MCQ 11 Mark
The mean and median of a distribution are $14$ and $15$ respectively. The value of mode is :
AnswerGiven mean $= 14$
Median $= 15$
Mode $= ?$
Empharical formula:
$\text{3Median = Mode + 2mean}$
$\text{Mode = 3Median $-$ 2mean}$
Mode $= 3 \times 15 - 2 \times 14$
$= 45 - 28$
$= 17$
$\therefore$ Value of mode $= 17$
View full question & answer→MCQ 21 Mark
The marks of $20$ students in a test were as follows : $\{5, 6, 8, 9, 10, 11, 11, 12, 13, 13, 14, 14, 15, 15, 15, 16, 16, 18, 19, 20\}$ The mode is :
AnswerTerms are : $\{5, 6, 8, 9, 10, 11, 11, 12, 13, 13, 14, 14, 15, 15, 15, 16, 16, 18, 19, 20.\}$
$15$ occurs most number of times $(3$ times$).$
Hence $,15$ is the mode of the marks.
View full question & answer→MCQ 31 Mark
Median $=?$
- ✓
$\text{l+}\begin{Bmatrix}\text{h}\times\frac{\Big(\frac{\text{N}}{2}-\text{cf}\Big)}{\text{f}}\end{Bmatrix}$
- B
$\text{l+}\begin{Bmatrix}\text{h}\times\frac{\Big(\text{cf}-\frac{\text{N}}{2}\Big)}{\text{f}}\end{Bmatrix}$
- C
$\text{l}-\begin{Bmatrix}\text{h}\times\frac{\Big(\frac{\text{N}}{2}-\text{cf}\Big)}{\text{f}}\end{Bmatrix}$
- D
AnswerCorrect option: A. $\text{l+}\begin{Bmatrix}\text{h}\times\frac{\Big(\frac{\text{N}}{2}-\text{cf}\Big)}{\text{f}}\end{Bmatrix}$
$\text{l+}\begin{Bmatrix}\text{h}\times\frac{\Big(\frac{\text{N}}{2}-\text{cf}\Big)}{\text{f}}\end{Bmatrix}$
View full question & answer→MCQ 41 Mark
In formula $\overline{\text{x}}=\text{a+h}\frac{\sum\text{f}_i\text{u}_i}{\sum\text{f}_i}$ for finding the mean,$\text{u}_\text{i}=$
- ✓
$\frac{\text{x}_\text{i}\text-{a}}{\text{h}}$
- B
$\frac{\text{a}-\text{x}_\text{i}}{\text{h}}$
- C
$\text{h}(\text{x}_\text{i}-\text{a})$
- D
$\frac{\text{x}_\text{i}+\text{a}}{\text{h}}$
AnswerCorrect option: A. $\frac{\text{x}_\text{i}\text-{a}}{\text{h}}$
In formula $\overline{\text{x}}=\text{a+h}\frac{\sum\text{f}_i\text{u}_i}{\sum\text{f}_i}$ for finding the mean,
$\text{u}_\text{i}=\frac{\text{x}_\text{i}-\text{a}}{\text{h}}.$
$a =$ assumed mean values
$\text{x}_\text{i}-\text{a}=$ deviation of each mid $-$ value from the assumed mean value
$h =$ class size
View full question & answer→MCQ 51 Mark
The relation between mean, mode and median is :
- A
Mode $= (3 \times$ mean$) − (2 \times$ median$)$
- ✓
Mode $= (3 \times$ median$) − (2 \times$ mean$)$
- C
Mode $= (3 \times$ mean$) − (2 \times$ mode$)$
- D
Mode $= (3 \times$ median$) − (2 \times $ mode$)$
AnswerCorrect option: B. Mode $= (3 \times$ median$) − (2 \times$ mean$)$
Mode $= (3 \times$ median$) − (2 \times$ mean$)$
View full question & answer→MCQ 61 Mark
AnswerMode is the most frequency value of observation or a class,
View full question & answer→MCQ 71 Mark
Choose the correct answer from the given four options: Consider the following distribution:
|
Marks obtained
|
Number of students
|
|
More than or equal to 0
|
63
|
|
More than or equal to 10
|
58
|
|
More than or equal to 20
|
55
|
|
More than or equal to 30
|
51
|
|
More than or equal to 40
|
48
|
|
More than or equal to 50
|
42
|
the frequency of the class 30-40 is:
Answera. 3
Solution:
|
Marks obtained
|
Number of students
|
|
0-10
|
(63 - 58) = 5
|
|
10-20
|
(58 - 55) = 3
|
|
20-30
|
(55 - 51) = 4
|
|
30-40
|
(51 - 48) = 3
|
|
40-50
|
(48 - 42) = 6
|
|
50...
|
42 = 42
|
Hence, frequency in the class interval 30-40 is 3
View full question & answer→MCQ 81 Mark
Each question consists of two statements, namely, Assertion $(A)$ and Reason $(R)$. For selecting the correct answer : use the following code:
Assertion $(A)$
Consider the following frequency distribution:
|
Class interval
|
$3-6$ |
$6-9$ |
$9-12$ |
$12-15$ |
$15-18$ |
$18-21$ |
|
Frequency
|
$2$ |
$5$ |
$21$ |
$23$ |
$10$ |
$12$ |
The mode of the above data is $12.4.$
Reason $(R)$
The value of the variable which occurs most often is the mode. - A
Both Assertion $(A)$ and Reason $(R)$ are true and Reason $(R)$ is a correct explanation of Assertion $(A)$.
- ✓
Both Assertion $(A)$ and Reason $(R)$ are true but Reason $(R)$ is a not a correct explanation of Assertion $(A).$
- C
Assertion $(A)$ is true and Reason $(R)$ is false.
- D
Assertion $(A)$ is false and Reason $(R)$ is true
AnswerCorrect option: B. Both Assertion $(A)$ and Reason $(R)$ are true but Reason $(R)$ is a not a correct explanation of Assertion $(A).$
Reason $(R)$ is true.
Maximum frequency $= 23$
Hence, modakl class is $12-15$
Now, mode $=\text{x}_\text{k}+\text{h}\Big\{\frac{(\text{f}_\text{k}-\text{f}_{\text{k}-1})}{(2\text{f}_\text{k}-\text{f}_{\text{k}-1}-\text{f}_{\text{k}+1})}\Big\}$
$=12+3\Big\{\frac{(23-21)}{(2(23)-21-10)}\Big\}$
$=12+3\times\frac{2}{15}$
$=12+0.4$
$=12.4$
Thus, Assertion $(A)$ and Reason $(R)$ are both true but Reason $(R)$ rs nor !he eereee explananoon of Assertion $(A).$
View full question & answer→MCQ 91 Mark
The marks obtained by $9$ students in Mathematics are $\{59, 46, 31, 23, 27, 40, 52, 35\}$ and $29$. The mean of the data is:
AnswerGiven : $\{59, 46, 31, 23, 27, 40, 52, 35\}$ and $29$
$\text{Mean}=\frac{\text{Sum of all observetion}}{\text{Number of observation}}$
$=\frac{342}{9}$
$=38$
View full question & answer→MCQ 101 Mark
Each question consists of two statements, namely, Assertion $(A)$ and Reason $(R)$. For selecting the correct answer : use the following code :
| Assertion $(A)$ |
Reason $(R)$ |
| If the median and mode of a frequency distribution are $150$ and $154$ respectively, then its mean is $148$. |
Mean, median and mode of a frequency distribution are related as : mode $= 3$ median $- 2$ mean |
- ✓
Both Assertion $(A) $ and Reason $(R)$ are true and Reason $(R)$ is a correct explanation of Assertion $(A).$
- B
Both Assertion $(A)$ and Reason $(R)$ are true but Reason $(R)$ is a not a correct explanation of Assertion $(A).$
- C
Assertion $(A)$ is true and Reason $(R)$ is false.
- D
Assertion $(A)$ is false and Reason $(R)$ is true.
AnswerCorrect option: A. Both Assertion $(A) $ and Reason $(R)$ are true and Reason $(R)$ is a correct explanation of Assertion $(A).$
Reason $(R)$ is true.
Using the relation given in $(R),$ we have
Median $= 150$
Mode $= 154$
Mode $= 3$ Median $- 2$ Mean
Hence, mean $=\frac{3\text{Median}-\text{Mode}}{2}$
$=\frac{3(150)-154}{2}=\frac{450-154}{2}$
$=\frac{296}{2}=148$
Thus, Assertion $(A)$ and Reason $(R)$ are true and Reason $(R)$ is a correct explanation of Assertion $(A).$
View full question & answer→MCQ 111 Mark
In formula $\bar{\text{x}}=\text{a}+\text{h}\big(\frac{\sum\text{f}_1\text{u}_\text{i}}{\sum\text{f}_\text{i}}\big)$,$\text{h}^,$ stands for:
Answera. class size
Solution:
In formula $\bar{\text{x}}=\text{a}+\text{h}\big(\frac{\sum\text{f}_1\text{u}_\text{i}}{\sum\text{f}_\text{i}}\big)\text{,h}^,$ stands for class size
$\text{Class size }=\frac{\text{Upper limit+Lower limit}}{2}$
View full question & answer→MCQ 121 Mark
Median divides the total frequency into $........... $ equal parts :
AnswerThe median of the data series is the middle term or the mean of the two middle terms.
Hence, it divides the data series or the frequency of terms into two equal halves.
View full question & answer→MCQ 131 Mark
In the given data if $n = 44, l = 400, cf = 8, h = 100, f = 20,$ then its median is :
Answer$=\text{l}+\frac{\frac{\text{n}}{2}-\text{c}}{\text{f}}\times \text{h}$
$=400+\frac{\frac{44}{2}-8}{20}\times100$
$=400+\frac{14}{20}\times100$
$=400+14\times5$
$=400+70$
$=470$
View full question & answer→MCQ 141 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 $= 145 – 125 = 20 \ (C)$ View full question & answer→MCQ 151 Mark
Choose the correct answer from the given four options : In the formula $\bar{\text{x}}=\text{a}+\frac{\sum\text{f}_\text{i}\text{d}_\text{i}}{\sum\text{f}_\text{i}},$ for finding the mean of grouped data $\mathrm{d}_{\mathrm{i}}$ 's are deviations from a of :
- A
Lower limits of the classes.
- B
Upper limits of the classes.
- ✓
Mid points of the classes.
- D
Frequencies of the class marks.
AnswerCorrect option: C. Mid points of the classes.
We know that, $d_i=x_i-a$
i.e., $\mathrm{d}_{\mathrm{i}}$ 's are the deviation from a of mid $-$ points of the classes.
View full question & answer→MCQ 161 Mark
Which of the following measures of central tendency is influenced by extreme values?
AnswerSince mean is the average of all observations, it is influenced by extreme values.
View full question & answer→MCQ 171 Mark
Value of the middle $-$ most observation $(s)$ is called :
AnswerTo find the Median, place the numbers in value order and find the middle number.
If there are two middle numbers, take the mean of the two numbers and this will be the median of the data set.
The middle most observation of a data series is called the median of the series.
View full question & answer→MCQ 181 Mark
The mode of $\{, 5, 6, 8, 5, 4, 8, 5, 6, x, 8\}$ is $8$. The value of $x$ is :
AnswerHere Observations $5$ and $8$ have the same frequency, i.e., $3$.
For $8$ be the mode of the data, it should have the maximum frequency.
$\therefore 8$ should repeat itself at least one more.
$\Rightarrow x$ should be $8.$
View full question & answer→MCQ 191 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 201 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 - A
$16.5$
- B
$18.5$
- C
$18$
- ✓
$17.5$
AnswerCorrect option: D. $17.5$
The given series is in indusive from.
comverting in to exclusive from and preparing cumulative frequency tabla, we get
| 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, $\text{N}=57$
$\Rightarrow\frac{\text{N}}{2}=28.5$
The cumulative frequency just greater than $28.5$ is $38$.
Hence, the median class is $11.5-17.5$
$\therefore$ Upper limit of the median class $= 17.5$ View full question & answer→MCQ 211 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 221 Mark
The mean of $'n\ ’$ observations is $\bar{\text{x}}$ If the first item is increased by $1,$ second by $2$ and so on, then the new mean is :
- A
$\bar{\text{x}}=\frac{\text{n}-1}{2}$
- B
$\bar{\text{x}}$
- C
$\bar{\text{x}}-\frac{\text{n}+1}{2}$
- ✓
$\bar{\text{x}}+\frac{\text{n}+1}{2}$
AnswerCorrect option: D. $\bar{\text{x}}+\frac{\text{n}+1}{2}$
Let terms be $\text{x}_1,\text{x}_2,\text{x}_3...\text{x}_\text{n}.$
$\therefore\text{Mean }\bar{\text{x}} $
$\Rightarrow\frac{\text{x}_1+\text{x}_2+\text{x}_3+...\text{x}_\text{n}}{\text{n}}$
$\Rightarrow\text{x}_1+\text{x}_2+\text{x}_3+...+\text{x}_\text{n}=\text{n.}\bar{\text{x}}$
New observation,are $\text{x}_1+1\text{x}_2+2\text{,x}_3+3,...\text{x}_\text{n}+\text{n.}$
$\therefore\text{New Mean }=\frac{\text{x}_1+1\text{x}_2+2\text{,x}_3+3,...\text{x}_\text{n}+\text{n.}}{\text{n}}$
$\frac{\text{n}.\bar{\text{x}}+\frac{\text{n(n + 1)}}{2}}{\text{n}}$
$=\bar{\text{x}}+\frac{\text{n+1}}{2}$
View full question & answer→MCQ 231 Mark
For the following distribution :
|
Class
|
$60-70$ |
$70-80$ |
$80-90$ |
$90-100$ |
$100-110$ |
|
Frequency
|
$13$ |
$10$ |
$15$ |
$8$ |
$11$ |
The lower limit of the modal class is : AnswerIn the given data, Maximum frequency is $15$.
Therefore, the modal class is $80 - 90.$
The lower limit of the modal class is $80.$
View full question & answer→MCQ 241 Mark
Mode of a data is given by :
- A
$\text{l}-\Big(\frac{\text{f}_1-\text{f}_0}{2\text{ f}_1-\text{f}_0-\text{f}_2}\Big)\times\text{h}$
- ✓
$\text{l}+\Big(\frac{\text{f}_1-\text{f}_0}{2\text{ f}_1-\text{f}_0-\text{f}_2}\Big)\times\text{h}$
- C
$\text{l}-\Big(\frac{\text{f}_0-\text{f}_1}{2\text{ f}_1-\text{f}_0-\text{f}_2}\Big)\times\text{h}$
- D
$\text{h}+\Big(\frac{\text{f}_1-\text{f}_0}{2\text{ f}_1-\text{f}_0-\text{f}_2}\Big)\times\text{l}$
AnswerCorrect option: B. $\text{l}+\Big(\frac{\text{f}_1-\text{f}_0}{2\text{ f}_1-\text{f}_0-\text{f}_2}\Big)\times\text{h}$
Mode of data is given by
$\text{l}+\Big(\frac{\text{f}_1-\text{f}_0}{2\text{ f}_1-\text{f}_0-\text{f}_2}\Big)\times\text{h}$
Where $l =$ lower limit of the modal class
$f_1 =$ frequency of the modal class
$f_0 =$ frequency of the class preceding the modal class
$f_2 =$ frequency of the class succeeding the modal class
$h =$ size of the class interval $($assuming all class sizes to be equal$)$
View full question & answer→MCQ 251 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 261 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 271 Mark
In the formula $\bar{\text{x}}=\text{a}+\frac{\sum\text{f}_\text{i}\text{d}_\text{i}}{\sum\text{f}_\text{i}},\text{d}_\text{i}$ represents :
- A
$\text{a}+\text{x}_\text{i}$
- ✓
$\text{d}_\text{i}=\text{x}_\text{i}-\text{a}$
- C
$\text{a}-\text{x}_\text{i}$
- D
$\text{x}_1+\text{a}$
AnswerCorrect option: B. $\text{d}_\text{i}=\text{x}_\text{i}-\text{a}$
represents $\text{x}_\text{i}-\text{a}.\text{ d}_\text{i}$ is the deviation from each mid value $''\text{x}''_\text{i}$ by an assumed mean value $''\text{a}''_\text{i}$.
View full question & answer→MCQ 281 Mark
If $\sum\text{f}_\text{i}\text{x}_\text{i}=625$ and $ \sum{\text{f}_\text{i}}=25$ then the value of $\bar{\text{x}}$ is :
Answer$\sum\text{f}_\text{i}\text{x}_\text{i}=625$ and $ \sum{\text{f}_\text{i}}=25$
View full question & answer→MCQ 291 Mark
If $\sum\text{f}_\text{i}\text{u}_\text{i}=29,\sum\text{f}_\text{i}=30,\text{a}=47.5$ and $ \text{h}=15, $ then the value of $\bar{\text{x}}$ is :
Answer$\bar{\text{x}}=\text{a}+\frac{\sum\text{f}_\text{i}\text{u}_\text{i}}{\sum\text{f}_\text{i}}\times\text{h}$
$=47.5+\frac{29}{30}\times15$
$=47.5+\frac{29}{2}$
$=47.5+14.5$
$=62$
View full question & answer→MCQ 301 Mark
If the 'less than type' ogive and 'more than type' ogive intersect each other at $(20.5, 15.5)$ then the median of the given data is :
- A
$5.5$
- B
$15.5$
- ✓
$20.5$
- D
$36.0$
AnswerCorrect option: C. $20.5$
Since the abscissa of the point of intersection of both ihe ogives gives the median, we have median : $20.5$
View full question & answer→MCQ 311 Mark
Answerb. 3 Median - 2 Mean
Solution:
Mode can be found by using the formula:
Mode = 3 Median - 2 Mean
View full question & answer→MCQ 321 Mark
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 :
AnswerMedian is given by the abscissa of the point of intersection of the Less than Type and More than Type cumulative frequency curves.
View full question & answer→MCQ 331 Mark
The marks obtained by $9$ students in Mathematics are $\{59, 46, 31, 23, 27, 44, 52, 40\}$ and $29$. The mean of the data is :
Answer$\text{Mean}=\frac{\text{Sum of all observation}}{\text{Number of observation}}$
$=\frac{59+46+31+23+27+44+52}{9}$
$=\frac{351}{9}$
$=39$
View full question & answer→MCQ 341 Mark
The mean of a discrete frequency distribution $x_i f_i ; ~i=1,2, ……, n$ 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 351 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 361 Mark
Choose the correct answer from the given four options: In the formula $\bar{\text{x}}=\text{a}+\text{h}\Big(\frac{\sum\text{f}_\text{x}\text{u}_\text{i}}{\sum\text{f}_\text{i}}\Big),$ for finding the mean of grouped frequency distribution, $\mathrm{u}_{\mathrm{i}}=$
- 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, $\bar{\text{x}}=\text{a}+\text{h}\Big(\frac{\sum\text{f}_\text{i}\text{u}_\text{i}}{\sum\text{f}_\text{i}}\Big)$
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 371 Mark
Which of the following is not a measure of central tendency?
AnswerRange is not a measure of central tendency.
View full question & answer→MCQ 381 Mark
Mode $=?$
- A
$\text{x}_\text{k}+\text{h}.\Big\{\frac{(\text{f}_{\text{k}-1}-\text{f}_\text{k})}{(2\text{f}_\text{k}-\text{f}_{\text{k}-1}-\text{f}_{\text{k}+1})}\Big\}$
- ✓
$\text{x}_\text{k}+\text{h}.\Big\{\frac{(\text{f}_\text{k}-\text{f}_{\text{k}-1})}{(2\text{f}_\text{k}-\text{f}_{\text{k}-1}-\text{f}_{\text{k}+1})}\Big\}$
- C
$\text{x}_\text{k}+\text{h}.\Big\{\frac{(\text{f}_\text{k}-\text{f}_{\text{k}-1})}{(\text{f}_\text{k}-2\text{f}_{\text{k}-1}-\text{f}_{\text{k}-1})}\Big\}$
- D
$\text{x}_\text{k}+\text{h}.\Big\{\frac{(\text{f}_\text{k}-\text{f}_{\text{k}-1})}{(\text{f}_{\text{k}}-\text{f}_{\text{k}-1}-2\text{f}_{\text{k}+1})}\Big\}$
AnswerCorrect option: B. $\text{x}_\text{k}+\text{h}.\Big\{\frac{(\text{f}_\text{k}-\text{f}_{\text{k}-1})}{(2\text{f}_\text{k}-\text{f}_{\text{k}-1}-\text{f}_{\text{k}+1})}\Big\}$
$\text{x}_\text{k}+\text{h}.\Big\{\frac{(\text{f}_\text{k}-\text{f}_{\text{k}-1})}{(2\text{f}_\text{k}-\text{f}_{\text{k}-1}-\text{f}_{\text{k}+1})}\Big\}$
View full question & answer→MCQ 391 Mark
The most frequent value in the data is known as :
AnswerThe most frequent value in the data is known as the Mode.
e.g let us consider the following data set : $\{3, 5, 7, 5, 9, 5, 8, 4\} $ the mode is $5$ since it occurs most often in data set.
View full question & answer→MCQ 401 Mark
Choose the correct answer from the given four options :
The times, in seconds, taken by $150$ atheletes to run a $110 m$ hurdle race are tabulated below :
|
Class
|
$13.8-14$ |
$14-14.2$ |
$14.2-14.4$ |
$14.4-14.6$ |
$14.6-14.8$ |
$14.8-15$ |
|
Frequency
|
$2$ |
$4$ |
$5$ |
$71$ |
$48$ |
$20$ |
The number of atheletes who completed the race in less then $14.6$ seconds is : AnswerThe number of atheletes who completed the race in less than $14.6$
$= 2 + 4 + 5 + 71 = 82$
View full question & answer→MCQ 411 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 $10 + 15 = 25. (b)$ View full question & answer→MCQ 421 Mark
A child says that the median of $\{3, 14, 18, 20, 5\} $ is $18$. What concept does the child missed about finding the median?
AnswerTo calculate the median of any data series.
The data series has to be arranged in the ascending order.
The child hasn't arranged the data series in ascending order.
View full question & answer→MCQ 431 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 441 Mark
To represent the less than type graphically, we plot the $............$ on the $x-$ axis:
AnswerTo represent ‘the less than type’ graphically, we plot the upper limits on the $x-$ axis.
e.g marks obtained by students are represented in grouped data as $\{(0 - 10), (10 - 20), (20 - 30), (30 - 40) ...\}$
only upper limits such as $\{10, 20, 30, 40...\}$ are taken for the $x-$ axis.
View full question & answer→MCQ 451 Mark
The median of first $8$ prime numbers is :
AnswerFirst $8$ prime numbers are are follows :
$2, 3, 5, 7, 11, 13, 17, 19$
$N = 8\ ($even$)$
$\therefore$ Median $=\frac{\Big(\frac{8}{2}\Big)^{\text{th}}\text{value}+\Big(\frac{8}{2}+1\Big)^{\text{th}}\text{value}}{2}$
$=\frac{4^{\text{th}}\text{value }+\ 5^{\text{th}}\text{Value}}{2}$
$=\frac{7+11}{2}$
$=\frac{18}{2}$
$=9$
View full question & answer→MCQ 461 Mark
If the less than ogive and the more than ogive intersect at $(32, 48),$ then the median of the data is :
AnswerIf the less than ogive and the more than ogive intersect at $(32, 48),$ then the median of the data is $32$.
Because on the graph, the point of the $x-$ axis, where less than ogive and more than ogive intersects, is the median.
Therefore, the Median of the data is $32.$
View full question & answer→MCQ 471 Mark
The mean of $2, 7, 6$ and $x$ is $15$ and the mean of $18, 1, 6, x$ and $y$ is $10$. What is the value of $y?$
AnswerMean of $2, 7, 6$ and $x = 5$
$\Rightarrow\frac{2+7+6+\text{x}}{4}=5$
$\Rightarrow15+\text{x}=20$
$\Rightarrow\text{x}=5$
Mean of $18, 1, 6, x$ and $y = 10$
$\Rightarrow\frac{18+1+6+\text{x}+\text{y}}{5}=10$
$\Rightarrow\frac{18+1+6+\text{5}+\text{y}}{5}=10$
$\Rightarrow30+\text{y}=50$
$\Rightarrow\text{y}=20$
Note : Question modified
View full question & answer→MCQ 481 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 491 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 501 Mark
If $\text{x}_\text{i}'\text{s}$ are the midpoints of the class intervals of grouped data, $\text{f}_\text{i}'\text{s}$ are the corresponding frequencies and is the mean, then$\sum(\text{f}_\text{i}\text{x}_\text{i}-\bar{\text{x}})$ is equal to $ -$
AnswerIf $\text{x}_i'\text{s}$ are the midpoints of the class intervals of grouped data, $\text{f}_i'\text{s}$ are the corresponding frequencies and is the mean,
then $\sum(\text{f}_\text{i}\text{x}_\text{i}-\bar{\text{x}})$is equal to $0$.
i.e the difference between the sum of product of frequencies and mid values of corresponding class intervals of the grouped data and the sum of their mean value is equal to zero.
View full question & answer→