MCQ 11 Mark
For any data, $Z+\bar{x}=71$ and $Z -\bar{x}=3$ then using inter-relation between mean, mode and median value of $M =$
View full question & answer→MCQ 21 Mark
For any frequency distribution, mode is $12$ more than the mean then mode is _________ more than the median.
View full question & answer→MCQ 31 Mark
For any frequency distribution $Z - M $=___________ . $X$ $( M -\bar{x})$
View full question & answer→MCQ 41 Mark
______ is the cumulative frequency of the class $40-50$ in following frequency distribution.
| Class |
$20-30$ |
$30-40$ |
$40-50$ |
$50-60$ |
$60-70$ |
| Frequency |
$10$ |
$15$ |
$15$ |
$20$ |
$10$ |
View full question & answer→MCQ 51 Mark
Out of the following which measure depends on the value of all the observations.
MCQ 61 Mark
Out of the following, which is not the measure of central tendency ?
MCQ 71 Mark
Mean $=25$ and mode $=25$ then median = ______________.
View full question & answer→MCQ 81 Mark
$6,7, x-2, x, 17$ and $20$ are six observations in an increasing order with median $16$ then $x=\ldots \ldots \ldots$
View full question & answer→MCQ 91 Mark
For a frequency distribution median is $13.2$ and mean is $15.3$ . Then using inter-relation between mean, mode, median we get mode
View full question & answer→MCQ 101 Mark
Modal class of the following frequency distribution is _________
| Class |
$0-20$ |
$20-40$ |
$40-60$ |
$60-80$ |
$80-100$ |
| Frequency |
$8$ |
$15$ |
$22$ |
$25$ |
$10$ |
- A
$0-20$
- B
$20-40$
- C
$40-60$
- ✓
$60-80$
AnswerCorrect option: D. $60-80$
$60-80$
View full question & answer→MCQ 111 Mark
For any data $\bar{x}=20, M=25$ than $Z=$ _______
View full question & answer→MCQ 121 Mark
Equation of mean $X =a+\frac{\Sigma f_i u_i}{\Sigma f_i}$ where $u_i=$ _______
- ✓
$\frac{x_i-a}{h}$
- B
$x_i-a$
- C
$x_i-h$
- D
$\frac{x_i-h}{a}$
AnswerCorrect option: A. $\frac{x_i-a}{h}$
$\frac{x_i-a}{h}$
View full question & answer→MCQ 131 Mark
For a data $\bar{X}=25$ and $M=20$ then $Z=$ _______
View full question & answer→MCQ 141 Mark
For a data, $\bar{X}=35$ and $M=35$ then $Z=$ _______ .
View full question & answer→MCQ 151 Mark
In mean equation $\overline{ X }=a+\frac{\Sigma f i d i}{\Sigma f_i} d i$ _______
- ✓
$x_i-a$
- B
$a-x_i$
- C
$f_i-a$
- D
$a-f_i$
AnswerCorrect option: A. $x_i-a$
$x_i-a$
View full question & answer→MCQ 161 Mark
For a given frequency distribution. if $\overline{x} = 15$ and z=15 then M = ________
View full question & answer→MCQ 171 Mark
Find the class marks of classes $10-20$ and $35-55$.
Answer(c) : Class mark of class $10-20=\frac{10+20}{2}=15$ Class mark of class 35-55 $=\frac{35+55}{2}=45$
View full question & answer→MCQ 181 Mark
If $d_i=x_i-13, \Sigma f_i d_i=30$ and $\Sigma f_i=120$, then mean $(\bar{x})$ is equal to
Answer(c) : Here, $d_i=x_i-13 \Rightarrow a=13 \quad\left[\because d_i=x_i-a\right]$ Also, $\sum f_i d_i=30$ and $\sum f_i=120$ Now, mean, $\bar{x}=a+\frac{\Sigma f_i d_i}{\Sigma f_i}=13+\frac{30}{120}=13.25$
View full question & answer→MCQ 191 Mark
The mean of first ten odd natural numbers is
Answer(b) : First ten odd natural numbers are 1, 3, 5, 7, 9, $11,13,15,17$ and 19 .
$\therefore \quad$ Required mean
$
=\frac{1+3+5+7+9+11+13+15+17+19}{10}=\frac{100}{10}=10
$
View full question & answer→MCQ 201 Mark
If the mean of first $n$ natural numbers is $\frac{5 n}{9}$, then $n$ is equal to
AnswerMean of first $n$ natural numbers
$=\frac{1+2+3+\ldots+n}{n}$
$\Rightarrow \frac{5 n}{9}=\frac{n(n+1)}{2 n}$
$[\because$ Sum of first $n$ natural numbers $=\frac{n(n+1)}{2}]$
$\Rightarrow 10 n^2=9 n^2+9 n $
$\Rightarrow n^2-9 n=0$
$\Rightarrow n(n-9)=0 $
$\Rightarrow n=9 \ (\because n \neq 0)$
View full question & answer→MCQ 211 Mark
If the mean of $x, x+3, x+6, x+9$ and $x+12$ is 10 , then $x$ equals
Answer(c) : Mean, $\bar{x}=\frac{\text { Sum of all observations }}{\text { Total number of observations }}$
$
\begin{array}{ll}
\Rightarrow & 10=\frac{x+(x+3)+(x+6)+(x+9)+(x+12)}{5} \quad \text { [Given] } \\
\Rightarrow & 50=5 x+30 \\
\Rightarrow & 5 x=20 \\
\Rightarrow & x=4
\end{array}
$
View full question & answer→MCQ 221 Mark
Four observations are 2, 4,6 and 8. The frequencies of the first three observations are 3,2 and 1 respectively. If the mean of the observations is 4 , then find the frequency of the fourth observation.
Answer(c) : Let the frequency of the fourth observation be $f$.
$
\begin{aligned}
& \text { Mean, } \bar{x}=\frac{x_1 f_1+x_2 f_2+x_3 f_3+x_4 f_4}{f_1+f_2+f_3+f_4} \\
\Rightarrow & 4=\frac{2 \times 3+4 \times 2+6 \times 1+8 \times f}{3+2+1+f}=\frac{6+8+6+8 f}{6+f} \\
\Rightarrow & 24+4 f=20+8 f \\
\Rightarrow & 4 f=4 \Rightarrow f=1
\end{aligned}
$
View full question & answer→MCQ 231 Mark
The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is $₹\ 18$. Find the missing frequency $f$.
| Daily pocket allowance $($in $₹)$ |
$11-13$ |
$13-15$ |
$15-17$ |
$17-19$ |
$19-21$ |
$21-23$ |
$23-25$ |
| Frequemcy |
$7$ |
$6$ |
$9$ |
$13$ |
$f$ |
$5$ |
$4$ |
AnswerThe frequency distribution table from the given data can be drawn as :
| Daily allowance $($in $₹)$ |
Class mark $\left(x_i\right)$ |
Number of children $\left(f_i\right)$ |
$f_i x_i$ |
| $11-13$ |
$12$ |
$7$ |
$84$ |
| $13-15$ |
$14$ |
$6$ |
$84$ |
| $15-17$ |
$16$ |
$9$ |
$144$ |
| $17-19$ |
$18$ |
$13$ |
$234$ |
| $19-21$ |
$20$ |
$f$ |
$20f$ |
| $21-23$ |
$22$ |
$5$ |
$110$ |
| $23-25$ |
$24$ |
$4$ |
$96$ |
| |
|
\sum f_i=f+44 |
\Sigma f_i x_i=20 f+752 |
Now, mean$, \bar{x}=\frac{\sum f_i x_i}{\sum f_i}$
$\Rightarrow 18=\frac{20 f+752}{f+44}$
$\Rightarrow 20 f+752=18 f+792 $
$\Rightarrow 20 f-18 f=792-752$
$\Rightarrow 2 f=40 $
$\Rightarrow f=\frac{40}{2}=20$
Hence, the fourth frequency$, f$ is $20 .$ View full question & answer→MCQ 241 Mark
The mean of the following data is
| class interval |
$0-10$ |
$10-20$ |
$20-30$ |
$30-40$ |
$40-50$ |
| Frequemcy |
$3$ |
$5$ |
$9$ |
$5$ |
$3$ |
AnswerThe frequency distribution table from the given data can be drawn as :
| Daily allowance $($in $₹)$ |
Class mark $\left(x_i\right)$ |
Number of children $\left(f_i\right)$ |
$f_i x_i$ |
| $11-13$ |
$12$ |
$7$ |
$84$ |
| $13-15$ |
$14$ |
$6$ |
$84$ |
| $15-17$ |
$16$ |
$9$ |
$144$ |
| $17-19$ |
$18$ |
$13$ |
$234$ |
| $19-21$ |
$20$ |
$f$ |
$20$f$ |
| $21-23$ |
$22$ |
$5$ |
$110$ |
| $23-25$ |
$24$ |
$4$ |
$96$ |
| |
|
$\sum f_i=f+44$ |
$\Sigma f_i x_i=$
$20 f+752$ |
Now, mean, $\bar{x}=\frac{\sum f_i x_i}{\sum f_i}$
$\Rightarrow 18=\frac{20 f+752}{f+44}$
$\Rightarrow 20 f+752=18 f+792$
$\Rightarrow 20 f-18 f=792-752$
$\Rightarrow 2 f=40$
$\Rightarrow f=\frac{40}{2}=20$
Hence, the fourth frequency, $f$ is $20 .$ View full question & answer→MCQ 251 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)]
View full question & answer→MCQ 261 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, $\bar{x}$
$=\left(x_1-\bar{x}\right)+\left(x_2-\bar{x}\right)+\left(x_3-\bar{x}\right)+\ldots+\left(x_n-\bar{x}\right)$
$=\left(x_1+x_2+x_3+\ldots+x_n\right)-(\bar{x}+\bar{x}+\bar{x} \ldots n \text { times })$
$=\frac{n\left(x_1+x_2+x_3+\ldots+x_n\right)}{n}-n \bar{x}=n \bar{x}-n \bar{x}=0$
View full question & answer→MCQ 271 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 281 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 291 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 301 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 311 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 321 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. $24.5$
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 331 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 $=\frac{1}{2}\left(\left(\frac{6}{2}\right)^{\text {th }} \text { term }+\left(\frac{6}{2}+1\right)^{\text {th }} \text { term }\right)$
$\Rightarrow 16=\frac{1}{2}\left(3^{\text {rd }} \text { term }+4^{\text {th }} \text { term }\right)$
$\Rightarrow 32=x-2+x$
$\Rightarrow 2 x=34 $
$\Rightarrow x=17$
View full question & answer→MCQ 341 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 351 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 361 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 371 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 381 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 391 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 401 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 411 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 421 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 431 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 441 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 451 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 461 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 471 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 481 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 491 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 501 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 $=10$
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→