Maths M.C.Q

Mean of $100$ items $= 64$
Sum of $100$ items $= 64 × 100 = 6400$
Correct sum $= (6400 + 36 + 90 - 26 - 9) = 6491$
Correct mean $=\frac{6491}{100}=64.91$

The given frequency varies from $14$ to $112.$
So the class intervals are:
$13-22, 23-32, 33-42, 43-52, 53-62, 63-72, 73-82, 83-92, 93-102, 103-112.$
Number of class interval $= 10.$

Class mark $=\frac{190+120}{2}=\frac{210}{2}=105$

Calculated sum $= 64 × 100 = 6400$
Correct sum of these numbers
$= 6400 + ($sum of correct term$) - ($sum of incorrect term$)$
$= 6400 + (36 + 90) - (26 + 9)$
$= 6400 + 36 + 90 - 26 - 9$
$= 6491$
Correct mean $=\frac{6491}{100}$
$=64.91$

The median is the middle score for a set of data that has been arranged in ascending or descending order of magnitude.
Mode in a list of numbers refers to the integers that occur most number of times.
For list $3, 3, 4$
Both median and mode are $3.$

The class width is the difference between the upper- or lower-class limits of consecutive classes.
In this case, class width equals to the difference between the lower limits of the first two classes.
Let, W be the class width
$W = 21 - 1 = 20$
So class width is $20$

Adjusted frequency for the class $25-45$ is
$10 - 8 = 2$

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, median is calculated as average of two middle number
$22=\frac{(\text{x}+1)+(\text{x}+3)}{2}$
$44=2\text{x}+4$
$40=2\text{x}$
$\text{x}=20$

Mid value $=\frac{\text{Lower}\ \text{limit}+\text{Upper}\ \text{limit}}{2}$
$\Rightarrow\text{m}=2\text{m}-\text{L}$
$\therefore$ Upper class boundry of the class $= 2m - L.$

Class size is the difference between two consecutive values of the class mark.
Here, the difference between two consecutive class mark is $6.$
i.e., $34 - 28 = 6$

Since, $3$ Median $= 2$ Mean $+$ Mode
$\therefore 3 × 33 = 2$ Mean $+ 45$
$⇒ 2$ Mean $= 99 - 45$
$⇒ 2$ Mean $= 54$
$⇒$ Mean $= 27$

Let sum of all the $30$ observations be $x.$
The mean is equal to the sum of all the values in the data set divided by the number of values in the data set.
$\frac{\text{x}}{30}=12$
$\text{x}=360$
$360-25=335$

In a histogram the class intervals or the groups are taken along the horizontal axis or $X−$axis.

A cumulative frequency distribution is the sum of the class and all classes below it in a frequency distribution.
Subtract the previous cumulative frequency (c.f.) from the cumulative frequency of the current class.
So frequency of the class interval $20 - 30$ is $14 - 5 = 9$

Given that, the lower class limit of a class-interval is l and the mid-point of the class is $m.$ Let $u$ be the upper class limit of the class-interval.
Therefore, we have
$\text{m}=\frac{\text{l+u}}{2}$
$⇒ l + u = 2m$
$⇒ u = 2m - l$
Thus the upper class limit of the class is $(2m - l).$
Hence, the correct choice is $(c).$

Mid-value $= 10$
$\Rightarrow\frac{\text{Upper limit + Lower limit }}{2}=10$
$⇒$ Upper limit $+$ Lower limit $= 20 .....(i)$
Also, Class length $= 6$
$⇒$ Upper limit $-$ lower limit $= 6 .....(ii)$
Subtracting $(ii)$ from $(i),$ we get
$2 ×$ Lower limit $= 14$
$⇒$ Lower limit $= 7$

We are given frequency distribution $15, 20, 25, 30 ....$
Class size $= 20 - 15 = 5$
Class marks $= 20$
Lower limit $=\Big(20-\frac{5}{2}\Big)$
$=\frac{35}{2}=17.5$
Upper limit $=\Big(20+\frac{5}{2}\Big)$
$=\frac{45}{2}=22.5$
Thus, the required class is $17.5-22.5.$

Class mark $=\frac{130+150}{2}=\frac{280}{2}=140$

Sum of all the recorded speeds is
$48 + 52 + 57 + 55 + 42 + 39 + 60 + 49 + 53 + 47 = 502$
Because of the error, $\frac{5\text{km}}{\text{h}}$ in each case
The sum increases by $50$ i.e., $552$
So the average speed of $10$ vehicle is $\frac{55.2\text{km}}{\text{h}}$

Arranging the marks in an ascending order,
We have:
$14, 14, 14, 14, 15, 15, 15, 15, 16, 17, 18, 19, 19, 20$
Clearly, $15$ occurs maximum number of times.
Hence, mode $= 15$

We know that algebraic sun of deviations from mean is zero.
$\sum\limits_{\text{a}=1}^\text{b} (\text{x}_\text{t}-\bar{\text{x}})=(\text{x}_1-\bar{\text{x}})+(\text{x}_2-\bar{\text{x}})+(\text{x}_3-\bar{\text{x}})+\ ...\ +(\text{x}_\text{n}-\bar{\text{x}})$
$= (\text{x}_1+\text{x}_2+\text{x}_3+... +\text{x}_\text{n})- \text{n}\bar{\text{x}}$
$\Rightarrow\sum\limits_{\text{t}-1}^\text{b}\text{x}_\text{i}-\text{n}\bar{\text{x}}=\text{n}\bar{\text{x}}-\text{n}\bar{\text{x}}=0$ $\bigg[\because\sum\limits_{\text{i}=1}^\text{n} \text{x}_\text{i}=\text{n}\bar{\text{x}}\bigg]$
Hence, $(b)$ is correct answer.

The mode in a list of numbers refers to the integers that occur most number of times.
In the given list both $8$ and $9$ occur two times.
So the value of $x$ will decide the mode
If $x = 8,$ then the mode will be $8$
If $x = 9,$ then the mode will be $9$
Hence, the difference between the two modes is $1.$

If each observation of the data is decreased by $8$ then their mean is also decreased by $8.$

A cumulative frequency distribution is the sum of the class and all classes below it in a frequency distribution.
Less than $30$ has the class interval $20-30.$ Frequency of this class interval will be corresponding to.

The mean is equal to the sum of all the values in the data set divided by the number of values in the data set.
$\frac{4+7+\text{x}+8+9+10}{5}=8$
$\frac{(38+\text{x})}{6}=8$
$\text{x}=48-38=10$

Mean of 5 observations $= 11$
$\text{Mean}=\frac{\text{Sum of all observations}}{\text{Total number of observation}}$
$\Rightarrow11=\frac{\text{x}+\text{x}+2+\text{x}+4+\text{x}+6+\text{x}+8}{5}$
$\Rightarrow11=\frac{5\text{x}+20}{5}$
$\Rightarrow55=5\text{x}+20$
$\Rightarrow5\text{x}=35$
$\Rightarrow\text{x}=7$

Given $\bigg(\frac{\text{x}+\frac{1}{\text{x}}}{2}\bigg)^2=(\text{M})^2$
Taking cube on both sides
$\bigg(\frac{\text{x}+\frac{1}{\text{x}}}{2}\bigg)^3=(\text{M})^3$
$\bigg(\text{x}+{\frac{1}{\text{x}}}\bigg)^3=(2\text{M})^3$
$\bigg(\text{x}^2+3\text{x}\times{\frac{1}{\text{x}}}\Big(\text{x}+\frac{1}{\text{x}}\Big)+\frac{1}{\text{x}^3}\bigg)=(2\text{M})^3$
$\bigg(\text{x}^3+{\frac{1}{\text{x}^3}}\bigg)={8\text{M}^3-3}\Big(\text{x}+\frac{1}{\text{x}}\Big)$
Divide by $2$ on both sides to get mean
$\frac{\bigg(\text{x}^3+{\frac{1}{\text{x}^3}}\bigg)}{2}={4\text{M}^3-\frac{3}{2}}\Big(\text{x}+\frac{1}{\text{x}}\Big)$
$\frac{\bigg(\text{x}^3+{\frac{1}{\text{x}^3}}\bigg)}{2}={4\text{M}^3}-{3\text{M}}$

$\frac{150}{1000}=\frac{3}{20}$

Class mark $=\frac{\text{Upper}\ \text{limit}+\text{Lower}\ \text{limit}}{2}$
$=\frac{120+100}{2}$
$=110$

The observation corresponding to class $310–330 (330$ not included in this interval$)$ are $310, 310, 320, 319, 318, 316,$ i.e., $6$ observations.

The number of times a particular item occurs in a given data is called the frequency of the item. Hence, the correct choice is $(b).$

$32$ is the smallest even integer.
So three consecutive even integers are $32, 34$ and $36$
The mean is equal to the sum of all the values in the data set divided by the number of values in the data set.
$\frac{32+34+36}{3}=34$

The marks obtained by the students are $81, 72, 90, 90, 86, 85, 92, 70, 71, 83, 89, 95, 85, 79$ and $62.$
The highest and lowest marks are $95$ and $62$ respectively. Therefore, the range of marks is
$95 - 62$
$= 33$
Hence, the correct option is $(d).$

Maximum value of the variate $= 32$
And the minimum value of the variate $= 6$
Range $=$ Maximum value of the variate-Minimum value of the variate $= 32 - 6 = 26$

if is the mean of n observations $x_1, x_2, x_3, x_4 \ldots x_n$
then algebraic sum of deviations $=\sum\limits^\text{n}_{\text{i}=0}\Big(\text{x}_\text{i}-{\overline{\text{X}}}\Big)$
$=\sum\limits^\text{n}_{\text{i}=0}\text{x}_\text{i}-\text{n}{\overline{\text{X}}}$
$=\text{n}\bigg(\frac{\sum^\text{n}_{\text{i}=0}\text{x}_\text{i}}{\text{n}}\bigg)-\text{n}\overline{\text{X}}$
$=\text{n}\overline{\text{X}}-\text{n}\overline{\text{X}}$
$=0$

If the less than ogive and the more than ogive intersect at $(32, 48)$, then the median of the data is $32.$ Because on the graph, the point of the $x-$axis, where less than ogive and more than ogive intersects, is the median. Therefore, the Median of the data is $32.$

Class mark $=\frac{\text{Upper limit + lower limit}}{2}=\frac{120+100}{2}=110$

Let the mean of the initial sequence is $x.$
Given that, after subtracting $53$ from each number, the difference between the means is $3.5$
So, $x - 53 = 3.5$
Mean of the number is $x = 53 + 3.5 = 56.5$

The mean of $50$ observations is $39.$
So sum of these $50$ observations is $50 × 39 = 1950$
After replacing the observation value $23$ by $43,$
Sum becomes $1970$
So the mean is $\frac{1970}{50}=39.4$

Since, $3$ Median $=$ Mode $+\ 2$ Mean
$\Rightarrow\text{Median}=\frac{\text{Mode}}{3}+\frac{2}{3}\text{Mean}$
$=\frac{\text{Mode}}{3}+\frac{2}{3}\text{Mean}-\frac{2}{3}\text{Mode}+\frac{2}{3}\text{Mode}$
$\text{Median}=\text{Mode}+\frac{2}{3}(\text{Mean - Mode})$

First, we arrange the given observations in ascending order as follows
$22, 34, 39, 45, 54, 54, 56, 68, 78$ and $84$
Here, total number of observation, $n = 10$
Since, $n$ is even, so we use the formula for median,
Median $=\frac{\big(\frac{\text{n}}{2}\big)^{-1}\text{observation +}\big(\frac{\text{n}}{2}+1\big)\text{th observation}}{2}$
$=\frac{\big(\frac{10}{2}\big)\text{ th observation +}\big(\frac{10}{2}+1\big)\text{th observation}}{2} [$put $n = 10]$
$=\frac{5\text{th observation $+ 6th$ obsevation}}{2}=\frac{54+54}{2}=\frac{108}{2}=54$
Hence, the median of given data is $54.$

The adjusted frequency for the class $25 - 45$ is
$= \frac{\text{Frequency of the class }}{\text{Class width }}\times \text{Minimum width }=\frac{8}{20} \times5=2$

Median $= 4$
Mode $= 2$
$\text{Mean}=\frac{2+2+4+5+12}{5}=\frac{25}{5}=5$
Hence, $($Mean $= 5) > ($Mode $= 2)$

The given data is $5, 4, 7, 5, 8, 8, 8, 5, 7, 9, 5, 7, 9, 10$ and $8$
Make the frequency table.

The given data is $7, 5, 13, x$ and $9.$ They are $5$ in numbers.
The mean is $\frac{7+5+13+\text{x}+9}{5}=\frac{34+\text{x}}{5}$
But, it is given that the mean is $10$. Hence, we have
$\frac{34+\text{x}}{5}=10$
$⇒ 34 + x = 50$
$⇒ x = 50 - 34$
$⇒ x = 16$

Add the values corresponding to the height of the bar before $40.$
$6 + 3 + 4 + 2 = 15$

Difference between the maximum & minimum value of the observation is called as range.
So, $32 - 6 = 26$

Sol. Width of each of the five continuous classes in a frequency distribution is $5.$
Lower class limit of the lowest class $= 10$
Upper class limit of the lowest class is $10 + 5 = 15$
So, the five continuous classes are
$10 - 15, 15 - 20, 20 - 25, 25 - 30, 30 - 35$
Hence, the upper-class limit of the height class is $35.$

The class width is the difference between the upper- or lower-class limits of consecutive classes. In this case, class width equals to the difference between the lower limits of the first two classes.
$w = 10 - 0$
So, the class width is $10$

The number $20$ is included in $20–30.$
Hence, $(b)$ is the correct answer.

As the class size $= 5,$ Class interval is got by subtracting and adding $2.5$ to $43.$

Let the lower limit $= k$
Mid-point $= m$
Upper limit $= l$
Mid-point $=\frac{\text{(upper limit + lower limit)}}{2}$
$\text{m}=\frac{(\text{k+l})}{2}$
$2\text{m}=\text{k}+\text{l}$
$\text{k}=2\text{m}-\text{l}$
Therefore, lower limit $= 2m - l$

The mean of the six numbers is $23.$
So the sum of six numbers is $23 × 6 = 138$
After excluding one number, the mean of the remaining numbers is $20.$
So the sum of five numbers is $20 × 5 = 100$
The difference between them is$138 - 100 = 38$

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, median is calculated as average of two middle number.
for the given example $30$ and $x$ are in the middle and median is $35.$
So,
$35=\frac{30+\text{x}}{2}$
$70=30+\text{x}$
$\text{x}=40$

The mean of five observations is $15$
So the sum of these five observations is $15 × 5 = 75$
The mean of first three observations is $14$
So the sum of the first three observations is $14 × 3 = 42$
So the sum of the last two numbers is $75 - 42 = 33$
The mean of the last three observations is $17.$
So sum of last three observations is $17 × 3 = 51$
So the middle number is $51 - 33 = 18.$

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

Class mark $=\frac{\text{( upper limit + Lower limit ) }}{2}$
$=\frac{50+60}{2}$
$=\frac{110}{2}$
So class mark is $55.$

The median is the middle score for a set of data that has been arranged in ascending or descending order of magnitude.
Since $2x + 1$ is in the middle of the arranged numbers, so it is median
Hence, $2x + 1 = 7$
Now since $7$ occurs more number of times then other numbers so mode of the list is $7.$

$1\ cm = 30\ km$
So for $75\ km$
$\frac{75}{30}=2.5\text{cm}$

Average is equal to the sum of all the values in the data set divided by the number of values in the data set.
Average $=\frac{\text{x}+\text{x}+\text{y}+\text{y}+\text{y}}{5}$
Average $=\frac{2\text{x}+3\text{y}}{5}$

Given that the mean of $7$ numbers is $81$. Let us denote the numbers by $X_1, \ldots . X_7$
If $\overline{\text{X}}$ be the mean of the n observations $X_1, \ldots X_n$, then we have
$\overline{\text{X}}=\frac{1}{2}\sum_{\text{i}=1}^{\text{n}}\text{x}_\text{i}$
$\Rightarrow\sum_{\text{i}=1}^{\text{n}}\text{x}_\text{i}=\text{n}\overline{\text{X}}$
Hence the sum of $7$ numbers is
$\sum_{\text{i}=1}^{7}\text{x}_\text{i}=7\times81=567$
If one number is discarded then the mean becomes $78$ and the total numbers become $6.$
Let the number discarded is $x.$
After discarding one number the sum becomes $567 - x $ and then the mean is
$\frac{567-\text{x}}{6}$
But it is given that after discarding one number the mean becomes $78.$
Hence we have
$\frac{567-\text{x}}{6}=78$
$\Rightarrow 567 - x = 468$
$\Rightarrow 567 = x + 468$
$\Rightarrow x = 468 = 567$
$\Rightarrow x = 567 - 468$
$\Rightarrow x = 99$
Thus the excluded number is $99.$

Minimum value $= 14$
Maximum value $= 112$
The classes are
$13 - 22, 23 - 32, 33 - 42, 43 - 52, 53 - 62, 63 - 72, 73 - 82, 83 - 92, 93 - 102$ and $103 - 112.$
The number of classes in the distribution will be $10.$

The mean of $50$ observations is $39.$
So sum of these $50$ observations is $50 × 39 = 1950$
After replacing the observation value $23$ by $43,$
Sum becomes $1970$
So the mean is $\frac{1970}{50}=39.4$

$\text{Mean}=\frac{\text{a}+\text{b}+\text{c}+\text{d}+\text{e}}{5}=28$
$⇒ a + b + c + d + e = 140 ...(1)$
Also, $\text{Mean}=\frac{\text{a}+\text{c}+\text{e}}{3}=24$
$⇒ a+ c + e = 72 ...(2)$
Subtracting equation $(2)$ from $(1),$ we have
$b + d = 68$
$\text{Mean}=\frac{\text{b}+\text{d}}{2}=\frac{68}{2}=34$

Mean of first $13$ observation $= 32$
$\therefore$ Sum of all first $13$ observation $= (32 \times 13) = 416$
Mean of last $13$ observation $= 40$
$\therefore$ Sum of all last $13$ observation $= (40 \times 13) = 520$
Mean of $25$ observation $= 36$
$\therefore$ Sum of all first $25$ observation $= (36 \times 25) = 900$
Hence, $13^{th}$ observation $= 416 + 520 - 900 = 36$

Since mean is equal to the sum of all the values in the data set divided by the number of values in the data set.
So increase in the value of each observation will also increase the mean (average) by $3.$

The median is the middle score for a set of data that has been arranged in ascending or descending order of magnitude.
So for numbers $1, 2, 3, ... 9$
$5$ is the median.

Arranging the numbers in ascending order, we have:
$31, 35, 36, 38, 40, 44, 45, 52, 55, 60$
Here, $n$ is $10,$ which is an even number. Thus, we have:
Median $=$ Mean of $\Big(\frac{\text{n}}{2}\Big)\text{th}$ obervation & $\Big(\frac{\text{n}}{2}+1\Big)\text{th}$ obervation Median Weight $=$ mean of the weights of $\Big(\frac{10}{2}\Big)\text{th}$ student & $\Big(\frac{10}{2}+1\Big)\text{th}$ student $=$ mean of the weights of $5th$ student & $6th$ student$=\frac{1}{2}(40+44)=42$
Hence, the median weight is $42kg.$

Required mean $=\frac{(\text{ax}_1+\text{ax}_2+_{\dots}+\text{ax}_\text{n})+(\frac{\text{x}_1}{\text{a}},\frac{\text{x}_2}{\text{a}},\ \dots,\frac{\text{x}_\text{n}}{\text{a}})}{2\text{n}}$
$=\frac{1}{2}\bigg[\frac{\text{a}(\text{x}_1+\text{x}_2+{\dots}+\text{x}_\text{n})}{\text{n}}+\frac{\frac{1}{2}(\text{x}_1+\text{x}_2+{\dots}+\text{x}_\text{n})}{\text{n}}\bigg]$
$=\frac{1}{2}\Big[\text{a}\bar{\text{x}}+\frac{1}{\text{a}}\bar{\text{x}}\Big]$
$=\Big[\text{a}+\frac{1}{\text{a}}\Big]\frac{\bar{\text{x}}}{2}$

$\text{Mean}=\frac{7+5+13+\text{x}+9}{5}=10$
$\Rightarrow34+\text{x}=50$
$\Rightarrow\text{x}=16$

The empirical relationship between the three measures of central tendencies is $3$ Median $=$ Mode $+\ 2$ Mean.
The relationship is as per observation. A distribution in which the values of mean, median and mode coincide $($i.e., mean $=$ median $=$ mode$)$ is called symmetrical
Distribution. conversely, when the values of mean, median, mode are not equal, the distribution is called asymmetrical or skewed.
Knowing any two values, the third can be computed by this formula
$3$ median $= 2$ mean $+$ mode
$2$ mean $= 3$ median $–$ mode
$\text{Mean}=\frac{1}{2}(3 \text{ median - mode})$

Difference between the maximum and minimum value of the observation is called as range.
So, $99 - 47 = 52$

For $1st$ class mark
Class mark $=\frac{\text{(upper limit +lower limit)}}{2}$
$1st$ class mark $= 212$
Add the class size to get the class mark of the next class
$2nd$ class mark $= 212 + 25$
$2nd$ class mark $= 237$
Hence, the class mark of first two intervals are $212$ and $237.$

The difference between the maximum and minimum value is called range.
Thus, $30 - 15 = 15$

Data: $0, 2, 2, 2, -3, 5, -1, 5, 5, -3, 6, 6, 5, 6$
Rearranging data in increasing order, we have
$-3, -3, -1, 0, 2, 2, 5, 5, 5, 5, 6, 6, 6$
Number of observations $= n = 14$ (even)
Now,
$\text{Median}=\frac{\Big(\frac{\text{n}}{2}\Big)^{\text{th}}\text{observation}+\Big(\frac{\text{n}+1}{2}\Big)^{\text{th}}\text{observation}}{2}$
$=\frac{7^{\text{th}}\text{observation}+8^{\text{th}}\text{observation}}{2}$
$=\frac{2+5}{2}$
$\Rightarrow\text{Median}=3.5$

The mean is equal to the sum of all the values in the data set divided by the number of values in the data set.
The first $4$ prime numbers are $2, 3, 5, 7$
So, mean is $\frac{17}{4}=4.25$

A cumulative frequency distribution is the sum of the class and all classes below it in a frequency distribution.
Less than $30$ has the class interval 20-30. Frequency of this class interval will be corresponding to

In frequency polygon $x$-axis repesents data value whereas $y$-axis is used represent the frequencies of the data.
We include one class below lowest value and one class above highest value with zero frequencies.
The graph touches the $x$-axis at these points.
So, one sides of the frequency polygon is $x$-axis.

The classes are $10-15, 15-20, 20-25, 25-30, 30-35$ so that upper limit of the highest class is $35$.

Difference between the maximum and minimum value of the observations is called as range.
Let, minimum value be $'x'$
$75 - x = 20$
So, $x = 55$

Add the values corresponding to the height of the bar before $40$.
$6 + 3 + 4 + 2 = 15$.

Prime numbers between $30$ and $40$ are $31$ and $37$.
Mean $=\frac{31+37}{2}$
Mean $=34$

Arrange the given data in ascending order.
$7, 7, 7, 8, 9, 10, 13$
For odd number $(n)$ of observation median $=\Big[\frac{(\text{n+1)}}{2}\Big]\text{th}$ term,
Here $n = 7$ so median $=\Big[\frac{{(7+1)}}{2}\Big]\text{th}$ term $= 4th$ term that is $8$,
Hence median $= 8$

Since, Mean – Mode + Mean + $2$ Mode = $3$ Median
$\Rightarrow $ Mean – Mode = $3$ Median – Mean – $2$ Mode
$= 3$ Median – Mean – $2 (3$ Median – $2$ Mean$)$
$= 3$ Median – Mean – $6$ Median + $4$ Mean
$= 3$ Mean – $3$ Median
$= 3$ (Mean – Median)

The observations in ascending order can be written as:
$3, 4, 4, 5, 6, 7, 7, 7, 12$
Median $=\Big(\frac{9+1}{2}\Big)\text{th}$ term $= 5th$ term $= 6$

Range of observations = Highest observation - Lowest observation
$= 95 - 62 = 33$

We first arrange the given data in ascending order as follows $14, 14, 14, 14, 15, 15, 15, 15, 15, 16, 17, 18, 19, 19, 20$.
From above, we see that $15$ occurs most frequently i.e., $5$ times.
Hence, the mode of the given data is $15$.

Since $x^{-}$and $y^{-}$are two numbers, though being means, their arithmetic mean is given by:
$\text{z}=\frac{\text{x and y}}{2}$

$3$ Median $= 2 \times $ Mean + Mode
$\Rightarrow 3 \times 20 = 2 \times $ Mean + $18$
$\Rightarrow 2 \times$ Mean $= 60 - 18 = 42$
$\Rightarrow $ Mean $= 21$

A cumulative frequency distribution is the sum of the class and all classes below it in a frequency distribution.
Subtract cumulative frequency of class more than $70$ from the next cummulative Frequency of class more than $69$.
$30 − 18 = 12$

$1\ cm - 30\ km$
So for $75\ km$
$\frac{75}{30}=2.5\text{cm}$

First, we arrange the given numbers in ascending order is,
$3, 4, 4, 5, 6, 7, 7, 7$ and $12$
Here, $n = 9$
Since, $n$ is odd, so we use the formula for median,
Now, Median $=\Big(\frac{\text{n}+1}{2}\Big)^{\text{th}}\text{observation}$
$=\Big(\frac{9+1}{2}\Big)^{\text{th}}\text{observation}$ [Put $n = 9$]
$=\Big(\frac{10}{2}\Big)^{\text{th}}\text{observation}$
$=5^{\text{th}}\text{observation}=6$

Let the numbers be $x_1, x_2 \ldots ., x_n$
Now the new numbers after decrasing every number by $8:\left(x_1-8\right),\left(x_2-8\right) \ldots\left(x_n-8\right)$
New mean $=\frac{(\text{x}_1 - 8)+(\text{x}_2 - 8)+...+(\text{x}_\text{n} - 8)}{\text{n}}$
$=\frac{\text{x}_1+\text{x}_2+...+\text{x}_\text{n}-\text{8n}}{\text{n}}$
$=\frac{\text{x}_1+\text{x}_2+...+\text{x}_\text{n}}{\text{n}}-8$
New mean $=$ mean $-8$
Hence, mean is decreased by $8.$

Tallies are usually marked in a bunch of $4$.
Hence, the correct option is $(b)$.

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 $(m - l)$.
Therefore, we have two equations
$\frac{\text{l+m}}{2}=15$
$\Rightarrow l + m = 30,$
$m - l = 4$
Subtracting the second equation from the first equation, we have
$(l + m) - (m - l) = 30 - 4$
$\Rightarrow l + m - m + l = 26$
$\Rightarrow 2l = 26$
$\Rightarrow l = 13$
Hence, the lower limit of the class is $13$. Thus, the correct choice is $(c)$.

upper limit - lower limit = class width
upper limit - lower limit $= 10$
$\frac{\text{(upper limit + lower limit)}}{2}=\text{mid}-\text{value}$
upper limit + lower limit $= 2 \times 60.5$
upper limit + lower limit $= 121$
By solving the above two equations, we get
upper limit $= 65.5$
Lower limit $= 55.5$

We have:
Maximum value $= 32$
Minimum value $= 6$
We know:
Range = Maximum value - minimum value
$= 32 - 6$
$= 26$

Arranging the points in an ascending order,
We have:
$3, 4, 4, 5, 6, 7, 7, 7, 12$
Here, $n = 9$, Which is odd
$\therefore\ $median score = value of $\frac{1}{2}(9+1)^\text{th}$ terms
= value of $\Big(\frac{1}{2}\times10\Big)^\text{th}$ term
= value of 5$^{th}$ term
= $6$

Clearly, lower limit of the class corrseponding to class mark $20$
$=\frac{\text{Class mark of precending class + 20}}{2}$
$=\frac{15+20}{2}=17.5$
Uppar limit of the class corresponding to the class mark $20$
$=\frac{\text{20 + Class mark of succeending class}}{2}$
$=\frac{20+25}{2}=\frac{45}{2}=22.5$
Hence the required class is $17.5 - 22.5$.

This is the continuous from of frequency distribution.
Here, the upper limit of each class is excluded, while the lower limit is included.
So, the number $20$ is included in the class interval $20-30$.

The mode in a list of numbers refers to the integers that occurs most number of times.
In the given list both $8$ and $9$ occur two times.
So the value of will decide the mode
If $x = 8$, then mode will be $8$
If $x = 9$, then mode will be $9$
Hence, differnce between two modes is $1$.

The mode in a list of numbers refers to the integers that occur most number of times.
In the given list $5$ occur three times and $7$ occurs three times.
For $7$ to be mode of the list it should occur more number of times than any other number
So $x$ should be $7$.
Mode $= 7$

Mean of $5$ observations $= 9$
$\text{Mean}=\frac{\text{Sum of all observations}}{\text{Total number of observation}}$
$\Rightarrow9=\frac{\text{x}+\text{x}+3+\text{x}+5+\text{x}+7+\text{x}+10}{5}$
$\Rightarrow9=\frac{5\text{x}+18}{5}$
$\Rightarrow45=5\text{x}+18$
$\Rightarrow5\text{x}=20$
$\Rightarrow\text{x}=4$
So, the mean of the last three observations
$=\frac{\text{x}+5+\text{x}+7+\text{x}+10}{3}$
$=\frac{4+5+4+7+4+10}{3}$
$=\frac{34}{3}$
$=11\frac{1}{3}$

Number of classes $= 9$
Lower limit of the lowest class $= 10.6$
Width of each class $= 2.5$
So, Upper limit of the lowest class $= 10.6 + 2.5 = 13.1$
Now, Upper limit of the lowest class + Width of each class = Upper limit of the next class
Thus, we have
Upper limit of the lowest class + $8 \times $ width of each class = Upper limit of the highest ($9$$^{th}$) class
Upper limit of the highest (9$^{th}$) class $= 13.1 + 8 \times 2.5 = 33.1$

Since $\bar{z}$ is the mean of $x_1, x_2, \ldots, x_n, y_1, y_2, \ldots, y_n$
$\overline{ z }=\frac{\left( x _1+ x _2+\ldots+ x _{ n }\right)+\left( y _1+ y _2+\ldots y _{ n }\right)}{2 n }$
Since $\overline{ x }$ is the mean of $x _1, x _2, \ldots, x _{ n }$,
$\overline{ x }=\frac{ x _1+ x _2+\ldots+ x _{ n }}{ n }$
$\Rightarrow x _1+ x _2+\ldots+ x _{ n }= n \overline{ x }$
Since $\bar{y}$ is the mean of $y_1, y_2, \ldots, y_n$,
$\overline{ y }=\frac{ y _1+ y _2+\ldots+ y _{ n }}{ n }$
$\Rightarrow y _1+ y _2+\ldots+ y _{ n }= n \overline{ y }$
so,
$\overline{ z }=\frac{ n \overline{ x }}{2 n }+\frac{ n \overline{ y }}{2 n }=\frac{ n \overline{ x }+ n \overline{ y }}{2 n }$
$=\frac{\overline{ x }+\overline{ y }}{2}$

Let $x_1, x_2, \ldots, x_n$ be the $n$ observation,
Then, old mean $\bar{\text{x}}_{\text{old}}=\frac{\sum\limits_{\text{i}=1}^\text{n}\text{x}_\text{i}}{\text{n}}\ \dots(\text{i})$
Now, adding $5$ in each obsevation, the new mean becomes
$\bar{\text{x}}_{\text{new}}=\frac{(\text{x}_1+5)+(\text{x}_2+5)+\ ...\ +(\text{x}_\text{n}+5)}{\text{n}}$
$\Rightarrow\ \bar{\text{x}}_{\text{new}}=\frac{(\text{x}_1+\text{x}_2+\ ...\ +\text{x}_\text{n})+5\text{n}}{}$
$\Rightarrow\ \bar{\text{x}}_{\text{new}}=\frac{\sum\limits_{\text{i}=1}^\text{n}\text{x}_\text{i}}{\text{n}}+5=\bar{\text{x}}_{\text{old}}+5$ [from Eq. $(i)$]
$\Rightarrow\ \bar{\text{x}}_\text{new}=\bar{\text{x}}_{\text{old}}+5$
Hence, the new mean is increased by $5$.

Adjusted Frequency $=\Big(\frac{\text{Frequncey of the class }}{\text{widthof the class}}\Big)\times5$
Therefore adjusted frequencey of $25-45=\frac{8}{20}\times5=2$

Bar graph is a pictorial representation of data variables($x$-axis versus it value on $y$-axis) height of the bar.
Hence width of bar has no significance in bar graph.

Mean of n natural numbers is $\frac{(\text{n+1})}{2}$ So
Mean of $100$ numbers $=\frac{100+1}{2}$
Mean of $100$ numbers $=\frac{100}{2}=50.5$

The abscissa of the point of intersection of the less than type and of the more than type ogives of a grouped data gives its Median. Since the point of intersection of the more than type ogive and less than type ogive gives the median on the $x$-axis.

$\text{Mean}=\overline{\text{X}}=\frac{\text{Sum of all observations}}{\text{Total number of observations}}$
$=\frac{\text{Sum of all observations}}{\text{n}}$
Now if k is aded to each observation
$\text{New Mean},\overline{\text{X'}}=\frac{\text{Sum of all observations}+\text{nk}}{\text{n}}$
$=\frac{\text{Sum of all observations}}{\text{n}}+\text{k}$
$\Rightarrow\overline{\text{X}'}=\overline{\text{X}}+\text{k}$

Class mark $=\frac{\text{upper limit+lower limit}}{2}$
$2 \times 9.5$ = upper limit + lower limit
$19$ = upper limit + lower limit
Class size = upper limit - lower limit
$6$ = upper limit - lower limit
Solving above two equations we get,
$2 \times $ upper limit $= 25$
upper limit $= 12.5$
Hence, lower lower limit $= 12.5 - 6 = 6.5$
So, the class interval is$ 6.5 - 12.5$.

First five prime numbers are $2, 3, 5, 7, 11$
Mean is $\frac{2+3+5+7+11}{5}$
$\frac{28}{5}=5.6$
Median is $5$
So the difference between them is $5.6 - 5 = 0.6$

Let $a, b, c, d$ and $e$ are five numbers, then
Mean $=\frac{\text{(a + b + c + d + e)}}{5}=30$
$\Rightarrow (a + b + c + d + e) = 150 ..... (1)$
Now Let the number excluded be a
Then new mean $=\frac{\text{(b + c + d + e)}}{4}=28$
$\Rightarrow (b + c + d + e) = 112$
Putting this value in $(1)$
$\Rightarrow a + 112 = 150$
$\Rightarrow a = 150 - 112 = 38$
$\therefore$ Excluded number $= 38$

Let the mean of the initial sequence is $x$.
Given that, after subtracting $53$ from each number, the difference between the means is $3.5$.
So, $x - 53 = 3.5$
Mean of the number is $x = 53 + 3.5 = 56.5$

The observation which occurs maximum number of times is called the mode of the given data.
Above given data has a maximum number of $5$, so the mode is $5$.

Let the lower limit be $x$.
Class interval $= 4$
Upper limit $= x + 4$
Now, mid - value of a class $=\frac{\text{x}+4+\text{x}}{2}=\text{x}+2=15(\text{given)}$
$x = 13 =$ lower limit

The given class intervals are $10-20, 20-30$. In these class intervals the value $20$ is lies in the class interval $20-30$.Hence, the correct choice is $(b)$.

We first arrange the given data in ascending order as follows $14, 14, 14, 14, 15, 15, 15, 15, 15, 16, 17, 18, 19, 19, 20$ From above, we see that $15$ occurs most frequently i.e., $5$ times.

Mean of $5$ observations $= 11$
We know:
Mean $=\frac{\text{Sum of all observations}}{\text{Total number of observations}}$
$\Rightarrow 11=\frac{\text{x}+\text{x}+2+\text{x}+4+\text{x}+6+\text{x}+8}{5}$
$\Rightarrow11=\frac{\text{5x+20}}{5}$
$\Rightarrow 5x + 20 = 55$
$\Rightarrow 5x = 35$
$\Rightarrow x = 7$

Mean of $100$ observation $= 50$
$\therefore$ Total of $100$ observation $= 100 \times 50 = 5000$.
If one of the observation $50$ is replaced by $150$ than total of new observation is $= 5000 - 50 + 150 = 5100$
Hence the mean of new $100$ observation $=\frac{5100}{100}=51$

Mean of $a x_1, a x_2, \ldots . a x_n$, is $a x$
Mean of $\frac{\text{x}_1}{\text{a}},\frac{\text{x}_2}{\text{a}},...\frac{\text{x}_n}{\text{a}},$is $\frac{1}{\text{a}}\text{x}$
So the their mean is $\Big(\text{a}+\frac{1}{\text{a}}\Big)\frac{\text{x}}{2}$

Sum of the terms $=\text{n}_1\bar{\text{x}_1}+\text{n}_2\bar{\text{x}}_2+\ _{\dots},+\text{n}_\text{n}\bar{\text{x}}_\text{n}$
Number of terms $=\text{n}_1+\text{n}_2+\ _{\dots}+\text{n}_\text{n}$
Required mean $=\frac{\sum\limits_{\text{i}=1}^\text{n}\text{n}_\text{i}\bar{\text{x}}_\text{i}}{\sum\limits_{\text{i}=1}^\text{n}\text{n}_\text{i}}$

Arranging the points in an ascending order,
We have:
$22, 34, 39, 45, 54, 56, 68, 78, 84$
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 $6^{th}$ terms
$=\frac{1}{2}(54+54)$
$\frac{1}{2}\times108$
$=54$