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 $x = 53 + 3.5 = 56.5$