MCQ 511 Mark
The mean of the data when$\sum\text{f}_\text{i}\text{d}_{\text{i}}=-500,\sum\text{f}_\text{i} =100 $ and $ \text{a}=125$ is :
- ✓
$120$
- B
$100$
- C
$-500$
- D
$125$
Answer$\text{Mean}=\text{a}+\frac{\sum\text{f}_\text{i}\text{d}_\text{i}}{\sum\text{f}_\text{i}}$
$=125+\frac{-500}{100}$
$=125-5=120$
View full question & answer→MCQ 521 Mark
If the mean of a data is $27$ and its median is $33$. Then, the mode is :
AnswerMean $= 27$
Median $= 33$
$\text{Mode = 3 median $-$ 2 Mean}$
$= 3 \times 33 - 2 \times 27$
$= 99 - 54$
$= 45$
View full question & answer→MCQ 531 Mark
The percentage of marks obtained by $100$ students in an examination are as follows :
|
Mark
|
$30-35$ |
$35-40$ |
$40-45$ |
$45-50$ |
$50-55$ |
$55-60$ |
$60-65$ |
|
Frequency
|
$14$ |
$16$ |
$18$ |
$23$ |
$18$ |
$8$ |
$3$ |
The median class is :
- ✓
$45 - 50$
- B
$50 - 55$
- C
$40 - 45$
- D
AnswerCorrect option: A. $45 - 50$
|
Mark
|
$30-35$ |
$35-40$ |
$40-45$ |
$45-50$ |
$50-55$ |
$55-60$ |
$60-65$ |
|
Frequency
|
$14$ |
$16$ |
$18$ |
$23$ |
$18$ |
$8$ |
$3$ |
|
Cumulative Frequency
|
$14$ |
$30$ |
$48$ |
$71$ |
$89$ |
$97$ |
$100$ |
Here $N = 100$
$\Rightarrow\frac{\text{N}}{2}=50$
Therefore, the median class is $45 - 50.$ View full question & answer→MCQ 541 Mark
The mean of the first $10$ multiples of $6$ is :
AnswerThe first $10$ multiples of $6$ are $\{6, 12, 18, 24, 30, 36, 42, 48, 54, 60\}$
$\therefore\text{Mean}=\frac{\text{Sum of first 10 multiples of 6}}{10}$
$=\frac{6+12+18+24+30+36+42+48+54+60}{10}$
$=\frac{330}{10}$
$=33$
View full question & answer→MCQ 551 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 561 Mark
Median of $\{15, 28, 72, 56, 44, 32, 31, 43\}$ and $51$ is $43:$
AnswerThe terms are : $\{15, 28, 72, 56, 44, 32, 31, 43\}$ and $51.$
Arranging them in ascending order : $\{15, 28, 31, 32, 43, 44, 51, 56, 72\}.$
Since the total number of terms is odd that is $9,$
therefore the median will be the middle term that is the 5th term which is $43$.
View full question & answer→MCQ 571 Mark
The mode of $\{4, 5, 6, 8, 5, 4, 6, 5, 6, x, 8\}$ is $6$. The value of $'x\ ’$ is :
AnswerHere Observation is $6$ has more frequency than that of other numbers.
$\therefore 6$ could repeat itself at least once more.
$\Rightarrow x$ should be $6$.
View full question & answer→MCQ 581 Mark
Mean $–$ Mode $=$
- A
$3($Mean $–$ Mode$)$
- B
$3($Median $–$ Mean$)$
- ✓
$3($Mean $–$ Median$)$
- D
$3($Mode $–$ Mean$)$
AnswerCorrect option: C. $3($Mean $–$ Median$)$
Since, Mean $– $ Mode $+$ Mean $+ \ 2$ Mode $= 3$ Median
$\Rightarrow $ Mean $–$ Mode $= 3$ Median $–$ Mean $– 2$ Mode
$= 3$ Median $–$ Mean $– 2 \ (3$ Median $–\ 2$ Mean$)$
$= 3$ Median $–$ Mean $– 6$ Median $+ \ 4$ Mean
$= 3$ Mean $– 3$ Median
$= 3($Mean $–$ Median$)$
View full question & answer→MCQ 591 Mark
The modal value is the value of the variate which divides the total frequency into two equal parts :
AnswerFalse Modal value is the value which occurs maximum number of times in the data.
View full question & answer→MCQ 601 Mark
For finding the mean by using the formula, $\bar{\text{x}}=\text{A}+\text{h}\Big(\frac{\sum\text{f}_\text{i}\text{u}_\text{i}}{\sum\text{f}_\text{i}}\Big),$ we have $\mathrm{u}_{\mathrm{i}}=?$
- A
$\frac{(\text{A}-\text{x}_\text{i})}{\text{h}}$
- ✓
$\frac{\text{(x}_{\text{i}}-\text{A})}{\text{h}}$
- C
$\frac{(\text{A}+\text{x}_\text{i})}{\text{h}}$
- D
$\text{h}(\text{x}_\text{i}-\text{A})$
AnswerCorrect option: B. $\frac{\text{(x}_{\text{i}}-\text{A})}{\text{h}}$
By formula method,
$\bar{\text{x}}=\text{A}+\text{h}\Big(\frac{\sum\text{f}_\text{i}\text{u}_\text{i}}{\sum\text{f}_\text{i}}\Big),$
where $\text{u}_\text{i}=\frac{(\text{x}_\text{i}-\text{A})}{\text{h}}$
View full question & answer→MCQ 611 Mark
$29.$ The arithmetic mean of the following data is $53.$
|
Class
|
$0-20$ |
$20-40$ |
$40-60$ |
$60-80$ |
$80-100$ |
|
Frequency
|
$12$ |
$15$ |
$32$ |
$P$ |
$13$ |
The value of $p$ is : Answer
|
Class
|
Mid $-$ value $(\text{x}_\text{i})$
|
Frequency $(\text{f}_\text{i})$
|
$\text{f}_\text{i}\text{x}_\text{i}$
|
| $0-20$ |
$10$ |
$12$ |
$120$ |
| $20-40$ |
$30$ |
$15$ |
$450$ |
| $40-60$ |
$50$ |
$32$ |
$1600$ |
$
60-80$ |
$70$ |
$p$ |
$70p$ |
| $80-100$ |
$90$ |
$13$ |
$1170$ |
| Total |
|
$\sum f\text{i}=72+\text{p}$
|
$\sum f_\text{i}\text{x}_\text{i}=3340+70\text{p}$
|
Mean$=\frac{\sum f_\text{i}\text{x}_\text{i}}{\sum f_\text{i}}$
$\Rightarrow53 =\frac{3340+70\text{p}}{72+\text{p}}$
$\Rightarrow53(72+\text{p})=3340+70\text{p}$
$\Rightarrow3340+70\text{p}=3816+53\text{p}$
$\Rightarrow+70\text{p}-53\text{p}=3816-3340$
$\Rightarrow17\text{p}=476$
$\Rightarrow\text{p}=28$ View full question & answer→MCQ 621 Mark
$\frac{\text{Upper class limit + Lower class limit}}{2}$
AnswerIn each class interval of grouped data, there are two limits or boundaries $($upper limit and lower limit$)$ while the mid $-$ value is equal to $\frac{\text{Upper class limit+Lower class limit}}{2}.$
These mid $-$ values are also known as Class mark.
View full question & answer→MCQ 631 Mark
If the mode of the data is $45$ and the median is $33,$ then the mean is :
AnswerSince $,3$ Median $= 2$ Mean $+$ Mode
$\therefore 3 \times 33 = 2$ Mean $+\ 45$
$\Rightarrow 2$ Mean $= 99 - 45$
$\Rightarrow 2$ Mean $= 54$
$\Rightarrow $ Mean $= 27$
View full question & answer→MCQ 641 Mark
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 :
AnswerThe 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.
View full question & answer→MCQ 651 Mark
The mean and mode of a frequency distribution are $28$ and $16$ respectively. The median is :
Answermean $= 28$
Mode $= 16$
Mode $= 3$ median $- 2$ mean
Hence, median $=\frac{\text{Mode + 2Mean}}{3}$
$=\frac{16+2(28)}{3}$
$=\frac{16+56}{3}$
$=\frac{72}{3}$
$=24$
View full question & answer→MCQ 661 Mark
For a symmetrical distribution :
AnswerCorrect option: B. Mean $=$ Median $=$ Mode
If a frequency distribution has a symmetrical frequency curve,
then mean, median and mode are equal.
However, an empirical relationship exists between mean, median and mode.
For moderately skewed data distribution.
For a symmetrical distribution Mean $=$ Median $=$ Mode
View full question & answer→MCQ 671 Mark
The cumulative frequency table is useful in determining the :
AnswerThe cumulative frequency table is useful in determining the medan.
View full question & answer→MCQ 681 Mark
If the mean and median of a set of number are $8.9$ and $9$ respectively, then the mode will be :
- A
$7.2$
- B
$8.2$
- ✓
$9.2$
- D
$10.2$
AnswerMean $= 8.9$
Median $= 9$
Mode $= 3$ median $- 2$ mean
$= 3 \times 9 - 2 \times 8.9$
$= 27 - 27.8$
$= 9.2$
View full question & answer→MCQ 691 Mark
The percentage of marks obtained by $100$ students in an examination are as follows :
|
Mark
|
$130-135$ |
$135-140$ |
$140-145$ |
$145-150$ |
$150-155$ |
$155-160$ |
$160-165$ |
|
Frequency
|
$14$ |
$16$ |
$18$ |
$23$ |
$18$ |
$8$ |
$3$ |
The cumulative frequency of the class interval $140-145$ is : Answer
|
Mark
|
$130-135$ |
$135-140$ |
$140-145$ |
$145-150$ |
$150-155$ |
$155-160$ |
$160-165$ |
|
Frequency
|
$14$ |
$16$ |
$18$ |
$23$ |
$18$ |
$8$ |
$3$ |
|
Cumulative Frequency
|
$14$ |
$30$ |
$48$ |
$71$ |
$89$ |
$97$ |
$100$ |
Therefore, the cumulative frequency of the class interval of $140-145$ is $48$.
View full question & answer→MCQ 701 Mark
The middle most value of the data is :
AnswerThe median of a set of data values is the middle most value when the data has been arranged in ascending order
i.e. from smallest value to the largest value.
View full question & answer→MCQ 711 Mark
The mean of the first $10$ natural odd numbers is :
AnswerThe first $10$ natural odd numbers are $\{1, 3, 5, 7, 9, 11, 13, 15, 17, 19\}$
$\therefore\text{Mean}=\frac{\text{sum of first 10 natural odd numbers}}{10}$
$=\frac{1+3+5+7+9+11+13+15+17+19}{10}$
$=\frac{100}{10}$
$=10$
View full question & answer→MCQ 721 Mark
The median of a frequency distribution is found graphically with the help of :
Answerogives are used to determine the median of a frequency distribution.
View full question & answer→MCQ 731 Mark
The arithmetic mean of a set of $40$ values is $65$. If each of the $40$ values is increased by $5,$ what will be the mean of the set of new values :
AnswerMean of $40$ values $= 65$
Total of $40$ values $= 65 \times 40 = 2600$
When each value is increased by $5,$
then the total of $40$ values will be $40 \times 5 = 200$ more than $2600$
$\therefore$ New total of $40$ values $= 2600 + 200 = 2800$
Now, New mean of $40$ values$=\frac{2800}{40}=70$
View full question & answer→MCQ 741 Mark
While computing the mean of the grouped data, we assume that the frequencies are :
- A
Evenly distributed over the classes.
- ✓
Centred at the class marks of the classes.
- C
Centred at the lower limits of the classes.
- D
Centred at the upper limits of the classes.
AnswerCorrect option: B. Centred at the class marks of the classes.
While computing the mean of the grouped data.
we assume that the frequencies are centred at the class marks of the classes.
View full question & answer→MCQ 751 Mark
Choose the correct answer from the given four options : The abscissa of the point of intersection of the less than type and of the more than type cumulative frequency curves of a grouped data gives its :
AnswerSince, the intersection point of less than ogiven and more than ogiven given the median on the abscissa.
View full question & answer→MCQ 761 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.
Hence, the correct option is $(d).$
View full question & answer→MCQ 771 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 781 Mark
Choose the correct answer from the given four options: If $x_i$ 's are the mid points of the class intervals of grouped data, $\mathrm{f}_{\mathrm{i}}$ 's are the corresponding frequencies and $\overline{\mathrm{x}}$ is the mean, then $\sum\left(f_{\mathrm{i}}-\overline{\mathrm{x}}\right)$ is equal to :
Answer$\because\ \ \ \bar{\text{x}}=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\text{n}}$
$\therefore\ \ \sum(\text{f}_\text{i}\text{ x}_\text{i}-\bar{\text{x}})$
$=\sum\text{f}_\text{i}\text{ x}_\text{i}-\sum\bar{\text{x}}$
$=\text{n}\bar{\text{x}}-\text{n}\bar{\text{x}}$
$=0$
View full question & answer→MCQ 791 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 801 Mark
The time in seconds, taken by $75$ athletes to run a $500m$ race are tabulated as below :
|
classes
|
$65-85$ |
$85-105$ |
$105-125$ |
$125-145$ |
$145-165$ |
$145-185$ |
$185-205$ |
|
Frequency
|
$4$ |
$5$ |
$18$ |
$20$ |
$17$ |
$7$ |
$4$ |
The number of athletes who completed the race in less than $125$ seconds is : Answer
|
Classes
|
$65-85$ |
$85-105$ |
$105-125$ |
$125-145$ |
$145-165$ |
$165-185$ |
$185-205$ |
|
Frequency
|
$4$ |
$5$ |
$18$ |
$20$ |
$17$ |
$7$ |
$4$ |
|
Cumulative Frequency
|
$4$ |
$9$ |
$27$ |
$47$ |
$64$ |
$71$ |
$7$ |
Therefore, the number of athletes who completed the race in less than $125$ seconds is $27$. View full question & answer→MCQ 811 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 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 821 Mark
Directions: In the following questions, a statement of assertion $(A)$ is followed by a statement of reason $(R).$ Mark the correct choice as :
Assertion : If the median and mode of a frequency distribution are $8.9$ and $9.2$ respectively, then its mean is $9$.
Reason : Mean, median and mode of a frequency distribution are related as : mode $= 3$ median $- 2$ mean
- A
Both assertion $(A)$ and reason $(R)$ are true and reason $(R)$ is the correct explanation of assertion $(A)$.
- B
Both assertion $(A)$ and reason $(R)$ are true but reason $(R)$ is not the correct explanation of assertion $(A).$
- C
Assertion $(A)$ is true but reason $(R)$ is false.
- ✓
Assertion $(A)$ is false but reason $(R)$ is true.
AnswerCorrect option: D. Assertion $(A)$ is false but reason $(R)$ is true.
Clearly, Reason is correct.
Using the relation given in Statement $-II,$ we have
$\Rightarrow 2$ Mean $= 3$ Median $–$ Mode
$\Rightarrow 2$ Mean $= 3 \times 8.9 - 9.2$
$\Rightarrow 2$ Mean $= 26.7 - 9.2$
$\Rightarrow 2$ Mean $= 17.5$
$\Rightarrow$ Mean $= 8.75$
View full question & answer→MCQ 831 Mark
Consider 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 class and the modal class is : Answer
| Class interval |
Frequency |
frequency |
| $0-10$ |
$10$ |
$10$ |
| $5-10$ |
$15$ |
$25$ |
| $10-15$ |
$15$ |
$37$ |
| $15-20$ |
$20$ |
$57$ |
| $20-25$ |
$9$ |
$66$ |
Here, $\text{N}=66$
$\Rightarrow\frac{\text{N}}{2}=33$
The cumulative frequency just greater than $33$ is $37$.
Hence, the median class is $10-15$
$\therefore$ Lower limit of median class $= 10$
Class having maximum frequency is the modal class.
Here, maximum frequency $= 20$
$\therefore$ Lower limit of modal class $= 15$
$\therefore$ Required sum $= 10 + 15 = 25$ View full question & answer→MCQ 841 Mark
The mean of the first $10$ prime numbers is :
- A
$1.29$
- ✓
$12.9$
- C
$12.9$
- D
$11.9$
AnswerCorrect option: B. $12.9$
The first $10$ prime numbers are $\{2, 3, 5, 7, 11, 13, 17, 19, 23, 29\}$
$\therefore\text{Mean}=\frac{\text{sum of first 10 prime numbers}}{10}$
$=\frac{ 2, 3, 5, 7, 11, 13, 17, 19, 23, 29}{10}$
$=\frac{129}{10}$
$=12.9$
View full question & answer→MCQ 851 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 861 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 $n = 7$
$\therefore\text{Median}=\frac{7+1}{2}\text{ term}=4\text{ term}$
$=3$
$\therefore\text{q}=3$
$\therefore\text{p}+\text{q}=4+3=7$
View full question & answer→MCQ 871 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 :
AnswerCorrect option: C. mid $-$ points of classes.
We know that, $\mathrm{d}_{\mathrm{i}}=\mathrm{x}_{\mathrm{i}}-\mathrm{a}$
i.e, $d_i$ 's are the deviation from a mid $-$ points of the classes.
View full question & answer→MCQ 881 Mark
The wickets taken by a bowler in $10$ cricket matches are $\{2, 6, 4, 5, 0, 3, 1, 3, 2, 3\}$. The mode of the data is :
AnswerIn the given data, the frequency of $3$ is more than those other wickets taken by a bowler.
Therefore, Mode of given data is $3$.
View full question & answer→MCQ 891 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 $= 60$
Median $=\frac{60}{3}=20$
View full question & answer→MCQ 901 Mark
A student got marks in $5$ subjects in a monthly test is given below : $\{2, 3, 4, 5, 6\}$. In these obtained marks $,4$ is the :
AnswerMarks obtained $= \{2, 3, 4, 5, 6\}$
Median is the middle term $= 4.$
Mean $=\frac{2+3+4+5+6}{5}=4.$
View full question & answer→MCQ 911 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 921 Mark
Choose the correct answer from the given four options : Consider the data :
| 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 : AnswerHere,
| 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 $125.$
Required difference $=$ Upper limit of median class $-$ Lower limit of modal class $= 145 - 125 = 20$ View full question & answer→MCQ 931 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 941 Mark
Consider the following frequency distribution :
| Class |
$0-10$ |
$10-20$ |
$20-30$ |
$30-40$ |
$40-50$ |
$50-60$ |
| Frequency |
$3$ |
$9$ |
$15$ |
$30$ |
$18$ |
$5$ |
The modal class is : - A
$10\ - 20$
- B
$20\ - 30$
- ✓
$30\ - 40$
- D
$50\ - 60$
AnswerCorrect option: C. $30\ - 40$
Class having maxiroom frequency is the modal class.
Here, maximum frequency $= 30$
Hence. the modal class is $30\ - 40.$
View full question & answer→MCQ 951 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 961 Mark
For the following distribution :
|
Mark
|
$60-70$ |
$70-80$ |
$80--90$ |
$90-100$ |
$100-110$ |
|
Frequency
|
$13$ |
$10$ |
$15$ |
$8$ |
$11$ |
The upper limit of the median class is : Answer
|
Mark
|
$60-70$ |
$70-80$ |
$80-90$ |
$90-100$ |
$100-110$ |
|
Frequency
|
$13$ |
$10$ |
$15$ |
$8$ |
$11$ |
|
Cumulative
|
$13$ |
$23$ |
$38$ |
$46$ |
$57$ |
Here $N = 57$
$\Rightarrow\frac{\text{N}}{2}=28.5$
$\therefore$ Median class is $80\ - 90$
The upper limit of the median class is $90.$ View full question & answer→MCQ 971 Mark
If $x_i$ s are the midpoints of the class intervals of a grouped data, $f_i$ s are the corresponding frequency and $\bar{\text{x}}$ is the mean, then $\sum\text{f}_{\text{i}}(\text{x}_\text{i}-\bar{\text{x}})=?$
AnswerFor a grouped data
$\sum\text{f}_\text{i}(\text{x}_\text{i}-\bar{\text{x}})=0$
View full question & answer→MCQ 981 Mark
The mean of the first $10$ composite numbers is :
- A
$12.2$
- B
$11.4$
- ✓
$11.2$
- D
$112$
AnswerCorrect option: C. $11.2$
The first $10$ composite numbers are $\{4, 6, 8, 9, 10, 12, 14, 15, 16, 18\}$
$\therefore\text{Mean}=\frac{\text{sum of first $10$ composite numbers}}{10}$
$=\frac{ 4, 6, 8, 9, 10, 12, 14, 15, 16, 18}{10}$
$=\frac{112}{10}$
$=11.2$
View full question & answer→MCQ 991 Mark
and the mean of the distribution is $3,$ then the value of $p$ is :
Answer$\text{Mean}=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}$
$\Rightarrow3=\frac{\text{3p+36}}{15}$
$\Rightarrow\text{3p}+36=45$
$\Rightarrow\text{3p}-45=36$
$\Rightarrow\text{p}=\frac{9}{3}$
$\Rightarrow\text{p}=3$
View full question & answer→MCQ 1001 Mark
For the following distribution $38$
| Below |
$10$ |
$20$ |
$30$ |
$40$ |
$50$ |
$60$ |
| Frequency |
$3$ |
$12$ |
$27$ |
$57$ |
$75$ |
$80$ |
The modal class is : - A
$50\ – 60$
- B
$20\ – 30$
- C
$40\ – 50$
- ✓
$30\ – 40$
AnswerCorrect option: D. $30\ – 40$
According to the question
|
Classes
|
$0-10$ |
$10-20$ |
$20-30$ |
$30-40$ |
$40-50$ |
$50-60$ |
|
Frequency
|
$3$ |
$9$ |
$15$ |
$30$ |
$18$ |
$5$ |
Here Maximum frequency is $30.$
Therefore, the modal class is $30 \ – 40.$ View full question & answer→