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 (C) 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
⇒ Mode – median = 2 median – 2 mean
⇒ 24 = 2 (median – mean)
⇒ 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 $x_i f_i ; i=1,2, \ldots . . ., n$ 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)
⇒ 12 = 3 (Median − Mean)
⇒ Median − Mean = 4 .....(1)
Again,
Mode = 3 Median − 2 Mean
⇒ 2 Mode = 6 Median − 4 Mean
⇒ Mode − Mean + Mode = 6 Median − 5 Mean
⇒ 12 + (Mode − Median) = 5 (Median − Mean)
⇒ 12 + (Mode − Median) = 20 [Using (1)]
⇒ 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. (b) View full question & answer→MCQ 101 Mark
If the mean of 6, 7, x, 8, y, 14 is 9, then:
AnswerMean 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:
Answer
|
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
- C
- ✓
AnswerHence, the correct option is (d).
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
- ✓
mid$-$points of classes.
- D
frequency of the class marks .
AnswerCorrect option: C. mid$-$points of classes.
We 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
⇒ 12 = 3 median – 2 × 24
⇒ 12 = 3 median - 48
⇒ 12 + 48 = 3 median
⇒ 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 $x _1, x _2, x _3, \ldots . . . ., x _{ n }$ 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 $x_1+1, x_2+2, x_3+3, \ldots . x_n+n$
$\text{New mean}=\frac{(\text{x}_1+1)+(\text{x}_2+2)+(\text{x}_3+3)+.....+(\text{x}_\text{n}+\text{n})}{\text{n}}$
$=\frac{\text{x}_1+\text{x}_2+\text{x}_3+.......+\text{x}_\text{n}+(1+2+3+...+\text{n})}{\text{n}}$
$=\frac{\text{n}\overline{\text{X}}+\frac{\text{n}(\text{n+1})}{2}}{\text{n}}$
$\overline{\text{X}}+\frac{\text{n}+1}{2}$
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 ≠ 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 $y _1, y _2, y _3, \ldots \ldots, y _{ k }$ 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:
AnswerGiven, 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 $\text{u}_\text{i}=\frac{\text{x}_\text{i}-25}{10}\sum\text{f}_\text{i}\text{u}_\text{i}=20,\sum\text{f}_\text{i}=100,$ 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 371 Mark
Which of the following cannot be determined graphically?
AnswerMean cannot be determind grafically,
View full question & answer→MCQ 381 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 $x _1, x _2, x _3, \ldots \ldots . x _{ n }$ are observations and $\overline{ X }$ 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 391 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 401 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 411 Mark
If the mean of observations $x_1, x_2, \ldots, x_n$ is $\bar{x}$, then the mean of $x_1+a, x_2+a_1, \ldots, x_n+a$ 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 $x_1, x_2, \ldots \ldots, x_n$ is $\bar{x}$
$\frac{x_1+x_2+x_3 \ldots \ldots .+x_n}{n}=\bar{x}$
$x_1+a+x_2+a+x_3+a+\ldots \ldots x_n+a$
$=x_1+x_2+x_3+\ldots x_n+n a$
$\therefore \text { Mean of }\left(x_1+x_2+x_3 \ldots . .+x_n\right)+n a$
$=\bar{x}+\frac{n a}{n}$
$=\bar{x}+a$
View full question & answer→MCQ 421 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 431 Mark
There are 16 observations arranged in increasing order of their values in a data. The median will be the value of:
- A
- B
- ✓
average of 8th and 9th observation
- D
average of 7th and 8th observation
AnswerCorrect option: C. average of 8th and 9th observation
View full question & answer→MCQ 441 Mark
The value of x for which the class mark of the class interval 30-x is 36 is
MCQ 451 Mark
The mean of five numbers is 15. If we include one more number, the mean of six numbers is 17. The included number is
MCQ 461 Mark
If the mean of first in natural numbers is $\frac{5 n}{9}$, then the value of n is
View full question & answer→MCQ 471 Mark
After an examination, a teacher wants to know the marks obtained by maximum number of the students in her class. She requires to calculate __________ of marks.
MCQ 481 Mark
If value of each observation in a data is increased by 2, then the median of the new data
MCQ 491 Mark
The middle most observation of every ery data arranged in order is called
MCQ 501 Mark
If every term of the statistical data consisting of n terms is decreased by 2, then the mean of the data:
MCQ 511 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
MCQ 521 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
MCQ 531 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
MCQ 541 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
MCQ 551 Mark
In the formula $\bar{X}=a+h\left(\frac{1}{N} \sum f_i u_i\right)$,for finding the mean of grouped frequency distribution $u_i=$
- A
$\frac{x_i+a}{h}$
- B
$h\left(x_i-a\right)$
- ✓
$\frac{x_i-a}{h}$
- D
$\frac{a-x_i}{h}$
AnswerCorrect option: C. $\frac{x_i-a}{h}$
View full question & answer→MCQ 561 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
View full question & answer→MCQ 571 Mark
If 35 is removed from the data: 30, 34, 35, 36, 37, 38, 39, 40, then the median increases by
MCQ 581 Mark
If $u_i=\frac{x_i-25}{10}, \Sigma f_i u_i=20, \Sigma f_i=100$, then $\bar{x}=$
View full question & answer→MCQ 591 Mark
Mean of a certain number of observations is $\bar{x}$.If each observation is divided by $m(m \neq 0)$ and increased by in, then the mean of new observation is
- ✓
$\frac{\bar{x}}{m}+n$
- B
$\frac{\bar{x}}{n}+m$
- C
$\bar{x}+\frac{n}{m}$
- D
$\bar{x}+\frac{m}{n}$
AnswerCorrect option: A. $\frac{\bar{x}}{m}+n$
View full question & answer→MCQ 601 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
$a \bar{x}$
- B
$\bar{x}-a$
- ✓
$\bar{x}+a$
- D
$\frac{\bar{x}}{a}$
AnswerCorrect option: C. $\bar{x}+a$
View full question & answer→MCQ 611 Mark
If the mean of first n natural number is 15, then n =
MCQ 621 Mark
If mode of a series exceeds its mean by 12, then mode exceeds the median by
MCQ 631 Mark
If the arithmetic mean of 7, 8, x, 11, 14 is x, then x =
MCQ 641 Mark
If the difference of mode and median of a data is 24, then the difference of median and mean is
MCQ 651 Mark
The mean of first n odd natural numbers is $\frac{n^2}{81}$, then n =
View full question & answer→MCQ 661 Mark
The mean of first n odd natural number is
- A
$\frac{n+1}{2}$
- B
$\frac{n}{2}$
- ✓
- D
$n^2$
View full question & answer→MCQ 671 Mark
The arithmetic mean and mode of a data are 24 and 12 respectively, then its median is
MCQ 681 Mark
If the mean of first n natural numbers is $\frac{5 n}{9}$, then $n=$
View full question & answer→MCQ 691 Mark
The mean of n observations is $\bar{x}$ . 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
$\bar{x}+(2 n+1)$
- ✓
$\bar{x}+\frac{n+1}{2}$
- C
$\bar{x}+(n+1)$
- D
$\bar{x}-\frac{n+1}{2}$
AnswerCorrect option: B. $\bar{x}+\frac{n+1}{2}$
View full question & answer→MCQ 701 Mark
If the mean of 6, 7, x, 8, y, 14 is 9, then
MCQ 711 Mark
If the mean of a frequency distribution is 8.1 and $\Sigma f_i x_i=132+5 k, \Sigma f_i=20$, then $k=$
View full question & answer→MCQ 721 Mark
The mean of 1, 3, 4, 5, 7, 4 is m. The numbers 3, 2, 2, 4, 3, 3, p have mean m - 1 and median q. Then, p + q =
MCQ 731 Mark
If the mode of the data: 16, 15, 17, 16, 15, x, 19, 17, 14 is 15, then x =
MCQ 741 Mark
If the mode of the data: 64, 60, 48, x, 43, 48, 43, 34 is 43, then x + 3 =
MCQ 751 Mark
The median of first 10 prime numbers is
MCQ 761 Mark
If the median of the data: 6, 7, x - 2 x, 17, 20, written in ascending order, is 16 Then x =
MCQ 771 Mark
If the median of the data: 24, 25, 26, x + 2 x + 3 30, 31, 34 is 27.5, then x =
MCQ 781 Mark
If the arithmetic mean of x, x + 3 x + 6 x + 9 and x + 12 is 10, the x =
MCQ 791 Mark
The mean of a discrete frequency distribution $x_i / f_i ; i=1,2, \ldots, n$ is given by
- ✓
$\frac{\sum f_i x_i}{\sum f_i}$
- B
$\frac{1}{n} \sum_{i=1}^n f_i x_i$
- C
$\frac{\sum_{i=1}^n f_i x_i}{\sum_{i=1}^n x_i}$
- D
$\frac{\sum_{i=1}^n f_i x_i}{\sum_{i=1}^n i}$
AnswerCorrect option: A. $\frac{\sum f_i x_i}{\sum f_i}$
View full question & answer→MCQ 801 Mark
The relationship between mean, median and mode for a moderately skewed distribution is
MCQ 811 Mark
If the mean of the following distribution is 2.6, then the value of y is
| Variable (x) : | 1 | 2 | 3 | 4 | 5 |
| Frequency : | 4 | 5 | y | 1 | 2 |
MCQ 821 Mark
One of the methods of determining mode is
MCQ 831 Mark
The mean of in 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. $\bar{X}+\frac{n+1}{2}$
View full question & answer→MCQ 841 Mark
View full question & answer→MCQ 851 Mark
The mode of a frequency distribution can be determined graphically from
MCQ 861 Mark
The median of a given frequency distribution is found graphically with the help of
MCQ 871 Mark
Which of the following cannot be determined graphically?
MCQ 881 Mark
For a frequency distribution, mean, median and mode are connected by the relation
MCQ 891 Mark
The arithmetic mean of 1, 2, 3, ..., n is
- ✓
$\frac{n+1}{2}$
- B
$\frac{n-1}{2}$
- C
$\frac{n}{2}$
- D
$\frac{n}{2}+1$
AnswerCorrect option: A. $\frac{n+1}{2}$
View full question & answer→MCQ 901 Mark
The algebraic sum of the deviations of a frequency distribution from its mean is
MCQ 911 Mark
Which of the following is not a measure of central tendency
MCQ 921 Mark
The mean of the data $x+a, x+2 a,(x+3 a), \ldots, x+(2 n+1) a$ is
- ✓
$x+(n+1) a$
- B
$x+(n-1) a$
- C
$x+(n+2) a$
- D
$x+n a$
AnswerCorrect option: A. $x+(n+1) a$
(a)
Let $\bar{X}$ be the required mean. Then,
$\bar{X}=\frac{(x+a)+(x+2 a)+(x+3 a)+\ldots+(x+(2 n+1) a)}{2 n+1}$
$\Rightarrow \quad \bar{X}=\frac{x(2 n+1)+a\{1+2+3+\ldots+(2 n+1)\}}{2 n+1}$
$\Rightarrow \quad \bar{X}=x+\frac{a}{2 n+1}\left\{\frac{2 n+1}{2}(1+2 n+1)\right\}=x+a(n+1)$
View full question & answer→MCQ 931 Mark
While finding the mean of the grouped data by using the formula $\bar{X}=a+\frac{1}{N} \sum f_1 d_i$ $d_i$ 's are the diviations from a of
- ✓
the mid-values of the classes
- B
lower limits of the classes
- C
upper limits of the classes
- D
frequencies of the class marks
AnswerCorrect option: A. the mid-values of the classes
(a)
In the given formula, $d_i=x_i-a$, where $x_i$ are the mid-values of the classes
View full question & answer→MCQ 941 Mark
If the mean and median of a data are 12 and 15 respectively, then its mode is
Answer(b)
We have, Mean = 12 and Median = 15
∴ Mode = 3 Median-2 Mean ⇒ Mode = 3 * 15 - 2 * 12 = 21 .
View full question & answer→MCQ 951 Mark
If the value of cach observation of a statistical data consisting of n observations is increased by 3, then the mean of the data
Answer(b)
Let $x_1, x_2, \ldots, x_n$ be $n$ values of a variable X and Y be a new variable taking values $y_1, y_2, \ldots, y_n$ such that $y_i=a x_1+b ; i=1,2, \ldots, n$. Then $\bar{Y}=a \bar{X}+b$.
Here, a = 1 and b = 3. Therefore, $\bar{Y}=\bar{X}+3$. Hence, the mean is increased by 3.
View full question & answer→MCQ 961 Mark
Which computing the mean of grouped data, it is assumed that the frequencies are
- A
centred at the lower limits of the classes
- B
centred at the upper limits of the classes
- ✓
centred at the class marks of the classes
- D
evenly distributed over all the classes.
AnswerCorrect option: C. centred at the class marks of the classes
(c)
The mean $\bar{X}$ of the grouped data is given by
$\bar{X}=\frac{1}{N} \sum f_1 x_1$ where $x_i^{\prime} s$ are the mid-values of the class-intervals.
Therefore, class frequencies are centred at the class marks of the classes.
View full question & answer→MCQ 971 Mark
The mean and median of the data a, b and c are 50 and 35 respectively, where a < b < c If c - a = 55 then b - a =
Answer(d)
It is given that a < b < c Therefore, median = b. But, it is given that the median is 35.
Therefore, b = 35.
Mean of a, b and c is 50.
$\therefore \quad \frac{a+b+c}{3}=50 \Rightarrow a+b+c=150 \Rightarrow a+35+c=150 \Rightarrow a+c=115$
Thus, we have c + a = 115 and c - a =55 Rightarrow a=30 and c = 85
Hence, b - a = 35 - 30 = 5
View full question & answer→MCQ 981 Mark
3 If the arithmetic mean of 2, 4, 6, 8, 3 and 7 is 5, then the arithmetic mean of 1002, 1004, 1006, 1008, 1003 and 1007 is
Answer(a)
It is given that AM of 2, 4, 6, 8, 3, 7 is 5. Therefore, by increasing each observation by 1000, the AM also increases by 1000. Hence, the AM of 1002, 1004, 1006, 1008, 1003 and 1007 is 1005.
View full question & answer→MCQ 991 Mark
If the arithmetic mean of first n natural numbers is 15, then n is equal to
Answer(c)
It is given that
$15=\frac{1+2+3+\ldots+n}{n} \Rightarrow 15=\frac{n(n+1)}{2 n} \Rightarrow 15=\frac{n+1}{2} \Rightarrow n+1=30 \Rightarrow n=29$
View full question & answer→MCQ 1001 Mark
For some data $x_1, x_2, \ldots, x_n$ with respective frequencies $f_1, f_2, \ldots, f_n$, the value of $\sum_{i=1}^n f_i\left(x_1-\bar{X}\right)$ is equal to
- A
$n \bar{X}$
- B
- C
$\Sigma f_1$
- ✓
$0$
Answer(d)
We have
$\bar{X}=\frac{1}{N} \sum_{i=1}^n f_1 x_1 \Rightarrow \sum_{i=1}^n f_1 x_i=N \bar{X}$
$\sum_{i=1}^n f_1\left(x_i-\bar{X}\right)=\sum_{i=1}^n f_i x_i-\sum_{i=1}^n f_i \bar{X}=N \bar{X}-\bar{X} \sum_{i=1}^n f_i=N \bar{X}-N \bar{X}=0$
View full question & answer→MCQ 1011 Mark
Answer(c)
From the given cumulative frequency distribution, we obtain the following frequency distribution:
| Marks: | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
Number
of students: | 3 | 9 | 15 | 30 | 18 | 5 |
We find that the class with maximum frequency is 30-40. It is the modal class.
View full question & answer→MCQ 1021 Mark
Mode for the following distribution is 22. If x < y < 10 then the value of y is
Class interval: | 5 | 8 | 10 | x | y | 30 |
Frequency (f): | 5 | 8 | 10 | x | y | 30 |
Answer(b)
It is given that x < y < 10. So, 20-30 is the modal class such that $l=20, f=10, f_1=8$ $f_2=x$ and $h=10$
$\therefore \quad$ Mode $=l+\frac{f-f_1}{2 f-f_1-f_2} \times h \Rightarrow 22=20+\frac{10-8}{20-8-x} \times 10 \Rightarrow 2=\frac{2}{12-x} \times 10 \Rightarrow 12-x=10 \Rightarrow x=2$
We have,
$N=30 \Rightarrow 5+8+10+x+y=30 \Rightarrow x+y=7 \Rightarrow y=5 \quad[\because x=2]$
View full question & answer→MCQ 1031 Mark
If the sum of 15 observations of a data is (434 + x) and the mean of the observation is x then x =
Answer(c)
It is given that the sum of 15 observations is (434 + x) and their mean is x.
∴ $x=\frac{434+x}{15} \Rightarrow 15 x=434+x \Rightarrow 14 x=434 \Rightarrow x=31$.
View full question & answer→MCQ 1041 Mark
If the difference of mode and median of a data is 24, then the difference of median and mean is
Answer(a)
We know that
Mode 3 Median - 2 Mean
⇒ Mode - Median = 2 (Median-Mean)
⇒ 24=2 (Median - Mean) [∴ Mode - Median = 24 (given)]
⇒ Median - Mean = 12
View full question & answer→MCQ 1051 Mark
If the mean of the following frequency distribution is 2.6, then the value of p is
| Variate (X): | 1 | 2 | 3 | 4 | 5 |
| Frequency (f): | 4 | 5 | y | 1 | 2 |
Answer(b)
It is given that
Mean = 2.6
$\Rightarrow \frac{1 \times 4+2 \times 5+3 \times y+4 \times 1+5 \times 2}{4+5+y+1+2}=2.6 \Rightarrow \frac{3 y+28}{12+y}=2.6 \Rightarrow 3 y+28=31.2+2.6 y \Rightarrow 0.4 y=3.2 \Rightarrow y=8$
View full question & answer→MCQ 1061 Mark
If the median of the data $\frac{x}{5}, x, \frac{x}{3}, \frac{2 x}{3}, \frac{x}{4}, \frac{2 x}{5}, \frac{3 x}{4}, x>0$ is 4 , then $x=$
Answer(b)
Arranging the given data in ascending order, we obtain: $\frac{x}{5}, \frac{x}{4}, \frac{x}{3}, \frac{2 x}{5}, \frac{2 x}{3}, \frac{3 x}{4}, x$
There are 7 observations. Therefore,
Median = Value of $\left(\frac{7+1}{2}\right)^{\text {th }}$ i.e. $4^{\text {th }}$ observation $\Rightarrow 4=\frac{2 x}{5} \Rightarrow x=10$
View full question & answer→MCQ 1071 Mark
If n is an even positive integer and $x>0, \quad x \neq 1$, then the median of the data a, $a, a x, a x^2, a x^3, \ldots, a x^n$ is
- A
$a x^{n-1}$
- B
$a x^{\frac{n}{2}-1}$
- ✓
$a x^{\frac{n}{2}}$
- D
$a x^{\frac{n}{2}+1}$
AnswerCorrect option: C. $a x^{\frac{n}{2}}$
(c)
Given observations are in ascending or descending order of magnitude according as $x>1$ or $x<1$. If n is even there are (n + 1) an odd number of observations. The median is the value of middle i.e.,$\left(\frac{n+1+1}{2}\right)^{t/h}=\left(\frac{n}{2}+1\right)^{th}$
∴ Median $=a x^{\frac{n}{2}+1-1}=a x^{\frac{n}{2}}$
View full question & answer→MCQ 1081 Mark
If for a data, Mean:Median = 9:8, then Median: Mode =
Answer(b)
We have, $\frac{\text { Mean }}{\text { Median }}+\frac{9}{8}=$ Mean $=\frac{9}{8}$ Median ...(i)
We know that
Mode $=3$ Median -2 Mean
⇒ Mode $=3$ Median $-\frac{9}{4}$ Median $=\frac{3}{4}$ Median ...[Using (i)]
Median : Mode $=4: 3$
View full question & answer→MCQ 1091 Mark
If a variable takes discrete values $x+4, x-\frac{7}{2}, x-\frac{5}{2}, x-3, x-2, x+\frac{1}{2}, x-\frac{1}{2}, x+5$,$x>0$,
- A
$x-\frac{5}{2}$
- B
$x-\frac{5}{3}$
- ✓
$x-\frac{5}{4}$
- D
$x-\frac{5}{6}$
AnswerCorrect option: C. $x-\frac{5}{4}$
(c)
Arranging the values of the variable in ascending order, we obtain
$x-\frac{7}{2}, x-3, x-\frac{5}{2}, x-2, x-\frac{1}{2}, x+\frac{1}{2}, x+4, x+5$
These are 8 observations.
Median = AM of 4th and 5th observation $=\frac{x-2+x-\frac{1}{2}}{2}=x-\frac{5}{4}$
View full question & answer→MCQ 1101 Mark
The arithmetic mean of x, x + 3, x + 6, x + 9 and x + 12 is
Answer(a)
Let $\bar{X}$ be the arithmetic mean of x, x + 3 x + 6 x + 9 and x + 12.
Then, $\bar{X}=\frac{x+(x+3)+(x+6)+(x+9)+(x+12)}{5}=\frac{5 x+30}{5}=x+6$.
View full question & answer→