Showing the 12 most recent of 18 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.If $R$ is the range, $\bar{x}$ is the arithmetic mean and $M$ is the mean deviation from the mean of the data $15, 28, 3, 1, 36, 10, 6, 21$, then $\dfrac{R - \bar{x}}{M} =$ (A) $1$ (B) $2$ (C) $4$ (D) $5$

   ›Reveal solutionSolution

   Compute the range, mean, and mean deviation about the mean directly from the eight data values, then form the ratio. Answer: $2$.

   Concept and Intuition

   This is a direct application of three basic descriptive-statistics definitions on a small dataset: range = max − min, arithmetic mean = sum/n, and mean deviation from the mean = average of the absolute deviations $|x_i-\bar x|$.

   Step-by-Step Solution

   1. Data: $15, 28, 3, 1, 36, 10, 6, 21$ ($n=8$).
   2. Range: $R = \max-\min = 36-1=35$.
   3. Sum $=15+28+3+1+36+10+6+21=120$, so $\bar x=\dfrac{120}{8}=15$.
   4. Absolute deviations from the mean: $|15-15|=0$, $|28-15|=13$, $|3-15|=12$, $|1-15|=14$, $|36-15|=21$, $|10-15|=5$, $|6-15|=9$, $|21-15|=6$. …

   🔒 Start your 14-day free trial to see the full solution →
2. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.The mean deviation about the mean for the following data is

   $x_i$	1	2	4	7
   $f_i$	3	2	4	1
   (A) $3$
   (B) $2$
   (C) $1.5$
   (D) $1.6$

   ›Reveal solutionSolution

   Compute the mean, then the frequency-weighted mean absolute deviation. Answer: $1.6$.

   Concept and Intuition

   Mean deviation about the mean measures the average absolute distance of the data values from their mean, weighted by how often each value occurs.

   Step-by-Step Solution

   1. Total frequency $N = 3+2+4+1 = 10$.
   2. Mean $\bar{x} = \dfrac{1(3)+2(2)+4(4)+7(1)}{10} = \dfrac{3+4+16+7}{10} = \dfrac{30}{10} = 3$.
   3. Absolute deviations $|x_i - \bar{x}|$: for $x=1$, $|1-3|=2$; $x=2$, $|2-3|=1$; $x=4$, $|4-3|=1$; $x=7$, $|7-3|=4$.
   4. Multiply by frequency and sum: $2(3)+1(2)+1(4)+4(1) = 6+2+4+4=16$. …

   🔒 Start your 14-day free trial to see the full solution →
3. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.The mean of 5 observations is 5. If three of the observations are 1, 2, 6 and the other two observations are such that each is greater than 5, then the mean deviation from the mean of the observations is (A) $2.8$ (B) $2.6$ (C) $2.5$ (D) $2.4$

   ›Reveal solutionSolution

   The two unknown observations' individual values don't matter — only their sum (found from the mean) and the fact that both exceed 5, which fixes the total absolute deviation.

   Concept and Intuition

   Mean deviation from the mean is $\dfrac1n\sum|x_i-\bar x|$. When we don't know two of the observations individually but know they're both above the mean, their contribution to $\sum|x_i-\bar x|$ is just $(\text{their sum})-2\times(\text{mean})$, since each $|x_i-5|=x_i-5$ when $x_i>5$.

   Step-by-Step Solution

   1. Mean $=5$ over 5 observations $\Rightarrow$ total sum $=25$.
   2. Known observations $1,2,6$ sum to $9$, so the other two, say $p,q$, satisfy $p+q=25-9=16$.
   3. Each of $p,q>5$, so $|p-5|+|q-5|=(p-5)+(q-5)=(p+q)-10=16-10=6$.
   4. Deviations of the known three: $|1-5|=4,\ |2-5|=3,\ |6-5|=1$, summing to $8$. …

   🔒 Start your 14-day free trial to see the full solution →
4. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.The following data represents the frequency distribution of 20 observations

   $x_i$	3	4	5	8	10	11
   $f_i$	$\alpha+2$	$(\alpha-1)^2$	4	$\alpha-1$	2	$\alpha$

   Then its mean deviation about the mean is (A) 3 (B) 2.4 (C) 2.7 (D) 2.9

   ›Reveal solutionSolution

   First pin down $\alpha$ using $\sum f_i=20$, then compute the mean and finally the mean deviation about the mean; the answer is 2.7.

   Concept and Intuition

   Mean deviation about the mean measures the average absolute distance of the data points from the mean, weighted by frequency. Before we can compute anything we must first determine the unknown $\alpha$ using the fact that the frequencies must add to the total number of observations (20 here).

   Step-by-Step Solution

   1. Find $\alpha$. Sum the frequencies:

   $(\alpha+2)+(\alpha-1)^2+4+(\alpha-1)+2+\alpha = 20$

   Expand $(\alpha-1)^2=\alpha^2-2\alpha+1$:

   $\alpha+2+\alpha^2-2\alpha+1+4+\alpha-1+2+\alpha = \alpha^2+\alpha+8$

   So $\alpha^2+\alpha+8=20 \Rightarrow \alpha^2+\alpha-12=0 \Rightarrow (\alpha+4)(\alpha-3)=0$.

   Thus $\alpha=3$ or $\alpha=-4$. Since $\alpha=-4$ makes $\alpha-1=-5<0$ (a frequency cannot be negative), we take $\alpha=3$.

   2. Frequencies with $\alpha=3$:

   $f_i:\ \alpha+2=5,\ (\alpha-1)^2=4,\ 4,\ \alpha-1=2,\ 2,\ \alpha=3$

   Check: $5+4+4+2+2+3=20$. ✓

   3. Compute the mean. $\sum f_ix_i = 5(3)+4(4)+4(5)+2(8)+2(10)+3(11)=15+16+20+16+20+33=120$ …

   🔒 Start your 14-day free trial to see the full solution →
5. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
   Q.The mean deviation about the mean for the following data is:

   Class Interval	0-2	2-4	4-6	6-8	8-10
   Frequency	1	3	4	1	2
   (A) 3
   (B) $\dfrac{20}{11}$
   (C) $\dfrac{40}{11}$
   (D) 2

   ›Reveal solutionSolution

   Standard grouped-data mean deviation about the mean; works out to $\dfrac{20}{11}$, option (B).

   Concept and Intuition

   For grouped data, each class is represented by its midpoint. Mean deviation about the mean measures the average absolute distance of the data (via midpoints) from the overall mean, weighted by frequency.

   Step-by-Step Solution

   1. Midpoints: $0\!-\!2\to1$, $2\!-\!4\to3$, $4\!-\!6\to5$, $6\!-\!8\to7$, $8\!-\!10\to9$.
   2. $N=\sum f = 1+3+4+1+2=11$.
   3. $\sum fx = 1(1)+3(3)+4(5)+1(7)+2(9)=1+9+20+7+18=55$.
   4. Mean $\bar x=\dfrac{55}{11}=5$.
   5. Absolute deviations $|x-\bar x|$: $4,2,0,2,4$. …

   🔒 Start your 14-day free trial to see the full solution →
6. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
   Q.The mean deviation from the median for the following data is: $x_i$: 9, 3, 7, 2, 5; $f_i$: 1, 6, 2, 8, 4 (A) $\dfrac{94}{21}$ (B) $\dfrac{12}{7}$ (C) $\dfrac{10}{7}$ (D) $\dfrac{100}{21}$

   ›Reveal solutionSolution

   Find the median from the cumulative frequency, then average the absolute deviations of each value from it, weighted by frequency.

   Concept and Intuition

   Mean deviation about the median measures the average absolute spread of data around its middle value. For grouped/discrete frequency data, first locate the median using cumulative frequencies, then compute $\frac{\sum f_i|x_i-\text{median}|}{\sum f_i}$.

   Step-by-Step Solution

   1. Arrange the data in increasing order of $x_i$ with their frequencies: $x=2\,(f=8)$, $x=3\,(f=6)$, $x=5\,(f=4)$, $x=7\,(f=2)$, $x=9\,(f=1)$.
   2. Total frequency $N = 8+6+4+2+1 = 21$.
   3. Cumulative frequencies: at $x=2$: $8$; at $x=3$: $14$; at $x=5$: $18$; at $x=7$: $20$; at $x=9$: $21$.
   4. Since $N=21$ is odd, the median is the $\frac{N+1}{2}=11$th observation. The 11th observation lies in the $x=3$ bracket (positions 9–14), so median $=3$. …

   🔒 Start your 14-day free trial to see the full solution →
7. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
   Q.If $m$ and $M$ denote the mean deviations about mean and about median respectively of the data 20, 5, 15, 2, 7, 3, 11 then the mean deviation about the mean of $m$ and $M$ is (A) $\dfrac{1}{7}$ (B) $\dfrac{38}{7}$ (C) $\dfrac{36}{7}$ (D) $\dfrac{37}{7}$

   ›Reveal solutionSolution

   Compute the mean deviation about the mean ($m$) and about the median ($M$) of the given 7 numbers, then treat $\{m,M\}$ itself as a 2-point dataset and find its mean deviation about its mean; the answer is $\dfrac{1}{7}$.

   Concept and Intuition

   Mean deviation about a central value $A$ is $\dfrac{1}{n}\sum|x_i-A|$. Here we must compute it twice for the original data (about mean, then about median) to get two numbers $m$ and $M$, and then apply the same idea to this new 2-element dataset $\{m,M\}$.

   Step-by-Step Solution

   1. Data: $20,5,15,2,7,3,11$, so $n=7$ and $\text{Mean}=\dfrac{20+5+15+2+7+3+11}{7}=\dfrac{63}{7}=9$.
   2. Absolute deviations from the mean: $11,4,6,7,2,6,2$. Sum $=38$, so $m=\dfrac{38}{7}$.
   3. Sort the data: $2,3,5,7,11,15,20$. The median (4th of 7 values) is $7$.
   4. Absolute deviations from the median: $13,2,8,5,0,4,4$. Sum $=36$, so $M=\dfrac{36}{7}$.
   5. Now find the mean deviation about the mean of the two-element set $\{m,M\}=\left\{\dfrac{38}{7},\dfrac{36}{7}\right\}$. Their mean is $\dfrac{1}{2}\left(\dfrac{38}{7}+\dfrac{36}{7}\right)=\dfrac{37}{7}$. …

   🔒 Start your 14-day free trial to see the full solution →
8. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
   Q.If the mean deviation about the mean is $m$ and variance is $\sigma^2$ for the following data, then $m + \sigma^2 =$

   x	1	3	5	7	9
   f	4	24	28	16	8
   (A) $8$
   (B) $7.2$
   (C) $\dfrac{28}{5}$
   (D) $6$

   ›Reveal solutionSolution

   Compute the mean, then the mean deviation about the mean, and the variance from the frequency table; their sum is $6$.

   Concept and Intuition

   For grouped/discrete frequency data, mean $=\Sigma fx/n$, mean deviation about the mean $=\Sigma f|x-\bar x|/n$, and variance $=\Sigma fx^2/n - \bar x^2$. These are direct formula applications once the table sums are computed correctly.

   Step-by-Step Solution

   1. $n=\Sigma f = 4+24+28+16+8=80$.
   2. $\Sigma fx = 1(4)+3(24)+5(28)+7(16)+9(8) = 4+72+140+112+72=400$. Mean $\bar x = 400/80=5$.
   3. $|x-\bar x|$: for $x=1,3,5,7,9$ these are $4,2,0,2,4$.
   4. $\Sigma f|x-\bar x| = 4(4)+24(2)+28(0)+16(2)+8(4) = 16+48+0+32+32=128$. So $m=128/80=1.6$.
   5. $\Sigma fx^2 = 1(4)+9(24)+25(28)+49(16)+81(8) = 4+216+700+784+648=2352$. …

   🔒 Start your 14-day free trial to see the full solution →
9. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
   Q.Mean deviation about the mean for the following data is: Class Interval: 0-6, 6-12, 12-18, 18-24, 24-30; Frequency: 1, 2, 3, 2, 1 (A) $5$ (B) $\dfrac{16}{3}$ (C) $6$ (D) $\dfrac{19}{3}$

   ›Reveal solutionSolution

   This tests mean deviation about the mean for grouped data; the answer is $\dfrac{16}{3}$.

   Concept and Intuition

   For grouped data, use the class midpoint as the representative value of each class. Mean deviation about the mean is the frequency-weighted average of the absolute deviations of each midpoint from the overall mean.

   Step-by-Step Solution

   1. Midpoints: $3,9,15,21,27$ for classes $0$–$6,6$–$12,12$–$18,18$–$24,24$–$30$; frequencies $1,2,3,2,1$; $\Sigma f=9$.
   2. $\Sigma fx=1(3)+2(9)+3(15)+2(21)+1(27)=3+18+45+42+27=135$.
   3. Mean $=\dfrac{135}{9}=15$.
   4. Absolute deviations from 15: $|3-15|=12,\ |9-15|=6,\ |15-15|=0,\ |21-15|=6,\ |27-15|=12$. …

   🔒 Start your 14-day free trial to see the full solution →
10. AP EAPCET 2023Set eng-2023-05-18-FN1 markMCQ
    Q.Which one of the following is false? (A) Mean deviation from the mean and the Mean deviation from the Median must always be equal. (B) The measure of variability which is a number independent of units, is called the coefficient of variation (C) The coefficient of variation is a relative measure of variation (D) Among two given data, the one having greater coefficient of variation is said to have more variability than the other.

    ›Reveal solutionSolution

    Mean deviation is minimised when measured from the median, not the mean, so the two are generally different — statement (A) is the false one.

    Concept and Intuition

    Among all measures $\sum|x_i-a|$, the value of $a$ that minimises the sum is the median (a classical result in statistics), while $a=\bar x$ minimises $\sum(x_i-a)^2$. So mean deviation from the median is always $\le$ mean deviation from the mean, with equality only in special (e.g. symmetric) distributions — not always.

    Step-by-Step Solution

    1. Statement (A) claims MD(mean) $=$ MD(median) always — this contradicts the minimising property of the median described above, so it is false in general.
    2. Statement (B): the coefficient of variation is indeed defined as a unit-independent relative measure — true.
    3. Statement (C): CV is explicitly a relative measure of dispersion — true. …

    🔒 Start your 14-day free trial to see the full solution →
11. AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
    Q.The mean deviation from the mean for the data $6,7,10,12,13,4,12,16$ is (A) $3.25$ (B) $3.52$ (C) $3.33$ (D) $2.35$

    ›Reveal solutionSolution

    Compute the mean, then average the absolute deviations from it. Answer: (A).

    Concept and Intuition

    Mean deviation about the mean measures the average absolute distance of data points from their mean — a direct measure of spread that treats deviations above and below the mean symmetrically (unlike variance, which squares them).

    Step-by-Step Solution

    1. Data: $6,7,10,12,13,4,12,16$ ($n=8$).
    2. Sum $=6+7+10+12+13+4+12+16=80$. Mean $=80/8=10$.
    3. Absolute deviations from mean: $|6-10|=4,\ |7-10|=3,\ |10-10|=0,\ |12-10|=2,\ |13-10|=3,\ |4-10|=6,\ |12-10|=2,\ |16-10|=6$. …

    🔒 Start your 14-day free trial to see the full solution →
12. AP EAPCET 2022Set eng-2022-07-04-AN1 markMCQ
    Q.If the mean of the data $p, 6, 6, 7, 8, 11, 15, 16$, is 3 times $p$, then the mean deviation of the data from its mean is (A) $2.25$ (B) $3.75$ (C) $4.4$ (D) $2.5$

    ›Reveal solutionSolution

    Solving for $p$ from the mean condition gives $p=3$, mean $=9$, and the mean deviation about the mean works out to $3.75$.

    Concept and Intuition

    Mean deviation about the mean is the average of the absolute differences between each data value and the mean; first the unknown value must be pinned down using the given mean condition.

    Step-by-Step Solution

    1. Sum of data: $p+6+6+7+8+11+15+16 = p+69$.
    2. Mean $=\frac{p+69}{8}$, and this equals $3p$: $\frac{p+69}{8}=3p \Rightarrow p+69=24p \Rightarrow 69=23p \Rightarrow p=3$.
    3. So the mean is $3p=9$, and the data is $3,6,6,7,8,11,15,16$. …

    🔒 Start your 14-day free trial to see the full solution →