given, we can first solve for the unknown $\lambda$, then compute the variance as the average squared deviation from that mean over all 8 data points.

    Step-by-Step Solution

    1. Known values: $7,8,9,7,8,7,8$ (seven of them, excluding $\lambda$). Sum $=7+8+9+7+8+7+8=54$.
    2. With $\lambda$ included there are 8 values, and mean $=8$, so total sum $=8\times8=64$.
    3. $\lambda=64-54=10$.
    4. Full data set: $7,8,9,7,8,7,10,8$.
    5. Deviations from mean (8): $-1,0,1,-1,0,-1,2,0$.
    6. Squared deviations: $1,0,1,1,0,1,4,0$; sum $=1+0+1+1+0+1+4+0=8$.
    7. Variance $=\dfrac{\sum(x-\bar x)^2}{n}=\dfrac88=1$.

    Common Mistakes

    • Forgetting to solve for $\lambda$ first and instead computing variance over only the 7 known values (wrong $n$ and wrong data set).
    • Using $n-1$ (sample variance) instead of $n$ (population variance) — this problem's phrasing implies population variance over all 8 given values.

    ✓Final answer

    The correct option is (D) — $1$.

    ANSWER: D

11. AP EAPCET 2021Set eng-2021-08-19-AN1 markMCQ
    Q.The mean and variance of '$n$' observations $x_{1}, x_{2}, x_{3}, \ldots, x_{n}$ are 5 and 0 respectively. If $\sum_{i=1}^{n}x_{i}^{2} = 400$, then the value of '$n$' is equal to (A) 80 (B) 25 (C) 20 (D) 16

    ›Reveal solutionSolution

    Zero variance means zero spread — every data value equals the mean — which converts the sum-of-squares condition directly into an equation for $n$.

    Concept and Intuition

    Variance measures deviation from the mean; a variance of exactly 0 is only possible if every single observation is identical to the mean value. This turns a statistics problem into simple algebra.

    Step-by-Step Solution

    1. Given mean $\bar x = 5$ and variance $=0$.
    2. Variance $=\dfrac{1}{n}\sum x_i^2 - \bar x^2 = 0 \Rightarrow \dfrac{1}{n}\sum x_i^2 = \bar x^2 = 25$.
    3. So $\sum x_i^2 = 25n$.
    4. Given $\sum x_i^2 = 400$, so $25n = 400 \Rightarrow n = 16$.

    Common Mistakes

    • Trying to use the population-vs-sample variance formula distinction — irrelevant here since variance is exactly 0 either way.
    • Forgetting variance $=\frac{1}{n}\sum x_i^2 - \bar x^2$ (not $\sum x_i^2/n$ alone).

    ✓Final answer

    The correct option is (D) — 16.

    ANSWER: D

12. AP EAPCET 2021Set eng-2021-08-19-FN1 markMCQ
    Q.The variance of the variates $112, 116, 120, 125, 132$ about their A.M is (A) $58.8$ (B) $60$ (C) $48.8$ (D) $61.8$

    ›Reveal solutionSolution

    Variance about the mean is the average of squared deviations from the mean; here it computes to $48.8$.

    Concept and Intuition

    Variance measures the average squared spread of data around its own mean. The procedure is always: find the mean, subtract it from each value, square each deviation, then average those squares.

    Step-by-Step Solution

    1. Sum of data: $112+116+120+125+132 = 605$. Mean $=\dfrac{605}{5}=121$.
    2. Deviations from mean: $112-121=-9$, $116-121=-5$, $120-121=-1$, $125-121=4$, $132-121=11$.
    3. Squares of deviations: $81, 25, 1, 16, 121$.
    4. Sum of squares: $81+25+1+16+121 = 244$.
    5. Variance $=\dfrac{244}{5}=48.8$.

    Common Mistakes

    • Arithmetic slip computing the mean (605/5 must be exactly 121).
    • Forgetting to divide by $n$ (using $n-1$ would give sample variance, not the variance "about their A.M" as intended here).

    ✓Final answer

    The correct option is (C) — $48.8$.

    ANSWER: C