MATHS M.C.Q

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
$=\frac{1}{2}(54+54)$
$\frac{1}{2}\times108$
$=54$

Given, mean of $x_1, x_2, \ldots . ., x_n$ is $\bar{\text{x}}$
$\therefore\ \sum\limits^\text{n}_{\text{i}=1}\text{x}_\text{i}=\text{n}\bar{\text{x}}$
Now, let the mean of $\Big(\text{ax}_1,\text{ax}_2,\ ....\text{ ax}_\text{n},\frac{\text{x}_1}{\text{a}},\frac{\text{x}_2}{\text{a}},\ ...., \ \frac{\text{x}_\text{n}}{\text{a}}\Big)$ is $\bar{\text{z}}$
Then, $\bar{\text{z}}=\frac{(\text{ax}_1,\text{ax}_2,\ ....\text{ ax}_\text{n})+\Big(\frac{\text{x}_1}{\text{a}},\frac{\text{x}_2}{\text{a}},\ ...., \ \frac{\text{x}_\text{n}}{\text{a}}\Big)}{\text{n}+\text{n}}$
$=\frac{\text{a}(\text{x}_1+\text{x}_2+\ ....\ +\text{x}_\text{n})+\frac{1}{\text{a}}(\text{x}_1+\text{x}_2+\ ....\ +\text{x}_\text{n})}{2\text{n}}$
$=\frac{\big(\text{a}+\frac{1}{\text{a}}\big)(\text{x}_1+\text{x}_2+\ ....\ +\text{x}_\text{n})}{2\text{n}}$
$=\frac{\big(\text{a}+\frac{1}{\text{a}}\big)\sum\limits^\text{n}_{\text{i}=1}\text{x}_\text{i}}{2\text{n}}$
$=\frac{\big(\text{a}+\frac{1}{\text{a}}\big)\cdot\text{n}\bar{\text{x}}}{2\text{n}}$
$=\frac{\big(\text{a}+\frac{1}{\text{a}}\big){\bar{\text{x}}}}{2}$

The number of classes is $9$ and the uniform class size is $2.5$.
The lower limit of the lower class (first class) is $10.6$.
Therefore, the upper limit of the last class is
$10.6 + (9 × 2.5)$
$= 10.6 + 22.5$
$= 33.1$
Hence, the correct choice is $(b)$.