Relation between mean, median and mode Questions in English

(C) The formula for calculating the mode for a continuous frequency distribution is given by:
$Mode = l + \left( \frac{f_m - f_{m-1}}{2f_m - f_{m-1} - f_{m+1}} \right) \times C$
where:
$l$ is the lower limit of the modal class,
$f_m$ is the frequency of the modal class,
$f_{m-1}$ is the frequency of the class preceding the modal class,
$f_{m+1}$ is the frequency of the class succeeding the modal class,
$C$ (or $i$) is the width of the class interval.
Thus,option $C$ is correct.

(C) The mode of a data set is the value that appears most frequently.
Given frequencies:
$4$ appears $3$ times.
$5$ appears $5$ times.
$6$ appears $6$ times.
$8$ appears $8$ times.
$10$ appears $7$ times.
Since the number $8$ has the highest frequency ($8$ times),the mode of the set is $8$.

(C) To find the mode,we count the frequency of each observation:
$0$ appears $3$ times.
$1$ appears $1$ time.
$2$ appears $2$ times.
$3$ appears $1$ time.
$5$ appears $1$ time.
$6$ appears $5$ times.
$7$ appears $2$ times.
Since the observation $6$ has the highest frequency ($5$ times),the mode is $6$.

(B) The mode is the value that appears most frequently in a data set.
From the given table,we observe the frequencies of the marks:
- For $4$ marks,frequency is $6$.
- For $5$ marks,frequency is $7$.
- For $6$ marks,frequency is $10$.
- For $7$ marks,frequency is $8$.
- For $8$ marks,frequency is $3$.
The maximum frequency is $10$,which corresponds to the mark $6$.
Therefore,the mode of the distribution is $6$.

(C) The empirical relationship between mean,median,and mode is given by: $\text{Mode} = 3 \text{median} - 2 \text{mean}$.
Rearranging the terms,we get: $2 \text{mean} = 3 \text{median} - \text{mode}$.
Dividing both sides by $2$,we get: $\text{mean} = \frac{1}{2} (3 \text{median} - \text{mode})$.
Comparing this with the given equation $\text{mean} = (3 \text{median} - \text{mode}) k$,we find that $k = \frac{1}{2}$.

(B) For a moderately asymmetrical distribution,the empirical relationship between mean,median,and mode is given by:
$\text{Mode} = 3 \times \text{Median} - 2 \times \text{Mean}$
Given:
$\text{Mode} = 7$
$\text{Mean} = 4$
Substituting the values into the formula:
$7 = 3 \times \text{Median} - 2 \times 4$
$7 = 3 \times \text{Median} - 8$
$7 + 8 = 3 \times \text{Median}$
$15 = 3 \times \text{Median}$
$\text{Median} = \frac{15}{3} = 5$
Therefore,the median is $5$.

(A) For a moderately asymmetrical (skewed) distribution,the empirical relationship between mean,median,and mode is given by:
Mode = $3 \times \text{median} - 2 \times \text{mean}$
Given:
Mode = $6 \lambda$
Mean = $9 \lambda$
Substituting these values into the formula:
$6 \lambda = 3 \times \text{median} - 2 \times (9 \lambda)$
$6 \lambda = 3 \times \text{median} - 18 \lambda$
$3 \times \text{median} = 6 \lambda + 18 \lambda$
$3 \times \text{median} = 24 \lambda$
Median = $\frac{24 \lambda}{3} = 8 \lambda$

(B) We know the empirical relationship between mean,median,and mode is given by:
Mode = $3 \times \text{Median} - 2 \times \text{Mean}$
Given,Mean = $21$ and Median = $22$.
Substituting these values into the formula:
Mode = $3(22) - 2(21)$
Mode = $66 - 42$
Mode = $24$
Therefore,the correct option is $B$.

(D) In a perfectly symmetrical distribution,such as the normal distribution,the mean,median,and mode coincide at the same central point.
Therefore,$M = M_d = M_0$.

(B) The Karl Pearson's coefficient of skewness $(S_k)$ is given by the formula: $S_k = \frac{\text{Mean} - \text{Mode}}{\sigma}$.
Given: $S_k = 0.32$,Mean $(M)$ = $39.6$,and $\sigma = 6.5$.
Substituting the values: $0.32 = \frac{39.6 - M_o}{6.5}$.
$39.6 - M_o = 0.32 \times 6.5 = 2.08$.
$M_o = 39.6 - 2.08 = 37.52$.
We know the empirical relationship: $\text{Mode} = 3 \times \text{Median} - 2 \times \text{Mean}$.
$37.52 = 3 \times \text{Median} - 2 \times 39.6$.
$37.52 = 3 \times \text{Median} - 79.2$.
$3 \times \text{Median} = 37.52 + 79.2 = 116.72$.
$\text{Median} = \frac{116.72}{3} \approx 38.9067$.
Rounding to the nearest provided option,the correct value is $38.81$.

(D) In a normal distribution,the curve is perfectly symmetrical about the center.
Due to this symmetry,the values of the mean,median,and mode coincide at the central peak of the distribution.
Therefore,for a normal distribution,$Mean = Median = Mode$.

(C) The empirical relationship between Mean,Median,and Mode is given by the formula: $\text{Mode} = 3 \times \text{Median} - 2 \times \text{Mean}$.
Rearranging the terms to solve for $\text{Mean}$:
$2 \times \text{Mean} = 3 \times \text{Median} - \text{Mode}$.
$\text{Mean} = \frac{1}{2}(3 \times \text{Median} - \text{Mode})$.
Comparing this with the given equation $\text{Mean} = x(3 \times \text{Median} - \text{Mode})$,we get $x = 1/2$.

(A) The empirical relationship between mean,median,and mode is given by:
$\text{Mode} = 3 \times \text{Median} - 2 \times \text{Mean}$
Given:
$\text{Mean} = 21$
$\text{Median} = 22$
Substituting the values into the formula:
$\text{Mode} = 3 \times 22 - 2 \times 21$
$\text{Mode} = 66 - 42$
$\text{Mode} = 24$

(C) The formula for Standard Deviation is $S.D. = \sqrt{\frac{\sum(x_i - \bar{x})^2}{n}}$.
The formula for Mean Deviation about mean is $M.D. = \frac{\sum|x_i - \bar{x}|}{n}$.
Let $|x_i - \bar{x}| = y_i$. Then $(x_i - \bar{x})^2 = |x_i - \bar{x}|^2 = y_i^2$.
Consider $(S.D.)^2 - (M.D.)^2 = \frac{\sum y_i^2}{n} - \left(\frac{\sum y_i}{n}\right)^2$.
This expression represents the variance of the values $y_i$,which is always non-negative. Since not all values are equal,the variance is strictly greater than zero.
Therefore,$(S.D.)^2 - (M.D.)^2 > 0$,which implies $(S.D.)^2 > (M.D.)^2$.
Since both $S.D.$ and $M.D.$ are non-negative,we conclude $S.D. > M.D.$ or $M.D. < S.D.$.

(D) For a moderately skewed distribution,the empirical relationship between the three measures of central tendency is given by the formula:
$\text{Mode} = 3 \times \text{Median} - 2 \times \text{Mean}$

(D) The modal class is the class with the highest frequency. Here,the class $30-40$ has the highest frequency of $45$.
Here,$l = 30$,$f_1 = 45$,$f_0 = 30$,$f_2 = 35$,and $h = 10$.
The formula for mode is:
$\text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$
Substituting the values:
$\text{Mode} = 30 + \left( \frac{45 - 30}{2(45) - 30 - 35} \right) \times 10$
$\text{Mode} = 30 + \left( \frac{15}{90 - 65} \right) \times 10$
$\text{Mode} = 30 + \left( \frac{15}{25} \right) \times 10$
$\text{Mode} = 30 + \left( \frac{3}{5} \right) \times 10 = 30 + 6 = 36$.

(B) Standard deviation is independent of the change of origin. Therefore,adding a constant $2$ to each term does not change the standard deviation.
For the median,if every term in a data set is increased by a constant $k$,the median also increases by $k$. Thus,the median increases by $2$.

(B) First,we convert the class intervals into continuous form by subtracting $0.5$ from the lower limit and adding $0.5$ to the upper limit.

Class	$0.5 - 10.5$	$10.5 - 20.5$	$20.5 - 30.5$	$30.5 - 40.5$	$40.5 - 50.5$
$f_i$	$5$	$7$	$8$	$6$	$4$

The modal class is $20.5 - 30.5$ as it has the highest frequency $f_1 = 8$.
Here,$\ell = 20.5$,$f_1 = 8$,$f_0 = 7$,$f_2 = 6$,and $h = 10$.
Mode $= \ell + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$
Mode $= 20.5 + \left( \frac{8 - 7}{2(8) - 7 - 6} \right) \times 10$
Mode $= 20.5 + \left( \frac{1}{16 - 13} \right) \times 10$
Mode $= 20.5 + \frac{10}{3} = 20.5 + 3.33 = 23.83$.

(C) We know the empirical relationship between mean,median,and mode is given by:
$\text{Mode} = 3 \times \text{Median} - 2 \times \text{Mean}$
Given:
$\text{Mode} = 18$
$\text{Mean} = 24$
Substituting the values into the formula:
$18 = 3 \times \text{Median} - 2 \times 24$
$18 = 3 \times \text{Median} - 48$
$18 + 48 = 3 \times \text{Median}$
$66 = 3 \times \text{Median}$
$\text{Median} = \frac{66}{3}$
$\text{Median} = 22$

(A) Statement $(1)$ is true: The mode of a grouped frequency distribution can be determined graphically using a histogram.
Statement $(2)$ is true: The median is affected by the change of scale. If each observation is multiplied by a constant $k$,the median is also multiplied by $k$.
Statement $(3)$ is false: Variance is independent of the change of origin,but it is $NOT$ independent of the change of scale. If each observation is multiplied by $k$,the variance becomes $k^2$ times the original variance.
Therefore,statements $(1)$ and $(2)$ are true.

(D) The median is a measure of central tendency that is affected by both the change of origin and the change of scale. If a variable $X$ is transformed to $Y = aX + b$,then the median of $Y$ is given by $M_Y = aM_X + b$,where $M_X$ is the median of $X$. Therefore,it is not independent of either origin or scale.

(B) The mode of a frequency distribution is the value of the observation that has the highest frequency.
Looking at the given table:
- For $x = 4$,$f = 6$
- For $x = 5$,$f = 7$
- For $x = 6$,$f = 10$
- For $x = 7$,$f = 8$
- For $x = 8$,$f = 3$
The highest frequency is $10$,which corresponds to the observation $x = 6$.
Therefore,the mode of the given distribution is $6$.

(C) The mode is the value that appears most frequently in a data set.
Given frequencies:
$4$ appears $3$ times
$5$ appears $4$ times
$6$ appears $5$ times
$7$ appears $8$ times
$8$ appears $7$ times
$9$ appears $6$ times
Since the number $7$ has the highest frequency of $8$,the mode of the series is $7$.

(D) For a slightly asymmetric distribution,the empirical relationship between mean,median,and mode is given by the formula:
$\text{Mode} = 3 \times \text{Median} - 2 \times \text{Mean}$
Given:
$\text{Mean} = 5$
$\text{Median} = 6$
Substituting these values into the formula:
$\text{Mode} = 3(6) - 2(5)$
$\text{Mode} = 18 - 10$
$\text{Mode} = 8$
Therefore,the correct option is $D$.

(D) For any frequency distribution,the relationship between Mean Deviation $(M.D.)$ about the mean and Standard Deviation $(S.D.)$ is given by the inequality $M.D. \le S.D.$
This is because the variance of the absolute deviations from the mean is always non-negative.
Specifically,for a set of values,the $S.D.$ is defined as $\sqrt{\frac{1}{N}\sum f_i(x_i - \bar{x})^2}$ and the $M.D.$ about mean is $\frac{1}{N}\sum f_i|x_i - \bar{x}|$.
By the properties of dispersion,the root mean square deviation is always greater than or equal to the mean absolute deviation,hence $M.D. \le S.D.$

(B) The mean square deviation of a set of numbers is minimum when taken about the arithmetic mean.
Given that the mean square deviation is minimum about $a_{51}$,it implies that the mean of the group is $a_{51}$.
Since the sequence $a_1, a_2, \dots, a_{101}$ is strictly decreasing $(a_i > a_{i+1})$,the median of the group is the middle term,which is $a_{51}$.
For a symmetric distribution or a sequence where the mean,median,and mode coincide,the mode is also $a_{51}$.

(C) The modal class is the class with the highest frequency. Here,the highest frequency is $11$,which corresponds to the class interval $15-20$.
Thus,$l = 15$,$h = 5$,$f_1 = 7$,$f_0 = 11$,and $f_2 = 2$.
The formula for mode is: $\text{Mode} = l + \left( \frac{f_0 - f_1}{2f_0 - f_1 - f_2} \right) \times h$
Substituting the values:
$\text{Mode} = 15 + \left( \frac{11 - 7}{2(11) - 7 - 2} \right) \times 5$
$\text{Mode} = 15 + \left( \frac{4}{22 - 9} \right) \times 5$
$\text{Mode} = 15 + \left( \frac{4}{13} \right) \times 5$
$\text{Mode} = 15 + \frac{20}{13} = \frac{195 + 20}{13} = \frac{215}{13}$

(B) The sum of absolute deviations is defined as $S = \sum |X_i - A|$.
It is a well-known property in statistics that the sum of absolute deviations $\sum |X_i - A|$ is minimized when $A$ is the median of the data set.
If $A$ is the mean,the sum of squared deviations $\sum (X_i - A)^2$ is minimized,but for absolute deviations,the median provides the minimum value.

(D) The empirical relationship between mean,median,and mode is given by the formula:
$\text{Mode} = 3 \times \text{Median} - 2 \times \text{Mean}$
Given that $\text{Mean} = 21$ and $\text{Median} = 22$,we substitute these values into the formula:
$\text{Mode} = 3(22) - 2(21)$
$\text{Mode} = 66 - 42$
$\text{Mode} = 24$

(A) The total frequency $N = a + b + 12 + 9 + 5 = a + b + 26$.
The mean is given by $\frac{\sum f_i x_i}{N} = \frac{a(3) + b(9) + 12(15) + 9(21) + 5(27)}{a + b + 26} = \frac{3a + 9b + 180 + 189 + 135}{a + b + 26} = \frac{3a + 9b + 504}{a + b + 26} = \frac{309}{22}$.
Cross-multiplying: $22(3a + 9b + 504) = 309(a + b + 26)$.
$66a + 198b + 11088 = 309a + 309b + 8034$.
$243a + 111b = 3054$. Dividing by $3$: $81a + 37b = 1018$ (Equation $1$).
For the median,since $\text{median} = 14$,the median class is $12-18$. Here $l = 12, f = 12, cf = a + b, h = 6$.
$\text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h = 12 + \left( \frac{\frac{a+b+26}{2} - (a+b)}{12} \right) \times 6 = 14$.
$12 + \frac{\frac{a+b+26 - 2a - 2b}{2}}{2} = 14 \Rightarrow \frac{26 - a - b}{4} = 2$.
$26 - a - b = 8 \Rightarrow a + b = 18$ (Equation $2$).
From $b = 18 - a$,substitute into Equation $1$: $81a + 37(18 - a) = 1018$.
$81a + 666 - 37a = 1018 \Rightarrow 44a = 352 \Rightarrow a = 8$.
Then $b = 18 - 8 = 10$.
Therefore,$(a - b)^{2} = (8 - 10)^{2} = (-2)^{2} = 4$. (Note: The provided options seem to contain a typo; the calculated value is $4$).

(C) The mean and the median of a data set are measures of central tendency.
They are not necessarily equal for all data sets.
However,they can be equal for certain symmetric distributions or specific data sets.
Therefore,they are sometimes equal.
Hence,option $C$ is correct.

(B) Given data: $1, 2, 2, 3, 3, 3, 4, 6$.
Mean $= \frac{1+2+2+3+3+3+4+6}{8} = \frac{24}{8} = 3$.
Median is the average of the $4^{th}$ and $5^{th}$ terms: $\frac{3+3}{2} = 3$.
Mode is the value with the highest frequency,which is $3$.
Since Mean $=$ Median $=$ Mode $= 3$,the mean deviations about these central tendencies will be identical.
Therefore,$\alpha = \beta = \gamma$.
Hence,option $(B)$ is correct.