MCQ 11 Mark
AnswerMode is the most frequency value of observation or a class,
View full question & answer→MCQ 21 Mark
Consider the following frequency distribution:
| Class: |
$65-85$ |
$85-105$ |
$105-125$ |
$125-145$ |
$145-165$ |
$165-185$ |
$185-205$ |
| Frequency: |
$4$ |
$5$ |
$13$ |
$20$ |
$14$ |
$7$ |
$4$ |
The difference of the upper limit of the median class and the lower limit of the modal class is: Answer
|
Class
|
Frequency
|
Cumulative Frequency
|
| $65-85$ |
$4$ |
$4$ |
| $85-105$ |
$5$ |
$9$ |
| $105-125$ |
$13$ |
$22$ |
| $125-145$ |
$20$ |
$42$ |
| $145-165$ |
$14$ |
$56$ |
| $165-185$ |
$7$ |
$63$ |
| $185-205$ |
$4$ |
$67$ |
Here, $\frac{\text{N}}{2}=\frac{67}{2}=33.5$
which lies in the interval $125-145.$
Hence, upper limit of median class is $145.$
Here, we see that the highest frequency is $20.$
which lies in $125-145.$
Hence, the lower limit of modal class is $125.$
$\therefore$ Required difference $=$ Upper limit of median class $–$ Lower limit of modal class
$= 145 – 125$
$=20$ View full question & answer→MCQ 31 Mark
In the formula $\bar{\text{x}}=\text{a}+\text{h}\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}\Big),$ for finding the mean of grouped frequency distribution $\text{u}_\text{i}=$
- A
$\frac{\text{x}_\text{i}+\text{a}}{\text{h}}$
- B
$\text{h}(\text{x}_\text{i}-\text{a})$
- ✓
$\frac{\text{x}_\text{i}-\text{a}}{\text{h}}$
- D
$\frac{\text{a}-\text{x}_\text{i}}{\text{h}}$
AnswerCorrect option: C. $\frac{\text{x}_\text{i}-\text{a}}{\text{h}}$
Given $\bar{\text{x}}=\text{a}+\text{h}\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}\Big)$
Above formula is a step deviation formula.
$\text{u}_\text{i}=\frac{\text{x}_\text{i}-\text{a}}{\text{h}}$
View full question & answer→MCQ 41 Mark
If the mode of the data : $16, 15, 17, 16, 15, x, 19, 17, 14$ is $15,$ then $x =$
AnswerMode of $16, 15, 17, 16, 15, x, 19, 17, 14$ is $15$
$\because$ By definition mode of a number which has maximum frequency which is $15$
$\therefore\text{x}=15$
View full question & answer→MCQ 51 Mark
If the difference of mode and median of a data is $24,$ then the difference of median and mean is:
AnswerDifference of mode and median $= 24$
Mode $= 3$ median $– 2$ mean
$\Rightarrow$ Mode $–$ median $= 2$ median $– 2$ mean
$\Rightarrow 24 = 2 ($ median $–$ mean$)$
$\Rightarrow$ Median $–$ mean $=\frac{24}{2}=12$
View full question & answer→MCQ 61 Mark
If the mean of first $n$ natural numbers is $\frac{5\text{n}}{9},$ then $n =$
AnswerGiven:
Mean of first $n$ natural number $=\frac{5\text{n}}{9}$
$\Rightarrow\frac{1+2+3+.......+\text{n}}{\text{n}}=\frac{5\text{n}}{9}$
$\Rightarrow\frac{\frac{\text{n}(\text{n}+1)}{2}}{\text{n}}=\frac{5\text{n}}{9}$
$\Rightarrow\frac{\text{n}+1}{2}=\frac{5\text{n}}{9}$
$\Rightarrow9\text{n+9}=10\text{n}$
$\Rightarrow\text{n}=9$
Hence, the correct option is $(c).$
View full question & answer→MCQ 71 Mark
The mean of a discrete frequency distribution $x_if_i$ ; $i= 1, 2, ……, n$ is given by:
- ✓
$\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}$
- B
$\frac{1}{\text{n}}\sum_\text{i=1}^\text{n}\text{f}_\text{i}\text{x}_\text{i}$
- C
$\frac{\frac{1}{\text{n}}\sum_\text{i=1}^\text{n}\text{f}_\text{i}\text{x}_\text{i}}{\frac{1}{\text{n}}\sum_\text{i=1}^\text{n}\text{x}_\text{i}}$
- D
$\frac{\frac{1}{\text{n}}\sum_\text{i=1}^\text{n}\text{f}_\text{i}\text{x}_\text{i}}{\frac{1}{\text{n}}\sum_\text{i=1}^\text{n}\text{i}}$
AnswerCorrect option: A. $\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}$
The mean of discrete frequency distribution $\frac{\text{x}_\text{i}}{\text{f}_\text{i}};\text{i}=1, 2, 3, .....\text{n},$ will be $\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}$
View full question & answer→MCQ 81 Mark
If mode of a series exceeds its mean by $12$, then mode exceeds the median by:
AnswerGiven: Mode $−$ Mean $= 12$
We know that
Mode $= 3$ Median $− 2$ Mean
$\therefore$ Mode $−$ Mean $= 3 ($ Median $−$ Mean$)$
$\Rightarrow 12 = 3 ($ Median $−$ Mean$)$
$\Rightarrow$ Median $−$ Mean $= 4 .....(1)$
Again,
Mode $= 3$ Median $− 2$ Mean
$\Rightarrow 2$ Mode $= 6$ Median $− 4$ Mean
$\Rightarrow$ Mode $−$ Mean $+$ Mode $= 6$ Median $− 5$ Mean
$\Rightarrow 12 + ($Mode $−$ Median$) = 5 ($Median $−$ Mean$)$
$\Rightarrow 12 + ($ Mode $−$ Median$) = 20 [$ Using $(1)]$
$\Rightarrow$ Mode $−$ Median $= 20 − 12 = 8$
Hence, the correct option is $(b).$
View full question & answer→MCQ 91 Mark
For the following distribution:
| Class: |
$0-5$ |
$5-10$ |
$10-15$ |
$15-20$ |
$20-25$ |
| Frequency: |
$10$ |
$15$ |
$12$ |
$20$ |
$9$ |
the sum of the lower limits of the median and modal class is: Answer
| Class |
Frequency |
Cumulative Frequency |
| $0-5$ |
$10$ |
$10$ |
| $5-10$ |
$15$ |
$25$ |
| $10-15$ |
$12$ |
$37$ |
| $15-20$ |
$20$ |
$57$ |
| $20-25$ |
$9$ |
$66$ |
Now,$\frac{\text{N}}{20}=\frac{66}{2}=33,$ which lies in the interval $10-15.$
Therefore, lower limit of the median class is $10.$
The highest frequency is $20$,
which lies in the interval $15-20.$
Therefore, lower limit of modal class is $15$.
Hence, required sum is $10 + 15 = 25.$ View full question & answer→MCQ 101 Mark
If the mean of $6, 7, x, 8, y, 14$ is $9,$ then:
- A
$x + y = 21$
- ✓
$x + y = 19$
- C
$x - y = 19$
- D
$v - y = 21$
AnswerCorrect option: B. $x + y = 19$
Mean of $6, 7, x, 8, y, 14$ is $9$
$\Rightarrow\frac{6+7+\text{x}+8+\text{y}+14}{6}=9\ (\text{n}=6)$
$\Rightarrow\frac{35+\text{x}+\text{y}}{6}=9$
$\Rightarrow35+\text{x}+\text{y}=54$
$\Rightarrow\text{x}+\text{y}=54-35=19$
View full question & answer→MCQ 111 Mark
If the median of the data: $24, 25, 26, x + 2, x + 3, 30, 31, 34$ is $27.5,$ then $x =$
AnswerThe given observations are $24, 25, 26, x + 2, x + 3, 30, 31, 34.$
Median $= 27.5$
Here, $n = 8$
$\text{Median}=\frac{\Big(\frac{\text{n}}{2}\Big)^\text{th}\text{term}+\Big(\frac{\text{n}}{2}+1\Big)^\text{th}\text{term}}{2}$
$27.5=\frac{4\text{th term}+5\text{th term}}{2}$
$27.5=\frac{(\text{x}+2)+(\text{x}+3)}{2}$
$27.5=\frac{2\text{x}+5}{2}$
$2\text{x}+5=55$
$2\text{x}=50$
$\text{x}=25$
Hence, the correct option is $(b).$
View full question & answer→MCQ 121 Mark
The mean of first $n$ odd natural numbers is $\frac{\text{n}^2}{81},$ then $n =$
AnswerThe first $n$ odd natural numbers are $1, 3, 5, ...., (2n − 1).$
$\therefore$ Mean of first $n$ odd natural numbers
$=\frac{1+3+5+.....+(2\text{n}-1)}{\text{n}}$
$=\frac{\frac{\text{n}}{2}(1+2\text{n}-1)}{\text{n}}\ [\text{S}_\text{n}$
$=\frac{\text{n}}{2}(\text{a}+\text{l})]$
$=\frac{2\text{n}}{\text{n}}$
$=\text{n}$
Now,
Mean of first $n$ natural numbers $=\frac{\text{n}^2}{81} ($Given$)$
$\therefore\text{n}=\frac{\text{n}^2}{81}$
$\Rightarrow\text{n}=81$
Hence, the correct option is $(b).$
View full question & answer→MCQ 131 Mark
For the following distribution:
| Below: |
$10$ |
$20$ |
$30$ |
$40$ |
$50$ |
$60$ |
| Number of students: |
$3$ |
$12$ |
$27$ |
$57$ |
$75$ |
$80$ |
the modal class is: - A
$10-20$
- B
$20-30$
- ✓
$30-40$
- D
$50-60$
AnswerCorrect option: C. $30-40$
|
Below
|
Class interval
|
cumulative Frequency
|
Frequency
|
| $10$ |
$0-10$ |
$3$ |
$3$ |
| $20$ |
$10-20$ |
$12$ |
$9$ |
| $30$ |
$20-30$ |
$27$ |
$15$ |
| $40$ |
$30-40$ |
$57$ |
$30$ |
| $50$ |
$40-50$ |
$75$ |
$18$ |
| $60$ |
$50-60$ |
$80$ |
$5$ |
Here, $N = 80.$
$\therefore\frac{\text{N}}{2}=40,$
which lines in the interval $30-40.$
Therefore, the modle class is $30-40.$
Hence, the correct answer is option $(c).$ View full question & answer→MCQ 141 Mark
The relationship between mean, median and mode for a moderately skewed distribution is:
- A
Mode $= 2$ Median $− 3$ Mean.
- B
Mode $=$ Median $− 2$ Mean.
- C
Mode $= 2$ Median $−$ Mean.
- ✓
Mode $= 3$ Median $−2$ Mean.
AnswerCorrect option: D. Mode $= 3$ Median $−2$ Mean.
Mode $= 3$ Median $−2$ Mean.
View full question & answer→MCQ 151 Mark
If the mode of the data: $64, 60, 48, x, 43, 48, 43, 34$ is $43,$ then $x + 3 =$
Answer
| Value |
$34$ |
$43$ |
$48$ |
$60$ |
$64$ |
$x$ |
| Frequency |
$1$ |
$2$ |
$2$ |
$1$ |
$1$ |
$1$ |
$x + 3 = 46$ It is given that the mode of the given date is $43.$
So, it is the value with the maximum frequency.
Now, this is possible only when $x = 43.$
In this case, the frequency of the observation $43$ would be $3.$
Hence,
$x + 3 = 46$
Hence, the correct option is $(c).$ View full question & answer→MCQ 161 Mark
The median of a given frequency distribution is found graphically with the help of:
AnswerThe median of a given frequency distribution is found graphically with the help of ‘Ogive’.
Hence, the correct option is $(d).$
View full question & answer→MCQ 171 Mark
The mean of $n$ observations is $\bar{\text{x}}$ If the first observation is increased by $1,$ the second by $2,$ the third by $3,$ and so on, then the new mean is:
- A
$\bar{\text{x}}+(2\text{n}+1)$
- ✓
$\bar{\text{x}}+\frac{\text{n}+1}{2}$
- C
$\bar{\text{x}}+(\text{n}+1)$
- D
$\bar{\text{x}}-\frac{\text{n}-1}{2}$
AnswerCorrect option: B. $\bar{\text{x}}+\frac{\text{n}+1}{2}$
Mean of $n$ observations $=\bar{\text{x}}$
Increasing first observation by $1$, second by $2,$ third by $3$ and so on,
$\therefore$ Sum of increased number $ =\frac{\text{n}(\text{n}+1)}{2}$
and $\text{mean}=\frac{\text{n}(\text{n}+1)}{2\times\text{n}}=\frac{\text{n}+1}{2}$
$\therefore$ New mean$=\bar{\text{x}}+\frac{\text{n}+1}{2}$
View full question & answer→MCQ 181 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.
The relation between mean, median and mode is:
View full question & answer→MCQ 191 Mark
The mean of $1, 3, 4, 5, 7, 4$ is $m.$ The number $3, 2, 2, 4, 3, 3, p$ have mean $m – 1$ and median $q.$ Then $p + q =$
AnswerMean of $1, 3, 4, 5, 7, 4$ is $m$
$\therefore\frac{1+3+4+5+7+4}{6}=\text{m}$
$\Rightarrow\frac{24}{6}=\text{m}$
$\Rightarrow\text{m}=4$
Mean of $3, 2, 2, 4, 3, 3, p$ is $m = 1$
$\Rightarrow\frac{3+2+2+4+3+3+p}{7}=\text{m}-1$
$\Rightarrow\frac{17+\text{p}}{7}=4-1$
$\Rightarrow\frac{17+\text{p}}{7}=3$
$\Rightarrow17+\text{p}=21$
$\Rightarrow\text{p}=21-17=4$
Median of $3, 2, 2, 4, 3, 3, p$ is $q$
$3, 2, 2, 4, 3, 3, 4$ is $q$
Arranging in order, we get $4, 4, 3, 3, 3, 2, 2$
Here $n = 7$
$\therefore\text{Median}=\frac{7+1}{2}\text{th}$ term $=4\text{th}$ term
$=3$
$\therefore\text{q}=3$
$\therefore\text{p}+\text{q}=4+3=7$
View full question & answer→MCQ 201 Mark
In the formula $\overline{\text{X}}=\text{a}+\frac{\sum\text{f}_\text{i}\text{d}_\text{i}}{\sum\text{f}_\text{i}}$ for finding the mean of grouped data di’s are deviations from a of:
- A
- B
- ✓
- D
frequency of the class marks .
AnswerWe know that, $d_i = x_i – a$
i .e , $d_i$‘s are the deviation from a mid$-$points of the classes.
View full question & answer→MCQ 211 Mark
The arithmetic mean and mode of a data are $24$ and $12$ respectively, then its median is:
AnswerArithmetic mean $= 24$
Mode $= 12$
$\therefore$ But mode $= 3$ median $– 2$ mean
$\Rightarrow 12 = 3$ median $– 2 \times 24$
$\Rightarrow 12 = 3$ median $- 48$
$\Rightarrow 12 + 48 = 3$ median
$\Rightarrow 3$ median $= 60$
$\text{Median}=\frac{60}{3}=20$
View full question & answer→MCQ 221 Mark
If the mean of first $n$ natural number is $15,$ then $n =$
AnswerMean of first $n$ natural number $= 15$
$\frac{\text{n}(\text{n}+1)}{2\text{n}}=15$
$\frac{\text{n+1}}{2}=15$
$\Rightarrow\text{n}+1=30$
$\text{n}=30-1=29$
View full question & answer→MCQ 231 Mark
The mode of a frequency distribution can be determined graphically from:
AnswerMode of frequency can be found graphically by an ogive,
View full question & answer→MCQ 241 Mark
If the mean of frequency distribution is $8.1$ and $\sum\text{f}_\text{i}\text{x}_\text{i}=132+5\text{k},\sum\text{f}_\text{i}=20,$ then $k =$
AnswerGiven:
$\sum\text{f}_\text{i}\text{x}_\text{i}=132+5\text{k},\sum\text{f}_\text{i}=20$ and mean $= 8.1.$
Then,
$\text{Mean}=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}$
$8.1=\frac{132+5\text{k}}{20}$
$162=132+5\text{k}$
$5\text{k}=30$
$\text{k}=6$
Hence, the correct option is $(d).$
View full question & answer→MCQ 251 Mark
The mean of n observation is $\bar{\text{X}}$. If the first item is increased by $1$, second by $2$ and so on, then the new mean is:
AnswerCorrect option: C. $\overline{\text{X}}+\frac{\text{n}+1}{2}$
Let $x_1, x_2, x_3, ........, x_n$ be the n observations.
$\text{Mean}=\overline{\text{X}}=\frac{\text{x}_1+\text{x}_2+......+\text{x}_\text{n}}{2}$
$\Rightarrow\text{x}_1+\text{x}_2+\text{x}_3+......\text{x}_\text{n}=\text{n}\overline{\text{X}}$
if the first item is increased by $1$, second by $2$ and so on.
Then, the new observations are $x_1 + 1, x_2 + 2, x_3 +3,.....x_n + n$
$\text{New mean}=\frac{(\text{x}_1+1)+(\text{x}_2+2)+(\text{x}_3+3)+.....+(\text{x}_\text{n}+\text{n})}{\text{n}}$
$=\frac{\text{x}_1+\text{x}_2+\text{x}_3+.......+\text{x}_\text{n}+(1+2+3+...+\text{n})}{\text{n}}$
$=\frac{\text{n}\overline{\text{X}}+\frac{\text{n}(\text{n+1})}{2}}{\text{n}}$
$\overline{\text{X}}+\frac{\text{n}+1}{2}$
View full question & answer→MCQ 261 Mark
The arithmetic mean of $1, 2, 3, ..., n$ is:
- ✓
$\frac{\text{n}+1}{2}$
- B
$\frac{\text{n}-1}{2}$
- C
$\frac{\text{n}}{2}$
- D
$\frac{\text{n}}{2}+1$
AnswerCorrect option: A. $\frac{\text{n}+1}{2}$
Arithmetic mean of $1, 2, 3, ......, n$
$=\frac{1+2+3+......+\text{n}}{\text{n}}$
$=\frac{\frac{\text{n}(\text{n}+1)}{2}}{\text{n}}$
$=\frac{\text{n}+1}{2}$
Hence, the correct option is $(a)$
View full question & answer→MCQ 271 Mark
If the arithmetic mean of $7, 8, x, 11, 14$ is $x,$ then $x =$
AnswerArithmetic mean of $7, 8, x, 11, 14,$ is $x$
$\Rightarrow\frac{7+8+\text{x}+11+14}{5}=\text{x}$
$\Rightarrow\frac{40+\text{x}}{5}=\text{x}$
$\Rightarrow40+\text{x}=5\text{x}$
$\Rightarrow5\text{x}-\text{x}=40$
$\Rightarrow4\text{x}=40$
$\Rightarrow\text{x}=\frac{40}{4}=10$
View full question & answer→MCQ 281 Mark
Which of the following is not a measure of central tendency?
AnswerStandard deviation is not a measure of central tendency.
Only mean, median and mode are measures.
View full question & answer→MCQ 291 Mark
The abscissa of the point of intersection of less than type and of the more than type cumulative frequency curves of a grouped data gives its:
AnswerThe less than ogive and more than ogive when drawn on the same graph intersect at a point.
From this point, if we draw a perpendicular on the $x-$axis, the point at which it cuts the $x-$axis gives us the median.
Thus, the abscissa of the point of intersection of less than type and of the more than type cumulative curves of a grouped data gives its median.
Hence, the correct answer is option $(b).$
View full question & answer→MCQ 301 Mark
If the median of the data : $6, 7, x – 2, x, 17, 20,$ written in ascending order, is $16$. Then $x =$
AnswerMedian of $6, 7, x - 2, x, 17, 20$ is $16$
Here $n = 6$
$\therefore\text{Median}=\frac{1}{2}\Big[\frac{\text{n}}{2}\text{th}+\Big(\frac{\text{n}}{2}+1\Big)\text{th}\Big]\text{term}$
$=\frac{1}{2}\Big[\frac{6}{2}+\Big(\frac{6}{2}+1\Big)\Big]\text{term}$
$=\frac{1}{2}(3\text{rd}+4\text{th})\text{term}$
$=\frac{1}{2}(\text{x}-2+\text{x})$
$=\frac{1}{2}(2\text{x}-2)=\text{x}-1$
$\therefore\text{x}-1=16$
$\Rightarrow\text{x}=16+1=17$
View full question & answer→MCQ 311 Mark
While computing mean of grouped data, we assume that the frequencies are:
- A
Evenly distributed over all the classes.
- ✓
Centred at the class marks of the classes.
- C
Centred at the upper limit of the classes.
- D
Centred at the lower limit of the classes.
AnswerCorrect option: B. Centred at the class marks of the classes.
We know that while computing the mean of a grouped data, the frequencies are centered at the class marks of the classes.
Hence, the correct answer is option $(b).$
View full question & answer→MCQ 321 Mark
Mean of a certain number of observation is. If each observation is divided by $m(m \neq 0)$ and increased by $n$, then the mean of new observation is:
- ✓
$\frac{\bar{\text{x}}}{\text{m}}+\text{n}$
- B
$\frac{\bar{\text{x}}}{\text{n}}+\text{m}$
- C
$\bar{\text{x}}+\frac{\text{n}}{\text{m}}$
- D
$\bar{\text{x}}+\frac{\text{m}}{\text{n}}$
AnswerCorrect option: A. $\frac{\bar{\text{x}}}{\text{m}}+\text{n}$
Let $y_1, y_2, y_3, ......, y_k$ be $k$ observations.
Mean of the observation $=\bar{\text{x}}$
$\Rightarrow\frac{\text{y}_1+\text{y}_2+\text{y}_3+...\text{y}_\text{k}}{\text{k}}=\bar{\text{x}}$
$\Rightarrow\text{y}_1+\text{y}_2+\text{y}_3+.....\text{y}_\text{k}=\text{k}\bar{\text{x}}\ .......(1)$
each observation is divided by m and increased by n, then the new observations are
$\frac{\text{y}_1}{\text{m}}+\text{n},\frac{\text{y}_2}{\text{m}}+\text{n},\frac{\text{y}_3}{\text{m}}+\text{n},.....,\frac{\text{y}_\text{k}}{\text{m}}+\text{n}$
$\therefore$ Mean of new observations
$=\frac{\Big(\frac{\text{y}_1}{\text{m}}+\text{n}\Big)+\Big(\frac{\text{y}_2}{\text{m}}+\text{n}\Big)+.....+\Big(\frac{\text{y}_\text{k}}{\text{m}}+\text{n}\Big)}{\text{k}}$
$=\frac{\Big(\frac{\text{y}_1}{\text{m}}+\frac{\text{y}_2}{\text{m}}+.....+\frac{\text{y}_\text{k}}{\text{m}}+\Big)+(\text{n}+\text{n}+.....+\text{n})}{\text{k}}$
$=\frac{\text{y}_1+\text{y}_2+....+\text{y}_\text{k}}{\text{mk}}+\frac{\text{nk}}{\text{k}}$
$=\frac{\text{k}\bar{\text{x}}}{\text{mk}}+\frac{\text{nk}}{\text{k}}$
$=\frac{\bar{\text{x}}}{\text{m}}+\text{n}$
View full question & answer→MCQ 331 Mark
One of the methods of determining mode is:
- A
Mode $= 2$ Median $– 3$ Mean.
- 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.
Mode $= 3$ Median $– 2$ Mean.
View full question & answer→MCQ 341 Mark
Consider the following frequency distribution:
| Class: |
$0-5$ |
$6-11$ |
$12-17$ |
$18-23$ |
$24-29$ |
| Frequency: |
$13$ |
$10$ |
$15$ |
$8$ |
$11$ |
The upper limit of the median class is: AnswerCorrect option: B. $17.5$
Given, classes are not continuous, so we make continuous by subtracting $0.5$ from lower limit and adding $0.5$ to upper limit of each class.
|
Class
|
Frequency
|
Cumulative Frequency
|
| $0.5-5.5$ |
$13$ |
$13$ |
| $5.5-11.5$ |
$10$ |
$23$ |
| $11.5-17.5$ |
$15$ |
$38$ |
| $17.5-23.5$ |
$8$ |
$46$ |
| $23.5-29.5$ |
$11$ |
$57$ |
Here,$\frac{\text{N}}{2}=\frac{57}{2}=28.5,$
which lies in the interval $11.5-17.5.$
Hence, the upper limit is $17.5.$ View full question & answer→MCQ 351 Mark
The mean of first $n$ odd natural number is:
- A
$\frac{\text{n}+1}{2}$
- B
$\frac{\text{n}}{2}$
- ✓
$\text{n}$
- D
$\text{n}^2$
AnswerCorrect option: C. $\text{n}$
Mean of first $n$ odd numbers Sum of first $n$ odd number
$=\frac{\text{n}}{2}[2\text{a}+(\text{n}+1)\text{d}]$
i.e. $1 + 3 + 5 + 7 + ........ n$ term
$=\frac{\text{n}}{2}[2\times1+(\text{n}+1)\times2]$
$($Here $a = 1, d = 2)$
$=\frac{\text{n}}{2}[2+2\text{n}-2]=\frac{\text{n}}{2}\times2\text{n}=\text{n}^2$
$\therefore\text{Mean}=\frac{\text{Sum of n terms}}{\text{n}}=\frac{\text{n}^2}{\text{n}}=\text{n}$
View full question & answer→MCQ 361 Mark
If the arithmetic mean of $x, x + 3, x + 6, x + 9,$ and $x + 12$ is $10,$ then $x =$
Answer$\text{Mean of x,x}+3,\text{x}+6,\text{x}+9,\text{x}+12=10$
$\Rightarrow\frac{\text{x}+\text{x}+3+\text{x}+6+\text{x}+9+\text{x}+12}{5}=10$
$\Rightarrow\frac{5\text{x}+30}{5}=10$
$\Rightarrow\text{x}+6=10$
$\Rightarrow\text{x}=10-6=4$
View full question & answer→MCQ 371 Mark
if $\text{u}_\text{i}=\frac{\text{x}_\text{i}-25}{10}\sum\text{f}_\text{i}\text{u}_\text{i}=20,\sum\text{f}_\text{i}=100,$ then $\bar{\text{x}}$
Answer$\text{u}_\text{i}=\frac{\text{x}_\text{i}-25}{10},\ \sum\text{f}_\text{i}\text{u}_\text{i}=20,\ \sum\text{f}_\text{i}=100$
Here assumed mean $= 25$
and class interval $(h) = 10$
$\therefore\ \bar{\text{x}}=\text{A}+\frac{\sum\text{f}_\text{i}\text{u}_\text{i}}{\sum\text{f}_\text{i}}\times\text{h}$
$=25+\frac{20}{100}\times10$
$=25+2=27$
View full question & answer→MCQ 381 Mark
Which of the following cannot be determined graphically?
AnswerMean cannot be determind grafically,
View full question & answer→MCQ 391 Mark
The algebraic sum of the deviations of a frequency distribution from its mean is:
- A
- B
- ✓
$0.$
- D
A non$-$zero number.
AnswerThe algebraic sum of the deviations of a frequency distribution from its mean is zero
Let $x_1, x_2, x_3, …… x_n$ are observations and $\overline{\text{X}}$ is the mean
$\therefore (\bar{\text{x}}-\text{x}_1)+(\bar{\text{x}}-\text{x}_2)+(\bar{\text{x}}-\text{x}_3)+ ....(\bar{\text{x}}-\text{x}_\text{n})$
$=\text{n}\bar{\text{x}}-(\text{x}_1+\text{x}_2+\text{x}_3+.....\text{x}_\text{n})$
$=\text{n}\bar{\text{x}}-\text{n}\bar{\text{x}}=0$
View full question & answer→MCQ 401 Mark
The median of first $10$ prime numbers is:
AnswerFirst $10$ prime numbers are $2, 3, 5, 7, 11, 13, 17, 19, 23, 29$
Here $n = 10$
$\therefore\text{Median}=\frac{1}{2}\Big[\frac{\text{n}}{2}\text{th}+\Big(\frac{\text{n}}{2}+1\Big)\text{th}\Big]\text{term}$
$=\frac{1}{2}\Big[\frac{10}{2}\text{th}+\Big(\frac{10}{2}+1\Big)\text{th}\Big]\text{term}$
$=\frac{1}{2}[5\text{th}+6\text{th}]\text{term}$
$=\frac{1}{2}[11+13]$
$=\frac{1}{2}\times24$
$=12$
View full question & answer→MCQ 411 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 421 Mark
If the mean of observations $x_1, x_2, …, x_n$ is $\bar{\text{x}},$ then the mean of $x_1 + a, x_2 + a,…, x_n + a$ 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 $x_1, x_2, ......, x_n$ is $\bar{\text{x}}$
$\frac{\text{x}_1+\text{x}_2+\text{x}_3.......+\text{x}_\text{n}}{\text{n}}=\bar{\text{x}}$
$x_1 + a + x_2 + a + x_3 + a +.....x_n + a$
$= x_1 + x_2 + x_3 +....x_n + na$
$\therefore$ Mean of $(x_1 + x_2 + x_3 .....+ x_n) + na$
$=\bar{\text{x}}+\frac{\text{na}}{\text{n}}=\bar{\text{x}}+\text{a}$
View full question & answer→MCQ 431 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 $\text{median}=37-36.5=0.5$
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