MCQ 1511 Mark
Choose the correct answer from the given four options : For the following distribution :
| Marks |
Number of students |
| Below $10$ |
$3$ |
| Below $20$ |
$12$ |
| Below $30$ |
$27$ |
| Below $40$ |
$57$ |
| Below $50$ |
$75$ |
| Below $60$ |
$80$ |
the modal class is :
- A
$10-20$
- B
$20-30$
- ✓
$30-40$
- D
$50-60$
AnswerCorrect option: C. $30-40$
Here,
| Marks |
Number of students |
Cumulative frequency |
| Below $10$ |
$3 = 3$ |
$3$ |
| $10-20$ |
$(12 - 3) = 9$ |
$12$ |
| $20-30$ |
$(27 - 12) = 15$ |
$27$ |
| $30-40$ |
$(57 - 27) = 30$ |
$57$ |
| $40-50$ |
$(75 - 57) = 18$ |
$75$ |
| $50-60$ |
$(80 - 75) = 5$ |
$80$ |
Here, we see that the highest frequency is $30.$
which lies in the inteval $30-40.$ View full question & answer→MCQ 1521 Mark
If the mean of the following distribution is $2.6,$ then the value of $y$ is :
| Varible $(x)$ |
$1$ |
$2$ |
$3$ |
$4$ |
$5$ |
| Fraquency |
$4$ |
$5$ |
$y$ |
$1$ |
$2$ |
AnswerMean $= 2.6$
|
Variable $(x)$
|
Frequency $(f)$
|
$fx$
|
| $1$ |
$4$ |
$4$ |
| $2$ |
$5$ |
$10$ |
| $3$ |
$y$ |
$3y$ |
| $4$ |
$1$ |
$4$ |
| $5$ |
$2$ |
$10$ |
|
Total
|
$12 + y$ |
$28 + 3y$ |
$\therefore\text{Mean}=\frac{\sum\text{fx}}{\sum\text{f}}$
$\Rightarrow2.6=\frac{28+3\text{y}}{12+\text{y}}$
$\Rightarrow2.6(12+\text{y})=28+3\text{y}$
$\Rightarrow31.2+2.6\text{y}=28+3\text{y}$
$\Rightarrow3\text{y}-2.6\text{y}=31.2-28$
$\Rightarrow0.4\text{y}=3.2$
$\Rightarrow4\text{y}=32$
$\Rightarrow\text{y}=\frac{32}{4}=8$
$\text{y}=8$ View full question & answer→MCQ 1531 Mark
Which of the following cannot be determined graphically?
AnswerMean cannot be determined graphically.
View full question & answer→MCQ 1541 Mark
The mean of 25 observations is 36. If the mean of the first 13 observations is 32 and that of the last 13 observations is 39, then the 13th observation is:
Answerd. 23
Solution:
Let terms be $x_1, x_2, x_3 \ldots x_{25}$
According to the questions,
$\frac{\text{x}_1,\text{x}_2,\text{x}_3...\text{x}_{25}}{25}=36$
$\Rightarrow{\text{x}_1,\text{x}_2,\text{x}_3...\text{x}_{25}}=900...\text{i}$
$\text{And}=\frac{\text{x}_1,\text{x}_2,\text{x}_3...\text{x}_{13}}{13}=32$
$\Rightarrow{\text{x}_1,\text{x}_2,\text{x}_3...\text{x}_{25}}=416...\text{ii}$
$\text{Also}=\frac{\text{x}_{13}+\text{x}_{14}+\text{x}_{15}+...\text{x}_{25}}{13}=39$
${\text{x}_{13}+\text{x}_{14}+\text{x}_{15}+...\text{x}_{25}}=507...\text{(iii)}$
Adding eq,(ii) and (iii),we get ,
$\Rightarrow{\text{x}_{13}+\text{x}_{14}+\text{x}_{15}+...\text{x}_{25}}+\text{x}_{13}=923$
$\Rightarrow900+\text{x}_{13}=923$
$\Rightarrow \text{x}_{13}=23$
View full question & answer→MCQ 1551 Mark
If $x_i$ is changed to $x_i+a$, then is changed to:
- A
$\frac{\overline{\text{x}}}{\text{a}}$
- ✓
$\overline{\text{x}}-\text{a}$
- C
$\overline{\text{x}}+\text{a}$
- D
$\text{a}\overline{\text{x}}$
AnswerCorrect option: B. $\overline{\text{x}}-\text{a}$
b. $\bar{\text{x}}-\text{a}$
Solution:
Let terms be $\text{x}_1,\text{x}_2,\text{x}_3,...,\text{x}_\text{n}$
$\therefore$ Mean ($\overline{\text{x}}$) $=\frac{\text{x}_1+\text{x}_2+\text{x}_3+..........+\text{x}_\text{n}}{\text{n}}$
New observations are $\text{x}_1+\text{a},\text{x}_2+\text{a},\text{x}_3+\text{a},...,\text{x}_\text{n}+\text{a}$
$\therefore$ New Mean $=\frac{\text{x}_1+\text{a}+\text{x}_2+\text{a}+\text{x}_3+\text{a+.........+}\text{x}_\text{n}+\text{a}}{\text{n}}$
$\frac{\text{x}_1+\text{x}_2+\text{x}_3+.......+\text{x}_\text{n}+\text{na}}{\text{n}}$
$=\overline{\text{x}}+\text{a}$
View full question & answer→MCQ 1561 Mark
The wickets taken by a bowler in $10$ cricket matches are $\{2, 6, 4, 5, 0, 2, 1, 3, 2, 3\}$. The mode of the data is :
AnswerGiven : $\{2, 6, 4, 5, 0, 2, 1, 3, 2, 3\}$
Here frequency of $2$ is more.
Therefore, the mode of given data is $2.$
View full question & answer→MCQ 1571 Mark
State true or false : The mode is the most frequently occurring observation :
AnswerThe observation occurring the most number of times or which has highest frequency is called the mode.
Thus, the given statement is true.
View full question & answer→MCQ 1581 Mark
Consider the following table :
|
Class interval
|
$10-14$ |
$14-18$ |
$18-22$ |
$22-26$ |
$26-30$ |
|
Frequency
|
$5$ |
$11$ |
$16$ |
$25$ |
$19$ |
The mode of the above data is : AnswerCorrect option: C. $24.4$
Maximum frequency $= 25$
Hence, modal class is $22-26$
Now, mode $=\text{x}_\text{k}+\text{h}\Big\{\frac{(\text{f}_\text{k}-\text{f}_{\text{k}-1})}{(2\text{f}_\text{k}-\text{f}_{\text{k}-1}-\text{f}_{\text{k}+1})}\Big\}$
$22+4\Big\{\frac{(25-16)}{(2(25)-16-19)}\Big\}$
$=22+4\times\frac{9}{15}$
$=22+2.4$
$=24.4$
View full question & answer→MCQ 1591 Mark
If $\sum\text{f}_\text{i}\text{u}_\text{i}=-7,\sum\text{f}_\text{i}=225$ and $\text{h} =50$ then the value of $\bar{\text{x}}$ is :
Answer$\bar{\text{x}}=\text{a}+\frac{\sum\text{f}_\text{i}\text{u}_\text{i}}{\sum\text{f}_\text{i}}$
$=225+\frac{-7}{25}\times50$
$=225-14$
$=211$
View full question & answer→MCQ 1601 Mark
The marks obtained by $9$ students in Mathematics are $\{59, 46, 30, 23, 27, 44, 52, 40\}$ and $29$. The median of the data is :
AnswerArranging the given data in ascending order : $\{23, 27, 29, 30, 40, 44, 46, 52, 59.\}$
Here $n = 9,$ which is odd.
$\therefore \text{Median}=\Big(\frac{\text{n}+1}{2}\Big)^\text{th}$
$=\Big(\frac{9+1}{2}\Big)^\text{th}\text{term}$
$=5^{th} \text{ term }$
$=40$
View full question & answer→MCQ 1611 Mark
If the median of the data $4, 7, x - 1, x - 3, 16, 25,$ written in ascending order, is $13$ then $x $ is equal to
AnswerData in ascending odhar :
$4, 7, x - 1, x - 3, 16, 25$
$N = 6\ ($even$)$
$\therefore$ median $=\frac{\Big(\frac{6}{2}\Big)^{\text{th}}\text{value}+\Big(\frac{6}{2}+1\Big)^{\text{th}}\text{value}}{2}$
$\therefore13=\frac{3^{\text{rd}}\text{value}+4^{\text{th}}\text{value}}{2}$
$\therefore26-2\text{x}-4$
$\therefore2\text{x}=30$
$\therefore\text{x}=15$
View full question & answer→MCQ 1621 Mark
In the given data if $n = 230, l = 40, cf = 76, h = 10, f = 65,$ then its median is :
Answer$\text{Median}=\text{l}+\frac{\frac{\text{n}}{2}-\text{cf}}{\text{f}}\times\text{h}$
$=40+\frac{\frac{230-76}{2}}{65}\times10$
$=40+\frac{115-76}{65}\times10$
$=40+\frac{39}{65}\times10$
$=40+\frac{390}{65}$
$=40+6$
$=46$
View full question & answer→MCQ 1631 Mark
The mean of the data when $\sum\text{f}_\text{i}\text{d}_\text{i}=435,\sum\text{f}_\text{i}=30$ and $\text{a}=47.5$ is :
AnswerMean$=\bar{\text{x}}=\text{a}+\frac{\sum\text{f}_\text{i}\text{d}_\text{i}}{\sum\text{f}_\text{i}}$
$=47.5+\frac{435}{30}$
$= 47.5+14.5=62$
View full question & answer→MCQ 1641 Mark
If the mean of observations $x_1, x_2, \ldots, x_n$ is $\bar{x},$ then the mean of $x_1+a, x_2+a, \ldots, x_n+a$ is
- A
$\text{a}\bar{\text{x}}$
- B
$\bar{\text{x}}-\text{a}$
- ✓
$\bar{\text{x}}+\text{a}$
- D
$\frac{\bar{\text{x}}}{\text{a}}$
AnswerCorrect option: C. $\bar{\text{x}}+\text{a}$
Meam of observations $x_1, x_2, \ldots . ., x_n$ is $\bar{x}$
$\frac{x_1+x_2+x_3 \ldots \ldots+x_n}{n}=\bar{x}$
$x_1+a+x_2+a+x_3+a+\ldots . x_n+a$
$=x_1+x_2+x_3+\ldots x_n+n a$
$\therefore$ Mean of $\left(x_1+x_2+x_3 \ldots . .+x_n\right)+n a$
$=\bar{x}+\frac{n a}{n}=\bar{x}+a$
View full question & answer→MCQ 1651 Mark
If $35$ is removed from the data : $\{30, 34, 35, 36, 37, 38, 39, 40\},$ then the median increases by :
AnswerGiven data $= \{30, 34, 35, 36, 37, 38, 39, 40\}$
Here $n = 8$ which is even
$\therefore\text{Median}=\frac{1}{2}\Big[\frac{\text{n}}{2}\text{th}+\Big(\frac{\text{n}}{2}+1\Big)\text{th}\Big]\text{term}$
$=\frac{1}{2}(4\text{th}+5\text{th term})$
$=\frac{1}{2}(36+37)=\frac{73}{2}=36.5$
After removing $35,$ then $n = 7$
$\therefore$ New $\text{median}=\frac{7+1}{2}\text{th term}=4\text{th term}=37$
$\therefore$ Increase in $\text{median}=37-36.5=0.5$
View full question & answer→MCQ 1661 Mark
Look at the frequency distribution table given below:
|
Class interval
|
35-45
|
45-55
|
55-65
|
65-75
|
|
Frequency
|
8
|
12
|
20
|
10
|
The median of the above distribution is:
Answerb. 57.5
Solution:
|
Class intervalClass interval
|
FrequencyFrequency
|
Cumulative frequencyCumulative frequency
|
|
35-45
|
8
|
8
|
|
45-55
|
12
|
20
|
|
55-65
|
20
|
40
|
|
65-75
|
10
|
50
|
Here, $\text{N}=50\Rightarrow\frac{\text{N}}{2}=25$
The cumulaove frequency just greater than 25 is 40.
Hence, median class is 55-65.
Now, median $=\text{l}+\begin{Bmatrix}\text{h}\times\frac{\Big(\frac{\text{N}}{2}-\text{cf}\Big)}{\text{f}}\end{Bmatrix}$
$=55+\Big\{10\times\frac{(25-20)}{20}\Big\}$
$=55+2.5$
$=57.5$ View full question & answer→MCQ 1671 Mark
$(x+2), x$ and $(x-1)$ are the frequencies of the numbers 12,15 and 20 respectively. If the mean of the distribution is 14.5 , the value of $x$ is
Answer(b) : We have, mean $=\frac{\sum f_i x_i}{\sum f_i}$
$ \begin{aligned} & \Rightarrow \frac{12(x+2)+15(x)+20(x-1)}{(x+2)+(x)+(x-1)}=14.5 \\ & \Rightarrow 2(47 x+4)=29(3 x+1) \\ & \Rightarrow 94 x+8=87 x+29 \Rightarrow 7 x=21 \Rightarrow x=3 \end{aligned} $
(43-47) : Let us consider the following table :| Class | Class marks $\left(x_i\right)$ | $d_i=x_i-A$ | Frequency $\left(f_i\right)$ | $f_i d_i$ |
| 30-40 | 35 | -20 | 80 | -1600 |
| 40-50 | 45 | -10 | 110 | -1100 |
| 50-60 | 55 = A | 0 | 120 | 0 |
| 60-70 | 65 | 10 | 70 | 700 |
| 70-80 | 75 | 20 | 40 | 800 |
| | | | $\Sigma f_i=420$ | $\sum f_i d_i=1200$ |
View full question & answer→MCQ 1681 Mark
A sequence $a, a+d, a+2 d, \ldots, a+n d$, has odd number of terms. Find its median.
- A
$a+\left(\frac{n-1}{2}\right) d$
- B
$a+\left(\frac{n}{2}-1\right) d$
- ✓
$a+\left(\frac{n+2}{2}\right) d$
- D
$a+\left(\frac{n}{2}+2\right) d$
AnswerCorrect option: C. $a+\left(\frac{n+2}{2}\right) d$
(c) : Clearly, the given sequence is an A.P. with first term $a$ and common difference is $d$. Given, sequence is $a, a+d, a+2 d, a+3 d, \ldots, a+n d$ As, there are odd number of terms, so the median is
$
\left(\frac{n+1+1}{2}\right)^{\text {th }} \text { term }=\left(\frac{n+2}{2}\right)^{\text {th }} \text { term }=a+\left(\frac{n+2}{2}\right) d
$
View full question & answer→MCQ 1691 Mark
For a certain frequency distribution, if $\Sigma f_i=50$ and $\Sigma f_i x_i=2550$, then what is the mean of the distribution?
Answer(b) : Mean of the distribution $=\frac{\sum f_i x_i}{\sum f_i}=\frac{2550}{50}=51$.
View full question & answer→MCQ 1701 Mark
If the mean and mode of a frequency distribution be $53.4$ and $55.2$ respectively, then find the median.
AnswerWe have$, 3$ Median $=$ Mode $+2$ Mean
$=55.2+2(53.4)=55.2+106.8=162$
$\Rightarrow$ Median $=162 / 3=54$
View full question & answer→MCQ 1711 Mark
The numbers $3,5,7$ and 9 have their respectively frequencies $x-2, x+2, x-3$ and $x+3$. If the mean is 6.5 , then the value of $x$ is
Answer$\begin{aligned} & \text {(c) : } \frac{3(x-2)+5(x+2)+7(x-3)+9(x+3)}{(x-2)+(x+2)+(x-3)+(x+3)}=6.5 \\ & \Rightarrow \quad \frac{24 x+10}{4 x}=6.5 \Rightarrow x=5\end{aligned}$
View full question & answer→MCQ 1721 Mark
If mean of ten consecutive odd numbers is $120$ , then the mean of first five odd numbers among them is
AnswerLet ten consecutive odd numbers be $2 x+1$, $2 x+3, \ldots, 2 x+19$
Hence, $(2 x+1)+(2 x+3)+\ldots+(2 x+19)=10 \times 120$
$\Rightarrow 20 x+100=1200$
$\Rightarrow x=\frac{1100}{20}=55$
$\therefore$ Mean of first five odd numbers
$=\frac{(2 x+1)+(2 x+3)+\ldots .+(2 x+9)}{5}$
$=\frac{10 \times 55+25}{5}=\frac{575}{5}=115$
$\text { 38. (c) }: \frac{3(x-2)+5(x+2)+7(x-3)+9(x+3)}{(x-2)+(x+2)+(x-3)+(x+3)}=6.5$
$\Rightarrow \frac{24 x+10}{4 x}=6.5$
$\Rightarrow x=5$
View full question & answer→MCQ 1731 Mark
The mean of $x_1, x_2, \ldots \ldots, x_n$ is $M$. If $x_i$, $i=1,2, \ldots \ldots, n$ is replaced by $5 x_i$, the mean becomes $M_1$, then $M_1$ is equal to
- ✓
$5 M$
- B
$M+5$
- C
$M+100$
- D
$10 M$
Answer(a) : Given that
$
\begin{array}{c}
\frac{\sum_{i=1}^n x_i}{n}=M \text { and } \frac{\sum_{i=1}^n 5 x_i}{n}=M_1 \\
\Rightarrow 5 \cdot \frac{\sum_{i=1}^n x_i}{n}=M_1 \Rightarrow 5 M=M_1
\end{array}
$
View full question & answer→MCQ 1741 Mark
If the mean of the following distribution is $2.6$, then the value of $y$ is
| Class interval $(x_i)$ |
$1$ |
$2$ |
$3$ |
$4$ |
$5$ |
| Frequency |
$4$ |
$5$ |
$y$ |
$1$ |
$2$ |
Answer$n=\Sigma f_i=4+5+y+1+2=12+y$.
$\Sigma f_i x_i=4+10+3 y+4+10=28+3 y.$
Mean $=\frac{\Sigma f_i x_i}{\Sigma f_i}$
$\Rightarrow 2.6=\frac{28+3 y}{12+y}$
$\Rightarrow 26(12+y)=10(28+3 y)$
$\Rightarrow y=8$
View full question & answer→MCQ 1751 Mark
The numbers are arranged in the descending order : $108,94,88,82, x+7, x-7,60,58,42, 39.$ If the median is $73 ,$ the value of $x$ is
AnswerNumber of terms $=10$
$\text { Median }=\frac{5^{\text {th }} \text { term }+6^{\text {th }} \text { term }}{2}$
$=\frac{x+7+x-7}{2} $
$\Rightarrow 73=\frac{2 x}{2}$
$\Rightarrow x=73$
View full question & answer→MCQ 1761 Mark
The mean of six numbers: $x-5, x-1, x, x+2, x+4$ and $x+12$ is $15 .$ Find the mean of first four numbers.
AnswerGiven,
$\frac{(x-5)+(x-1)+x+(x+2)+(x+4)+(x+12)}{6}=15$
$\Rightarrow 6 x+12=90 $
$\Rightarrow 6 x=78 $
$\Rightarrow x=13$
Now, mean of first four numbers
$=\frac{(x-5)+(x-1)+x+(x+2)}{4}$
$=\frac{4 x-4}{4}$
$=x-1$
$=13-1$
$=12$
Hence, the required mean is $12 .$
View full question & answer→MCQ 1771 Mark
Mean of $20$ observations is $15$ . If each observation is multiplied by $\frac{2}{3}$, then the mean of new observations is
AnswerLet $x_1, x_2, x_3, \ldots \ldots, x_{20}$ be the $20$ observations such that,
$\frac{x_1+x_2+x_3+\ldots \ldots \ldots+x_{20}}{20}=15$
$\Rightarrow x_1+x_2+x_3+\ldots . .+x_{20}=300$
$\Rightarrow \frac{2}{3}\left(x_1+x_2+\ldots \ldots . .+x_{20}\right)=\frac{2}{3} \times 300$
$\Rightarrow \frac{2}{3} x_1+\frac{2}{3} x_2+\ldots \ldots . .+\frac{2}{3} x_{20}=200$
$\Rightarrow \frac{\frac{2}{3} x_1+\frac{2}{3} x_2+\ldots \ldots .+\frac{2}{3} x_{20}}{20}=\frac{200}{20}=10$
View full question & answer→MCQ 1781 Mark
The mean, mode and median of the observations, 7, 7, 5, 7 and $x$ are the same. Then the observation $x$ is
Answer(b) : Mean $=\frac{7+7+5+7+x}{5}=\frac{26+x}{5}$
Here, Mode $=7$
It is given that, Mean $=$ Mode $=$ Median
$\Rightarrow \quad \frac{26+x}{5}=7 \Rightarrow 26+x=35 \Rightarrow x=9$
View full question & answer→MCQ 1791 Mark
The frequency distribution table given below:
Ankit and Ankita observed and said the following:
Ankit : Median $=58.5$
Ankita : Median $=57.5$
Which of them is/are correct?| Class interval | 35-45 | 45-55 | 55-65 | 65-75 |
| Frequency | 8 | 12 | 20 | 10 |
Answer(b) : The cumulative frequency distribution table from the given data can be drawn as :| Class interval | Frequency | Cumulative frequancy |
| 35-45 | 8 | 8 |
| 45-55 | 12 | 20 |
| 55-65 | 20 | 40 |
| 65-75 | 10 | 50 |
Here, $n=50 \Rightarrow \frac{n}{2}=25$, which lies in class interval 55-65. $\therefore$ Median $=l+\left(\frac{\frac{n}{2}-c . f .}{f}\right) \times h=55+\left(\frac{25-20}{20}\right) \times 10=57.5$ View full question & answer→MCQ 1801 Mark
The mean of $n$ observations is $\bar{X}$. If the first item is increased by $1 ,$ second by $2$ and so on, then the new mean is
- A
$\bar{X}+n$
- B
$\bar{X}+\frac{n}{2}$
- ✓
$\bar{X}+\frac{n+1}{2}$
- D
AnswerCorrect option: C. $\bar{X}+\frac{n+1}{2}$
Mean of $n$ observations $=\bar{X}$
Sum of the observations $=n \bar{X}$
The first term is increased by $1 ,$ second by $2$ and so on.
The new sum
$=n \bar{X}+(1+2+\ldots \ldots+n)=n \bar{X}+\frac{n(n+1)}{2}$
$\therefore$ New mean $=\left[n \bar{X}+\frac{n(n+1)}{2}\right] \div n=\bar{X}+\frac{(n+1)}{2}$
View full question & answer→MCQ 1811 Mark
One of the properties of mode is
- A
- ✓
It is not affected by greatest and least values
- C
- D
Difference of greatest and least values
AnswerCorrect option: B. It is not affected by greatest and least values
(b) : Since mode is that value among observations which occurs most often. It is not affected by greatest and least values.
View full question & answer→MCQ 1821 Mark
Extreme value of a given data
Answer(b) : Since median is the value of the middle most item, extreme values do not affect the median.
View full question & answer→MCQ 1831 Mark
Consider the following frequency distribution:
The upper limit of the median class is
| Class interval | 1-7 | 8-14 | 15-21 | 22-28 | 29-35 |
| Frequency | 3 | 10 | 5 | 8 | 12 |
Answer(d): The cumulative frequency distribution table from the given data can be drawn as :| Class interval | Frequency | Cumulative frequancy |
| 0.5-7.5 | 3 | 3 |
| 7.5-14.5 | 10 | 13 |
| 14.5-21.5 | 5 | 18 |
| 21.5-28.5 | 8 | 26 |
| 28.5-35.5 | 12 | 38 |
$
\therefore n=38 \Rightarrow \frac{n}{2}=\frac{38}{2}=19
$
The cumulative frequency just greater than 19 is 26 . So, the median class is $21.5-28.5$ and its upper limit is 28.5 . View full question & answer→MCQ 1841 Mark
The correct formula for finding the mode of a grouped frequency distribution is
- A
Mode $=h+\left(\frac{f_1-f_0}{2 f_1-f_0-f_2}\right) \times l$
- B
Mode $=f_1+\left(\frac{f_1-f_0}{2 h-f_1-f_2}\right) \times l$
- C
$\quad$ Mode $=l-\left(\frac{f_1-f_0}{2 f_1-f_0-f_2}\right) \times h$
- ✓
Mode $=l+\left(\frac{f_1-f_0}{2 f_1-f_0-f_2}\right) \times h$
AnswerCorrect option: D. Mode $=l+\left(\frac{f_1-f_0}{2 f_1-f_0-f_2}\right) \times h$
(d) : The correct formula for finding the mode of a grouped frequency distribution is
$
l+\left(\frac{f_1-f_0}{2 f_1-f_0-f_2}\right) \times h
$
View full question & answer→MCQ 1851 Mark
The mean of 15 numbers is 25 . If each number is multiplied by 4 , mean of the new numbers is
Answer(b) : Mean of 15 numbers $=25$
Sum of the numbers $=25 \times 15=375$
Since each number is multiplied by 4 the new sum will be $4 \times 375=1500$
$\therefore \quad$ Mean of new numbers $=\frac{1500}{15}=100$
View full question & answer→MCQ 1861 Mark
The mean of $1,2,3,4$ .............., $n$ is given by
- A
$\frac{n(n+1)}{2}$
- B
$\frac{(n+1)}{4}$
- C
$\frac{n}{2}$
- ✓
$\frac{(n+1)}{2}$
AnswerCorrect option: D. $\frac{(n+1)}{2}$
(d): Sum of the numbers $1,2,3, \ldots, n=n \frac{(n+1)}{2}$
$\therefore \quad$ Mean $=\frac{n(n+1)}{2} \div n=\frac{n+1}{2}$
View full question & answer→MCQ 1871 Mark
If mode of a series exceeds its mean by 12 , then mode exceeds the median by
Answer(b) : Let the mean of the series be $x$.
$\therefore \quad$ Mode $=x+12$
Now, 3 Median $=$ Mode +2 Mean $=x+12+2 x=3 x+12$
$\Rightarrow$ Median $=\frac{1}{3}(3 x+12)=x+4$
Hence, the mode exceeds the median by $x+12-(x+4)$ i.e., 8 .
View full question & answer→MCQ 1881 Mark
The mean and mode of a frequency distribution are 28 and 16 respectively. The median is
Answer(c) : We know that, Mode $=3$ Median -2 Mean
$\Rightarrow 3$ Median $=$ Mode +2 Mean
$\Rightarrow 3$ Median $=16+2 \times 28 \Rightarrow$ Median $=72 / 3=24$
View full question & answer→MCQ 1891 Mark
For a frequency distribution, mean, median and mode are connected by the relation
- A
Mode $=3$ Mean -2 Median
- B
Mode $=2$ Median -3 Mean
- ✓
Mode $=3$ Median -2 Mean
- D
Mode $=3$ Median +2 Mean
AnswerCorrect option: C. Mode $=3$ Median -2 Mean
(c) : The relation between three measures of central tendency is given as Mode $=3$ Median -2 Mean
View full question & answer→MCQ 1901 Mark
The median class for the following data is
| Class interval | 20-40 | 40-60 | 60-80 | 80-100 |
| Frequency | 10 | 12 | 20 | 22 |
- A
$20-40$
- B
$40-60$
- ✓
$60-80$
- D
$80-100$
AnswerCorrect option: C. $60-80$
(c) : The cumulative frequency distribution table from the given data can be drawn as :| Class interval | Frequency | Cumulative frequancy |
| 20-40 | 10 | 10 |
| 40-60 | 12 | 22 |
| 60-80 | 20 | 42 |
| 80-100 | 22 | 64 |
Here, $n=64 \Rightarrow \frac{n}{2}=32$
So, the median class is $60-80$. View full question & answer→MCQ 1911 Mark
Find the class mark of the modal class in the following distribution.
| Class interval | Frequency |
| 40-50 | 10 |
| 50-60 | 25 |
| 60-70 | 28 |
| 70-80 | 12 |
| 80-90 | 10 |
| 90-100 | 15 |
Answer(c) : The class $60-70$ is a modal class as it has highest frequency.
$
\therefore \text { Class mark }=\frac{60+70}{2}=65
$
View full question & answer→MCQ 1921 Mark
If the median of the data: $6,7, x-2, x, 17, 20$ written in ascending order, is $16$ . Then $x$ is equal to
AnswerHere, $n=6$, which is even.
$\therefore$ Median $=\frac{1}{2}((\frac{6}{2})^{\text {th }}$ term $+(\frac{6}{2}+1)^{\text {th }}$ term$)$
$\Rightarrow 16=\frac{1}{2}(3^{\text {rd }}$ term $+4^{\text {th }}$ term$)$
$\Rightarrow 32=x-2+x$
$\Rightarrow 2 x=34 $
$\Rightarrow x=17$
View full question & answer→MCQ 1931 Mark
Consider the following table : The mode of the above data is
| Class interval |
$10-14$ |
$14-18$ |
$18-22$ |
$22-26$ |
$26-30$ |
| Frequency |
$5$ |
$11$ |
$16$ |
$25$ |
$19$ |
AnswerCorrect option: C. $24.4$
Here, the maximum frequency is $25$ and the corresponding modal class is $22-26.$
$\therefore l=22, h=4, f_1=25, f_0=16$ and $f_2=19$
$\therefore$ Mode $=l+\left(\frac{f_1-f_0}{2 f_1-f_0-f_2}\right) \times h$
$=22+\left(\frac{25-16}{50-16-19}\right) \times 4$
$=22+\frac{9}{15} \times 4$
$=22+\frac{12}{5}$
$=24.4$
View full question & answer→MCQ 1941 Mark
The calculation for mode for the following distribution is given below:
Here, the maximum frequency is 32 and the corresponding modal class is $30-40$.
$\therefore \quad l=30, h=10, f_0=12, f_1=32, f_2=20($ Step 1$)$
$\therefore \quad$ Mode $=l+\left(\frac{f_1-f_0}{2 f_1-f_0-f_2}\right) \times h \quad$ (Step 2)
$=30+\left(\frac{32-12}{64-12-20}\right) \times 10=30+\frac{25}{4}=36.25$ (Step 3$)$
In which step is there an error in solving?| Marks | Number of students |
| 0-10 | 6 |
| 10-20 | 10 |
| 20-30 | 12 |
| 30-40 | 32 |
| 40-50 | 20 |
View full question & answer→MCQ 1951 Mark
The modal class of data given below is $10-15$, then| Class interval | 0-5 | 5-10 | 10-15 | 15-20 | 20-25 |
| Frequency | 8 | 6 | $f$ | 4 | 3 |
- A
$f<9$
- ✓
$f \geq 9$
- C
$f>9$
- D
$f<3$
AnswerCorrect option: B. $f \geq 9$
(b) : The class having maximum frequency is called modal class. Here, the modal class is $10-15$. Therefore, its frequency must be greater than all other frequencies.
$
\therefore f \geq 9
$
View full question & answer→MCQ 1961 Mark
The frequency of the class succeeding the modal class in the following frequency distribution is
| Class interval | Frequency |
| 10-15 | 3 |
| 15-20 | 7 |
| 20-25 | 16 |
| 25-30 | 12 |
| 30-35 | 9 |
| 35-40 | 5 |
| 40-45 | 3 |
Answer(d) : The modal class is $20-25$ as it has maximum frequency. So, the class succeeding the modal class is 25-30 with frequency 12.
View full question & answer→MCQ 1971 Mark
Life time of electric bulbs are given in the following frequency distribution.
Find the class mark of the modal class.
| Life time (in hours) | 250-300 | 300-350 | 350-400 | 400-450 | 450-500 |
| Number of bulbs | 5 | 14 | 21 | 12 | 10 |
Answer(b) : The maximum frequency is 21 and its corresponding class is $350-400$. So, the modal class is 350-400.
Its class mark is $\frac{350+400}{2}$ i.e., 375 .
View full question & answer→MCQ 1981 Mark
A frequency distribution is given below :
Rahul and Suresh observed the table and said the following :
Rahul : Modal Class is 30-40.
Suresh : Modal Class is 50-60.
Which of them is/are correct?
| class interval | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
| Frequemcy | 3 | 9 | 15 | 30 | 18 | 5 |
Answer(a): The class $30-40$ has maximum frequency. So, the modal class is $30-40$.
View full question & answer→MCQ 1991 Mark
The algebraic sum of all the deviations of all the observations from their mean is always
- ✓
$0$
- B
$+ve$
- C
$-ve$
- D
equal to the number of observations.
AnswerThe algebraic sum of all the deviations
i.e., differences of all the observations, say $x_1, x_2, \ldots, x_n$ from their mean, $\bar{x}$
$=\left(x_1-\bar{x}\right)+\left(x_2-\bar{x}\right)+\left(x_3-\bar{x}\right)+\ldots+\left(x_n-\bar{x}\right)$
$=\left(x_1+x_2+x_3+\ldots+x_n\right)-(\bar{x}+\bar{x}+\bar{x} \ldots n$ times $)$
$=\frac{n\left(x_1+x_2+x_3+\ldots+x_n\right)}{n}-n \bar{x}$
$=n \bar{x}-n \bar{x}$
$=0$
View full question & answer→MCQ 2001 Mark
The mean of $n$ observations $x 1, x 2, x 3, \ldots, x n$ is $\bar{x}$. If each observation is multiplied by $p$, then the mean of the new observations is
- A
$\frac{\bar{x}}{p}$
- ✓
$p \bar{x}$
- C
$\bar{x}$
- D
$p+\bar{x}$
AnswerCorrect option: B. $p \bar{x}$
(b): $\because \bar{x}$ is the mean of given observations.
$
\therefore \quad \bar{x}=\frac{x_1+x_2+\ldots+x_n}{n}
$Multiplying each observation by $p$, we get new observations as: $p x_1, p x_2, p x_3, \ldots, p x_n$
$\therefore \quad$ Mean of new observations
$
=\frac{p x_1+p x_2+p x_3+\ldots+p x_n}{n}=p\left(\frac{x_1+x_2+\ldots+x_n}{n}\right)
$
$
=p \bar{x}
$
[Using (i)]
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