MATHS M.C.Q

The median is the middle score for a set of data that has been arranged in ascending or descending order of magnitude.
For even number of observations, the median is calculated as an average of two middle numbers.
For the given example $30$ and $x$ are in the middle and the median is $35$.
So,
$35=\frac{30+\text{x}}{2}$
$70=30+\text{x}$
$\text{x}=40$

Frequency polygon is the line graph plotted with class marks on $x$-axis & frequency of the class on $y$-axis.

Let, Vihaan obtains $x$ marks in the fourth test.
So,
$\frac{92+85+78+\text{x}}{4}=87$
$\frac{255+\text{x}}{4}=87$
$255 + x = 348$
$x = 348 - 255$
$x = 93$ marks

The given data is $0, 2, 2, 2, -3, 5, -1, 5, 5, -3, 6, 6, 5$ and $6$.
Arranging the given data in ascending order, we have
$-3, -3, -1, 0, 2, 2, 2, 5, 5, 5, 5, 6, 6, 6$
Here, the number of observation $n = 14$, which is an even number.
Hence, the median is
$\frac{\Big(\frac{\text{n}}{2}\Big)^\text{th}\text{ observation }+{\Big(\frac{\text{n}}{2}+1\Big)}^\text{th}\text{ observation }}{2}$
$=\frac{\Big(\frac{14}{2}\Big)^\text{th}\text{ observation }+{\Big(\frac{14}{2}+1\Big)}^\text{th}\text{ observation }}{2}$
$=\frac{7^\text{th}\text{ observation }+8^\text{th}\text{ observation }}{2}$
$=\frac{2+5}{2}$
$=\frac{7}{2}$
$=3.5$

Class mark is the mid-value of each class interval.
For class interval $15 - 20$.
Class mark $=\frac{15+20}{2}=\frac{35}{2}=17.5$

In statistics, the mode in a list of numbers refers to the integers that occur most number of times.
For the set of numbers, $9$ occurs three times i.e., more than any other number in the list.

We have, $\bar{\text{x}}=\frac{\sum\text{x}_\text{i}}{\text{n}}$
$\Rightarrow50=\frac{\sum\text{x}_\text{i}}{100}$
$\Rightarrow\sum\text{x}_\text{i}=50\times100=5,000$
If one of the observation which was 50 is resplaced by 150, then
$\sum\text{x}_\text{i}=5,000-50+150=5100$
Then, the resulting mean $=\frac{5100}{100}=51$

For the data $2, 2, 2, 2, 4$ of $5$ numbers, we have
Mean $=\frac{2+2+2+2+4}{5}=\frac{12}{5}=2.4$
Median $=\Big(\frac{5+1}{2}\Big)^\text{th}$ Value $=$ $3rd$ Value $= 2$
Since, $2$ occurs maximum number of times, Mode $= 2$
Mean $\neq$ Median
For the data $1, 3, 3, 3, 5$ of $5$ numbers, we have
Mean $=\frac{1+3+3+3+5}{5}=\frac{15}{5}=3$
Median $=\Big(\frac{5+1}{2}\Big)^\text{th}$ Value $= 3^{rd}$ Value $= 3$
Since, $3$ occurs maximum number f times, Mode $= 3$
Mean $=$ Median $=$ Mode

Let $x_1, x_2, x_3, x_4$, and $x_5$ be five numbers and one of the excluded number be $x_5$,
Given, mean of the numbers $=30$
$\Rightarrow\frac{\text{x}_1+\text{x}_2+\text{x}_3+\text{x}_4+\text{x}_5}{5}=30$
$\Rightarrow {\text{x}_1+\text{x}_2+\text{x}_3+\text{x}_4+\text{x}_5}=150$
$\Rightarrow {\text{x}_1+\text{x}_2+\text{x}_3+\text{x}_4}=150-\text{x}_5$
On dividing both sides by 4, we get
$\frac{\text{x}_1+\text{x}_2+\text{x}_3+\text{x}_4}{4}=\frac{150-\text{x}_5}{4}\ \dots(\text{i})$
Given, mean of the numbers $= 28$
$\therefore\ \frac{150-\text{x}_5}{4}=28$ [from Eq. $(i)$]
$\Rightarrow 150 -\text{x}_5=112$
$\Rightarrow \text{x}_5=150 -112$
$\Rightarrow\text{x}_5=38$
Hence, the excluded number is $38$.

Given that:
$\frac{\text{x}+(\text{x+3})+(\text{x+5})+(\text{x+7})+(\text{x+10})}{5}=9$
$\frac{5\text{x}+25}{5}=9$
$5x + 25 = 45$
$5x = 20$
$x = 4$
So the last three numbers are $9, 11, 14$
So there mean is $11\frac{1}{3}$

Let us take n observations $X_1, \ldots, X_n$
If $\overline{\text{X}}$ be the mean of the n observations then we have
$\overline{\text{X}}=\frac{1}{\text{n}}\sum_{\text{i}=1}^{\text{n}}\text{X}_{\text{i}}$
$\Rightarrow\sum_{\text{i}=1}^{\text{n}}\text{X}_{\text{i}}=\text{n}\overline{\text{X}}$
Multiply a constant k to each of the observations then observations becomes $kXi . \ldots \ldots kX_{N}$
If $\overline{\text{Y}}$ be the mean of the new observations, then we have
$\overline{\text{Y}}=\frac{1}{\text{n}}\sum_{\text{i}=1}^{\text{n}}\text{kX}_{\text{i}}$
$=\frac{\text{k}}{\text{n}}\sum_{\text{i}=1}^{\text{n}}\text{X}_{\text{i}}$
$={\text{k}}\cdot=\frac{1}{\text{n}}\sum_{\text{i}=1}^{\text{n}}\text{X}_{\text{i}}$
$=\text{k}\overline{\text{X}}$

Let $x$ and $y$ be the uppel and lower and lower class limit in a frequency distribution.
Now, mid value of a class $\frac{(\text{x}+\text{y})}{2}=10$ [given]
$\Rightarrow x + y = 20 ...(i)$
Also, given that, width of class$ x - y = 6 ...(ii)$
On adding Eqs. $(i)$ and $(ii)$, we get
$2x = 20 + 6$
$\Rightarrow 2x = 26$
$\Rightarrow x = 13$
On putting $x = 13$ in Eq. $(i)$, we get
$13 + y = 20$
$\Rightarrow y = 7$
Hence, the lower limit of the class is $7$.

Arranging the points in an ascending order,
We have:
$0, 13, 15, 20, 27, 29, 31, 34, 43, 50, 56,$
Here, $n = 11$, Which is odd
$\therefore\ $median score = value of $=\frac{1}{2}(11+1)^\text{th}$ term
$=$ value of $\Big(\frac{1}{2}\times12\Big)^\text{th}$ term
$=$ value of $6^{th}$ term
$= 29$

To represent ‘the less than type’ graphically, we plot the upper limits on the $x$-axis. e.g. marks obtained by students are represented in grouped data as $(0-10), (10-20), (20-30), (30-40) ...$ only upper limits such as $10, 20, 30, 40 ...$ are taken for the $x$-axis

Given ,$\frac{\text{x}+\frac{1}{\text{x}}}{2}=\text{M}$
Taking square on both sides
$\bigg(\frac{\text{x}+\frac{1}{\text{x}}}{2}\bigg)^2=(\text{M})^2$
$\bigg(\text{x}+{\frac{1}{\text{x}}}\bigg)^2=(2\text{M})^2$
$\bigg(\text{x}^2+2+{\frac{1}{\text{x}^2}}\bigg)=(2\text{M})^2$
$\bigg(\text{x}^2+{\frac{1}{\text{x}}}\bigg)^2={4\text{M}^2-2}$
Divide by $2$ on both sides to get mean
$\bigg(\frac{\text{x}^2+\frac{1}{\text{x}^2}}{2}\bigg)^2=2\text{M}^2-1$

Where, $l$ = lower limit of the median class
$F$ = frequency of the median class
$CF$ = cumulative frequency of the class preceding the median class
$N$ = number of observations
$H$ = size of the class interval (assuming all class sizes to be equal)

Given, Mean of $25$ observations $= 36$
$\therefore$ Sum of $25$ observations $= 36 \times 25 = 900$
Now, the mean of first $13$ observations $= 32$
$\therefore$ Sum of first $13$ observations $= 13 \times 32 = 416$
and the Mean of last $13$ observations $= 40$
$\therefore$ Sum of last $13$ observations $= 40 \times 13 = 520$
So, $13\ th$ observation = (Sum of last $13$ observations + Sum of first $13$ observations) - (Sum of $25$ observations)
$= (520 + 416) - 900 = 936 - 900 = 36$
Hence, the $13\ th$ observation is $36$.

Given that the mean of $a, b, c, d$ and $e$ is $28$. They are $5$ in numbers.
Hence, we have
$\frac{\text{a}+\text{b}+\text{c}+\text{d}+\text{e}}{5}=28$
$\Rightarrow\frac{(\text{a}+\text{c}+\text{e})+(\text{b}+\text{d})}{5}=28$
$\Rightarrow\frac{(\text{a}+\text{c}+\text{e})}{5}+\frac{(\text{b}+\text{d})}{5}=28$
But, it is given that the mean of $a, c$ and $e$ is $24$. Hence, we have
$\Rightarrow\frac{(\text{a}+\text{c}+\text{e})}{5}=24$
$\Rightarrow\text{a}+\text{c}+\text{e}=72$
Then We have
$\frac{72}{5}+\frac{(\text{b}+\text{d})}{5}=28$
$\Rightarrow\frac{\text{b}+\text{d}}{5}=28-\frac{72}{5}$
$\Rightarrow\frac{\text{b}+\text{d}}{5}=28-14.4$
$\Rightarrow\frac{\text{b}+\text{d}}{5}=13.6$
$\Rightarrow\text{b}+\text{d}=68$
$\Rightarrow\frac{\text{b}+\text{d}}{2}=\frac{68}{2}$
$\Rightarrow\frac{\text{b}+\text{d}}{2}=34$
Hence, the mean of $b$ and $d$ is $34$.

Arranging the points in an ascending order,
We have:
$31, 35, 36, 38, 40, 44, 45, 52, 55, 60$
Here, $n = 10$, Which is even
$\therefore\ $median = mean of $\Big(\frac{10}{2}\Big)^\text{th}$ and $\Big(\frac{10}{2}+1\Big)^\text{th}$ terms
$=$ mean of $\Big(\frac{10}{2}\Big)^\text{th}$ and $\Big(\frac{12}{2}\Big)^\text{th}$ term
$=$ mean of $5^{th}$ and $6^{th}$ term
$=\frac{1}{2}(40+44)$
$=\frac{1}{2}\times84$
$=42\text{kg}$

Let $l$ and $m$ respectively be the lower and upper limits of the class.
Then the mid-value of the class is $\frac{\text{l+m}}{2}$ and the class-size is $(l - m)$.
Given that the mid-value of the class is $42$ and the class-size is $10$.
Therefore, we have two equations
$\frac{\text{l+m}}{2}=42$
$\Rightarrow l + m = 84,$
$m - l = 10$
Adding the above two equations, we have
$(l + m) + (m - l) = 84 + 10$
$\Rightarrow l + m + m - l = 94$
$\Rightarrow 2m = 94$
$\Rightarrow l = 37$
Substituting the value of $m$ in the first equation, we have
$l + 47 = 84$
$\Rightarrow l = 84 - 47$
$\Rightarrow l = 37$
Hence, the upper and lower limits of the class are $47$ and $37$ respectively.
Thus, the correct choice is $(a)$.