Mean and Median Questions in English

(B) The sum of the first $n$ natural numbers is given by the formula $S_n = \frac{n(n+1)}{2}$.
The arithmetic mean is defined as the sum of the numbers divided by the count of numbers,which is $n$.
Therefore,$\text{Arithmetic Mean} = \frac{S_n}{n} = \frac{\frac{n(n+1)}{2}}{n} = \frac{n+1}{2}$.

(B) The mean of a series is the sum of the terms divided by the number of terms.
Given series: $a, a + nd, a + 2nd$.
Number of terms = $3$.
Mean = $\frac{a + (a + nd) + (a + 2nd)}{3}$
Mean = $\frac{3a + 3nd}{3}$
Mean = $a + nd$.

(C) The mean of a set of observations is given by the sum of observations divided by the total number of observations.
Given observations are $3, 4, x, 7, 10$ and the mean is $6$.
Total number of observations $n = 5$.
Sum of observations $= 3 + 4 + x + 7 + 10 = 24 + x$.
Mean $= \frac{\text{Sum of observations}}{n} = \frac{24 + x}{5}$.
Given that the mean is $6$,we have:
$6 = \frac{24 + x}{5}$
$30 = 24 + x$
$x = 30 - 24$
$x = 6$.

(C) Let the set of numbers be $x_1, x_2, \dots, x_n$. The mean is given by $\bar{x} = \frac{\sum_{i=1}^{n} x_i}{n}$,which implies $\sum_{i=1}^{n} x_i = n\bar{x}$.
If each number is multiplied by $\lambda$,the new set of numbers is $\lambda x_1, \lambda x_2, \dots, \lambda x_n$.
The new mean is $\frac{\sum_{i=1}^{n} \lambda x_i}{n} = \frac{\lambda \sum_{i=1}^{n} x_i}{n}$.
Substituting $\sum x_i = n\bar{x}$,we get the new mean as $\frac{\lambda (n\bar{x})}{n} = \lambda \bar{x}$.

(A) The mean (arithmetic mean) of $n$ discrete observations $y_1, y_2, \ldots, y_n$ is defined as the sum of all observations divided by the total number of observations.
Mathematically,this is expressed as:
$\text{Mean} = \frac{y_1 + y_2 + \ldots + y_n}{n} = \frac{\sum_{i=1}^n y_i}{n}$.

(B) The formula for the mean using the assumed mean method is given by:
${M_g} = a + \frac{1}{{\sum {f_i}}}(\sum {f_i}{d_i})$
where $a$ is the assumed mean.
Comparing this with the given equation ${M_g} = x + \frac{1}{{\sum {f_i}}}(\sum {f_i}{d_i})$,we get $x = a$.
Therefore,$x$ represents the assumed mean.

(A) Given that the mean of the numbers $27 + x, 31 + x, 89 + x, 107 + x, 156 + x$ is $82$.
$\frac{(27 + x) + (31 + x) + (89 + x) + (107 + x) + (156 + x)}{5} = 82$
$\frac{410 + 5x}{5} = 82$
$82 + x = 82$
$x = 0$
Now,we need to find the mean of $130 + x, 126 + x, 68 + x, 50 + x, 1 + x$.
Mean $= \frac{(130 + x) + (126 + x) + (68 + x) + (50 + x) + (1 + x)}{5}$
Mean $= \frac{375 + 5x}{5} = 75 + x$
Since $x = 0$,the mean is $75 + 0 = 75$.

(D) Given that the arithmetic mean of $x_1, x_2, ..., x_n$ is $\bar{x}$,we have $\bar{x} = \frac{\sum_{i=1}^{n} x_i}{n}$,which implies $\sum_{i=1}^{n} x_i = n \bar{x}$.
The required mean of the numbers $ax_1 + b, ax_2 + b, ..., ax_n + b$ is given by:
$\text{Mean} = \frac{(ax_1 + b) + (ax_2 + b) + ... + (ax_n + b)}{n}$
$= \frac{a(x_1 + x_2 + ... + x_n) + nb}{n}$
$= a \left( \frac{\sum_{i=1}^{n} x_i}{n} \right) + \frac{nb}{n}$
$= a \bar{x} + b$

(C) Let the $n$ observations be $x_1, x_2, \dots, x_n$.
The reciprocals of these observations are $\frac{1}{x_1}, \frac{1}{x_2}, \dots, \frac{1}{x_n}$.
The mean of these reciprocals is $\frac{\frac{1}{x_1} + \frac{1}{x_2} + \dots + \frac{1}{x_n}}{n}$.
The reciprocal of this mean is $\frac{n}{\frac{1}{x_1} + \frac{1}{x_2} + \dots + \frac{1}{x_n}}$.
This expression is the definition of the Harmonic Mean $(H.M.)$ of the $n$ observations.

(D) The mean $\bar{x}$ is given by the formula $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$.
Here,the values $x_i$ are $1, \frac{1}{2}, \frac{1}{3}, \dots, \frac{1}{n}$ and their corresponding frequencies $f_i$ are $1, 2, 3, \dots, n$.
Sum of products $\sum f_i x_i = (1 \times 1) + (2 \times \frac{1}{2}) + (3 \times \frac{1}{3}) + \dots + (n \times \frac{1}{n}) = 1 + 1 + 1 + \dots + 1$ ($n$ times) $= n$.
Sum of frequencies $\sum f_i = 1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2}$.
Therefore,the mean $\bar{x} = \frac{n}{\frac{n(n+1)}{2}} = \frac{2n}{n(n+1)} = \frac{2}{n+1}$.

(B) Let the sum of the first $10$ observations be $S_1$ and the sum of the remaining $30$ observations be $S_2$.
Given that the average of the first $10$ observations is $4.5$,we have:
$S_1 = 10 \times 4.5 = 45$
Given that the average of the remaining $30$ observations is $3.5$,we have:
$S_2 = 30 \times 3.5 = 105$
The total sum of all $40$ observations is $S = S_1 + S_2 = 45 + 105 = 150$.
The average of the whole group is $\frac{S}{40} = \frac{150}{40} = \frac{15}{4}$.

(A) The sum of the percentages in $3$ subjects is $75 + 80 + 85 = 240$.
Let the percentage in the fourth subject be $x$,where $0 \le x \le 100$.
The average of $4$ subjects is $\frac{240 + x}{4} = 60 + \frac{x}{4}$.
Since the minimum possible value for $x$ is $0$,the minimum average is $\frac{240 + 0}{4} = 60\%$.
Therefore,the average cannot be less than $60\%$.

(B) Given,$\frac{\Sigma x_i}{50} = 38$,therefore $\Sigma x_i = 50 \times 38 = 1900$.
After discarding two numbers $55$ and $45$,the new sum of the remaining $48$ numbers is $\Sigma x_{new} = 1900 - (55 + 45) = 1900 - 100 = 1800$.
The new number of observations is $n = 50 - 2 = 48$.
The new $A.M.$ is $\frac{1800}{48} = 37.5$.

(B) Let the average marks of the girls be $x$.
Total number of students = $100$.
Number of boys = $70$,so number of girls = $100 - 70 = 30$.
Total marks of the class = $100 \times 72 = 7200$.
Total marks of boys = $70 \times 75 = 5250$.
Total marks of girls = $7200 - 5250 = 1950$.
Average marks of girls = $\frac{1950}{30} = 65$.

(B) We know that the mean $\bar x = \frac{\sum_{i=1}^n x_i}{n}$,which implies $\sum_{i=1}^n x_i = n\bar x$.
The mean of the new set of numbers ${x_i} + 2i$ is given by:
$\text{Mean} = \frac{\sum_{i=1}^n (x_i + 2i)}{n} = \frac{\sum_{i=1}^n x_i + 2\sum_{i=1}^n i}{n}$
Substituting the known values:
$= \frac{n\bar x + 2 \times \frac{n(n+1)}{2}}{n}$
$= \frac{n\bar x + n(n+1)}{n}$
$= \bar x + (n + 1)$.

(D) Total number of students = $40 + 35 + 45 + 42 = 162$.
Total marks obtained = $(40 \times 50) + (35 \times 60) + (45 \times 55) + (42 \times 45)$.
$= 2000 + 2100 + 2475 + 1890 = 8465$.
Overall average of marks per student = $\frac{8465}{162} = 52.25$.

(C) The mean weight of $7$ students is $55 \ kg$.
Total weight of $7$ students $= 55 \times 7 = 385 \ kg$.
Sum of the weights of $6$ students $= 52 + 58 + 55 + 53 + 56 + 54 = 328 \ kg$.
Weight of the seventh student $= 385 - 328 = 57 \ kg$.

(B) The $Arithmetic \ mean$ is the most suitable measure of central tendency for evaluating the intelligence of students based on their marks because it considers every individual score in the data set,providing a comprehensive average that reflects the overall performance of the group.

(B) To find the $Median$,arrange the observations in ascending or descending order and identify the middle value.
If the number of observations is odd,the middle value is the $Median$.
If the number of observations is even,the $Median$ is the mean of the two middle values.
Therefore,the central value of a data set is defined as the $Median$.

(D) The median divides a data set into two equal halves.
In statistical terms:
$1$. The $50^{th}$ percentile $(P_{50})$ represents the value below which $50\%$ of the data falls,which is the median.
$2$. The $5^{th}$ decile $(D_5)$ represents the $50^{th}$ percentile,which is the median.
$3$. The $2^{nd}$ quartile $(Q_2)$ represents the $50^{th}$ percentile,which is the median.
$4$. The lower quartile $(Q_1)$ represents the $25^{th}$ percentile,which is not equal to the median.
Therefore,the correct option is $(d)$.

(A) Step $1$: Arrange the given data in ascending order: $6, 8, 9, 10, 11, 12, 14$.
Step $2$: Count the number of observations $(n)$. Here,$n = 7$,which is an odd number.
Step $3$: The formula for the median when $n$ is odd is the value of the ${\left( \frac{n+1}{2} \right)}^{th}$ term.
Step $4$: Substitute $n = 7$ into the formula: Median = ${\left( \frac{7+1}{2} \right)}^{th}$ term = ${\left( \frac{8}{2} \right)}^{th}$ term = $4^{th}$ term.
Step $5$: The $4^{th}$ term in the ordered sequence is $10$. Therefore,the median is $10$.

(A) The median divides a data set into two equal parts,representing the $50^{th}$ percentile.
Similarly,the second quartile ${Q_2}$ divides the data into two halves,the fifth decile ${D_5}$ represents the $50\%$ mark of the data,and the $50^{th}$ percentile ${P_{50}}$ is by definition the value below which $50\%$ of the observations fall.
Therefore,all these measures are equivalent: $M = {Q_2} = {D_5} = {P_{50}}$.

(C) For a symmetrical distribution,the median is the average of the first and third quartiles.
${Median} = \frac{{Q_1 + Q_3}}{2}$
Given ${Q_1} = 25$ and ${Q_3} = 45$.
${Median} = \frac{{25 + 45}}{2} = \frac{{70}}{2} = 35$.

(D) First,arrange the given $8$ values in ascending order:
$\alpha - 4, \alpha - 3.5, \alpha - 3, \alpha - 2.5, \alpha - 2, \alpha - 0.5, \alpha + 0.5, \alpha + 5$
Since the number of observations $n = 8$ (which is even),the median is the average of the $4^{th}$ and $5^{th}$ terms.
$4^{th} \text{ term} = \alpha - 2.5 = \alpha - \frac{5}{2}$
$5^{th} \text{ term} = \alpha - 2$
$\text{Median} = \frac{(\alpha - \frac{5}{2}) + (\alpha - 2)}{2}$
$\text{Median} = \frac{2\alpha - 4.5}{2} = \alpha - 2.25 = \alpha - \frac{9}{4}$

(C) The upper quartile $(Q_3)$ for a discrete distribution is given by the size of the $\left[ 3\frac{(n + 1)}{4} \right]^{th}$ item.
First,calculate the total frequency $(n = \Sigma f)$:
$n = 2 + 4 + 5 + 8 + 7 + 3 + 2 = 31$.
Substituting $n = 31$ into the formula:
$Q_3 = \text{Size of } \left[ 3\left( \frac{31 + 1}{4} \right) \right]^{th} \text{ item}$.

(D) Given that the number of observations $n = 9$.
The median of $9$ observations is the $\left( \frac{9+1}{2} \right)^{th} = 5^{th}$ observation.
Let the ordered observations be $x_1 < x_2 < x_3 < x_4 < x_5 < x_6 < x_7 < x_8 < x_9$.
The median is $x_5 = 20.5$.
If the largest $4$ observations $(x_6, x_7, x_8, x_9)$ are increased by $2$,the new set of observations is $x_1, x_2, x_3, x_4, x_5, (x_6+2), (x_7+2), (x_8+2), (x_9+2)$.
Since $x_5$ is smaller than $x_6$,and $x_6 < x_6+2$,the order of the first $5$ observations remains unchanged.
Therefore,the $5^{th}$ observation remains $x_5$,which is $20.5$.
Thus,the median remains the same as that of the original set.

(A) The mean is considered the most stable measure of central tendency because it incorporates every observation in a given distribution,making it less susceptible to sampling fluctuations compared to the median or mode.

(C) The arithmetic mean is calculated by summing all the observations and dividing by the total number of observations.
Because it incorporates the value of every single data point,the inclusion of an extreme value (an outlier) significantly shifts the sum,thereby changing the mean.
In contrast,the median and mode are positional or frequency-based measures that are much more robust against extreme values.
Therefore,the arithmetic mean is the most affected by extreme observations.

(B) The mean deviation of a set of observations about a point $k$ is minimized when $k$ is the median of the observations.
For a set of $n$ observations arranged in ascending order,the median is the middle term.
Here,$n = 101$,which is an odd number.
The position of the median is given by $\frac{n + 1}{2} = \frac{101 + 1}{2} = 51$.
Thus,the median is the $51^{st}$ observation,which is ${x_{51}}$.

(B) The initial average $M$ is given by $M = \frac{x_1 + x_2 + x_3 + ... + x_n}{n}$.
This implies the sum of the $n$ numbers is $nM = x_1 + x_2 + x_3 + ... + x_n$.
When $x_n$ is replaced by $x'$,the new sum of the numbers becomes $S' = (x_1 + x_2 + x_3 + ... + x_{n-1} + x_n) - x_n + x'$.
Substituting the sum $nM$ for the original numbers,we get $S' = nM - x_n + x'$.
The new average is the new sum divided by $n$,which is $\frac{nM - x_n + x'}{n}$.

(B) Arranging the data in ascending order of magnitude,we obtain:

Height (in $cm$)	$150$	$152$	$154$	$155$	$156$	$160$	$161$
Number of students	$8$	$4$	$3$	$7$	$3$	$12$	$4$
Cumulative frequency	$8$	$12$	$15$	$22$	$25$	$37$	$41$

Here,the total number of items $N = 41$,which is an odd number.
Hence,the median is the $\left(\frac{N + 1}{2}\right)^{th}$ item.
Median = $\left(\frac{41 + 1}{2}\right)^{th} = 21^{st}$ item.
From the cumulative frequency table,we find that the $21^{st}$ item corresponds to a height of $155$ $cm$ (since all items from $16^{th}$ to $22^{nd}$ are equal to $155$).
Therefore,the median is $155$ $cm$.

(C) Let the set of $n$ observations be $x_1, x_2, \dots, x_n$.
The mean is given by $\bar{x} = \frac{\sum x_i}{n}$.
When each observation $x_i$ is divided by $\alpha$ and increased by $10$,the new observations are $y_i = \frac{x_i}{\alpha} + 10$.
The new mean $\bar{y}$ is:
$\bar{y} = \frac{\sum y_i}{n} = \frac{\sum (\frac{x_i}{\alpha} + 10)}{n}$
$= \frac{1}{\alpha} \cdot \frac{\sum x_i}{n} + \frac{\sum 10}{n}$
$= \frac{1}{\alpha} \cdot \bar{x} + \frac{10n}{n}$
$= \frac{\bar{x}}{\alpha} + 10 = \frac{\bar{x} + 10\alpha}{\alpha}$.

(A) The median is the middle value of an ordered data set. For $21$ observations,the median is the $11^{th}$ term when arranged in ascending order.
Since only the observations greater than the median are increased,the $11^{th}$ term (the median itself) remains unchanged.
Therefore,the new median remains $40$.

(D) The sum of the first three terms is $3 \times 14 = 42$.
The sum of the next two terms is $2 \times 18 = 36$.
The sum of all five terms is $42 + 36 = 78$.
The mean of all five terms is $\frac{78}{5} = 15.6$.
Thus,the required mean is $15.6$.

(C) The sum of $n$ observations is $n\bar{x}$.
The sum of $n - 4$ observations is given as $K$.
The sum of the remaining $4$ observations is $n\bar{x} - K$.
Therefore,the mean of the remaining $4$ observations is $\frac{n\bar{x} - K}{4}$.

(B) First,arrange the given observations in ascending order:
$6, 8, 9, 10, 11, 12, 14$
The number of terms $n = 7$,which is odd.
The formula for the median of an odd number of observations is the $\left(\frac{n+1}{2}\right)^{\text{th}}$ term.
Median $= \left(\frac{7+1}{2}\right)^{\text{th}}$ term $= 4^{\text{th}}$ term.
The $4^{\text{th}}$ term in the ordered sequence is $10$.

(A) The mean $\bar{x}$ of a frequency distribution is given by the formula $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$.
First,calculate $\sum f_i x_i$:
$\sum f_i x_i = (3 \times 1) + (6 \times 2) + (9 \times 3) + (12 \times 4)$
$= 3 + 12 + 27 + 48 = 90$.
Next,calculate $\sum f_i$:
$\sum f_i = 1 + 2 + 3 + 4 = 10$.
Finally,calculate the mean:
$\bar{x} = \frac{90}{10} = 9$.

(C) Total number of students $N = 100$.
Mean marks of all students $\bar{X} = 72$.
Total marks of all students $= 100 \times 72 = 7200$.
Number of boys $n_1 = 70$.
Mean marks of boys $\bar{X}_1 = 75$.
Total marks of boys $= 70 \times 75 = 5250$.
Number of girls $n_2 = 100 - 70 = 30$.
Total marks of girls $= 7200 - 5250 = 1950$.
Mean marks of girls $\bar{X}_2 = \frac{1950}{30} = 65$.

(A) First,arrange the values in ascending order:
$\alpha - \frac{7}{2}, \alpha - 3, \alpha - \frac{5}{2}, \alpha - 2, \alpha - \frac{1}{2}, \alpha + \frac{1}{2}, \alpha + 4, \alpha + 5$
Since the number of observations $n = 8$ (which is even),the median is the average of the $4^{th}$ and $5^{th}$ terms.
Median $= \frac{1}{2} [(\alpha - 2) + (\alpha - \frac{1}{2})] = \frac{1}{2} [2\alpha - \frac{5}{2}] = \alpha - \frac{5}{4}$.

(A) The mean of the first $n$ squares is given by $\frac{\sum_{i=1}^{n} i^2}{n} = \frac{n(n+1)(2n+1)}{6n} = \frac{(n+1)(2n+1)}{6}$.
Given that the mean is $\frac{46n}{11}$,we have:
$\frac{(n+1)(2n+1)}{6} = \frac{46n}{11}$
$11(2n^2 + 3n + 1) = 276n$
$22n^2 + 33n + 11 = 276n$
$22n^2 - 243n + 11 = 0$
Factoring the quadratic equation:
$(n - 11)(22n - 1) = 0$
Since $n$ must be a positive integer,$n = 11$.

(C) First,arrange the observations in ascending order and calculate the cumulative frequency $(c.f.)$:

$x_i$	$3$	$6$	$7$	$10$	$12$	$15$
$f_i$	$3$	$4$	$13$	$2$	$8$	$10$
$c.f.$	$3$	$7$	$20$	$22$	$30$	$40$

Here,$N = 40$,which is even.
The median is given by $\frac{(\frac{N}{2})^{th} \text{ term} + (\frac{N}{2} + 1)^{th} \text{ term}}{2}$.
Median $= \frac{20^{th} \text{ term} + 21^{st} \text{ term}}{2}$.
From the table,the $20^{th}$ term is $7$ and the $21^{st}$ term is $10$.
Median $= \frac{7 + 10}{2} = \frac{17}{2} = 8.5$.

(B) The sum of $50$ observations $= 36 \times 50 = 1800$.
After removing two observations $30$ and $42$,the sum of the remaining $48$ observations is:
Sum $= 1800 - (30 + 42) = 1800 - 72 = 1728$.
The required mean $= \frac{1728}{48} = 36$.

(B) The given observations are $29, 32, 48, 50, x, x + 2, 72, 78, 84, 95$.
Total number of observations $n = 10$,which is an even number.
For an even number of observations,the median is the arithmetic mean of the $\left(\frac{n}{2}\right)^{th}$ and $\left(\frac{n}{2} + 1\right)^{th}$ terms.
Here,the $5^{th}$ term is $x$ and the $6^{th}$ term is $x + 2$.
Median $= \frac{x + (x + 2)}{2} = 63$.
$\frac{2x + 2}{2} = 63$.
$x + 1 = 63$.
$x = 62$.

(B) The sequence is $^{2n}C_0, ^{2n}C_1, ^{2n}C_2, \dots, ^{2n}C_n$.
The number of terms in the sequence is $n + 1$.
Since $n$ is even,$n + 1$ is odd.
The median of a sequence with an odd number of terms $N$ is the term at position $\frac{N+1}{2}$.
Here,$N = n + 1$,so the median is the term at position $\frac{(n+1)+1}{2} = \frac{n+2}{2} = \frac{n}{2} + 1$.
The term at position $\frac{n}{2} + 1$ in the sequence $^{2n}C_0, ^{2n}C_1, \dots, ^{2n}C_n$ is $^{2n}C_{n/2}$.

(A) Total marks obtained in three subjects out of $300$ is $75 + 80 + 85 = 240$.
If the marks of a fourth subject are added,the total marks are out of $400$.
To find the minimum possible average,we assume the student scores $0$ in the fourth subject.
Minimum average percentage $= \frac{240 + 0}{400} \times 100 = \frac{240}{400} \times 100 = 60\%$.

(A) Step-$1$: Identify the median position.
For $n = 9$ observations (where $n$ is odd),the median is the $\frac{n+1}{2}^{\text{th}}$ term.
Median $= \frac{9+1}{2} = 5^{\text{th}}$ term.
Step-$2$: Analyze the effect of the change.
The four largest observations are the $6^{\text{th}}, 7^{\text{th}}, 8^{\text{th}}$,and $9^{\text{th}}$ terms.
Since these terms are greater than the $5^{\text{th}}$ term,increasing them by $2$ does not change the value or the position of the $5^{\text{th}}$ term.
Step-$3$: Conclusion.
As the $5^{\text{th}}$ term remains unchanged,the median of the new set remains the same as the original median,which is $20.5$.

(A) The mean $\bar{x}$ is calculated using the formula $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$.
First,calculate $\sum f_i = 12 + 18 + 27 + 20 + 17 + 6 = 100$.
Next,calculate $\sum f_i x_i = (12 \times 5) + (18 \times 15) + (27 \times 25) + (20 \times 35) + (17 \times 45) + (6 \times 55)$.
$\sum f_i x_i = 60 + 270 + 675 + 700 + 765 + 330 = 2800$.
Therefore,$\bar{x} = \frac{2800}{100} = 28$.

(B) The mean of the first $n$ natural numbers is given by the formula $\frac{n + 1}{2}$.
Given that the mean is $\frac{n + 7}{3}$,we set up the equation:
$\frac{n + 1}{2} = \frac{n + 7}{3}$
Cross-multiplying,we get:
$3(n + 1) = 2(n + 7)$
$3n + 3 = 2n + 14$
$3n - 2n = 14 - 3$
$n = 11$

(D) The mean is calculated using the formula $\text{Mean} = \frac{\sum x_i f_i}{\sum f_i}$.
Given values $x_i = \{1, \frac{1}{2}, \frac{1}{3}, \dots, \frac{1}{n}\}$ and frequencies $f_i = \{1, 2, 3, \dots, n\}$.
Sum of products $\sum x_i f_i = (1 \times 1) + (\frac{1}{2} \times 2) + (\frac{1}{3} \times 3) + \dots + (\frac{1}{n} \times n) = 1 + 1 + 1 + \dots + 1$ ($n$ times) $= n$.
Sum of frequencies $\sum f_i = 1 + 2 + 3 + \dots + n = \frac{n(n + 1)}{2}$.
Therefore,$\text{Mean} = \frac{n}{\frac{n(n + 1)}{2}} = \frac{2n}{n(n + 1)} = \frac{2}{n + 1}$.