M.C.Q (1 Marks)

MCQ 11 Mark

Mode is:

A
Least frequent value.
B
Middle most value.
✓
Most frequent value.
D
None of these.

Answer

Correct option: C.

Most frequent value.

Mode is the most frequency value of observation or a class,

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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:

A
0
B
19
✓
20
D
38

Answer

Correct option: C.

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)

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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
$\frac{\text{a}-\text{x}_\text{i}}{\text{h}}$

Answer

Correct 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.
$\text{u}_\text{i}=\frac{\text{x}_\text{i}-\text{a}}{\text{h}}$

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MCQ 41 Mark

If the mode of the data : 16, 15, 17, 16, 15, x, 19, 17, 14 is 15, then x =

✓
15
B
16
C
17
D
19

Answer

Correct option: A.

Mode 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
$\therefore\text{x}=15$

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MCQ 51 Mark

If the difference of mode and median of a data is 24, then the difference of median and mean is:

✓
12
B
24
C
8
D
36

Answer

Correct option: A.

Difference of mode and median = 24
Mode = 3 median – 2 mean
⇒ Mode – median = 2 median – 2 mean
⇒ 24 = 2 (median – mean)
⇒ Median – mean $=\frac{24}{2}=12$

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MCQ 61 Mark

If the mean of first n natural numbers is $\frac{5\text{n}}{9},$ then n =

A
5
B
4
✓
9
D
10

Answer

Correct option: C.

Given:
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).

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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
$\frac{\frac{1}{\text{n}}\sum_\text{i=1}^\text{n}\text{f}_\text{i}\text{x}_\text{i}}{\frac{1}{\text{n}}\sum_\text{i=1}^\text{n}\text{i}}$

Answer

Correct 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 $\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}$

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MCQ 81 Mark

If mode of a series exceeds its mean by 12, then mode exceeds the median by:

A
4
✓
8
C
6
D
10

Answer

Correct option: B.

Given: 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).

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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:

A
15
✓
25
C
30
D
35

Answer

Correct option: B.

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)

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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

Answer

Correct option: B.

x + y = 19

Mean of 6, 7, x, 8, y, 14 is 9
$\Rightarrow\frac{6+7+\text{x}+8+\text{y}+14}{6}=9\ (\text{n}=6)$
$\Rightarrow\frac{35+\text{x}+\text{y}}{6}=9$
$\Rightarrow35+\text{x}+\text{y}=54$
$\Rightarrow\text{x}+\text{y}=54-35=19$

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MCQ 111 Mark

If the median of the data: 24, 25, 26, x + 2, x + 3, 30, 31, 34 is 27.5, then x =

A
27
✓
25
C
28
D
30

Answer

Correct option: B.

The 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).

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MCQ 121 Mark

The mean of first n odd natural numbers is $\frac{\text{n}^2}{81},$ then n =

A
9
✓
81
C
27
D
18

Answer

Correct option: B.

The 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).

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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

Answer

Correct 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).

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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.

Answer

Correct option: D.

Mode = 3 Median −2 Mean.

Hence, the correct option is (d).

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MCQ 151 Mark

If the mode of the data: 64, 60, 48, x, 43, 48, 43, 34 is 43, then x + 3 =

A
44
B
45
✓
46
D
48

Answer

Correct option: C.

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).

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MCQ 161 Mark

The median of a given frequency distribution is found graphically with the help of:

A
Histogram.
B
Frequency curve.
C
Frequency polygon.
✓
Ogive.

Answer

Correct option: D.

Ogive.

The median of a given frequency distribution is found graphically with the help of ‘Ogive’.
Hence, the correct option is (d).

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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
$\bar{\text{x}}-\frac{\text{n}-1}{2}$

Answer

Correct 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$=\bar{\text{x}}+\frac{\text{n}+1}{2}$

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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.

Answer

Correct option: C.

Mode = 3 Median – 2 Mean.

The relation between mean, median and mode is:

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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 =

A
4
B
5
C
6
✓
7

Answer

Correct option: D.

Mean 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
$\therefore\text{Median}=\frac{7+1}{2}\text{th term}=4\text{th term}$
$=3$
$\therefore\text{q}=3$
$\therefore\text{p}+\text{q}=4+3=7$

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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
lower limits of classes.
B
upper limits of classes.
✓
mid$-$points of classes.
D
frequency of the class marks .

Answer

Correct 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.

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MCQ 211 Mark

The arithmetic mean and mode of a data are 24 and 12 respectively, then its median is:

A
25
B
18
✓
20
D
22

Answer

Correct option: C.

Arithmetic 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
$\text{Median}=\frac{60}{3}=20$

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MCQ 221 Mark

If the mean of first n natural number is 15, then n =

A
15
B
30
C
14
✓
29

Answer

Correct option: D.

Mean of first n natural number = 15
$\frac{\text{n}(\text{n}+1)}{2\text{n}}=15$
$\frac{\text{n+1}}{2}=15\Rightarrow\text{n}+1=30$
$\text{n}=30-1=29$

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MCQ 231 Mark

The mode of a frequency distribution can be determined graphically from:

A
Histogram.
B
Frequency polygon.
✓
Ogive.
D
Frequency curve.

Answer

Correct option: C.

Ogive.

Mode of frequency can be found graphically by an ogive,

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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 =

A
3
B
4
C
5
✓
6

Answer

Correct option: D.

Given:
$\sum\text{f}_\text{i}\text{x}_\text{i}=132+5\text{k},\sum\text{f}_\text{i}=20$ 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).

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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:

A
$\overline{X}+\text{n}$
B
$\overline{\text{X}}+\frac{\text{n}}{2}$
✓
$\overline{\text{X}}+\frac{\text{n}+1}{2}$
D
None of these.

Answer

Correct 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 $x_1+1, x_2+2, x_3+3, \ldots . x_n+n$
$\text{New mean}=\frac{(\text{x}_1+1)+(\text{x}_2+2)+(\text{x}_3+3)+.....+(\text{x}_\text{n}+\text{n})}{\text{n}}$
$=\frac{\text{x}_1+\text{x}_2+\text{x}_3+.......+\text{x}_\text{n}+(1+2+3+...+\text{n})}{\text{n}}$
$=\frac{\text{n}\overline{\text{X}}+\frac{\text{n}(\text{n+1})}{2}}{\text{n}}$
$\overline{\text{X}}+\frac{\text{n}+1}{2}$

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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
$\frac{\text{n}}{2}+1$

Answer

Correct 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)

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MCQ 271 Mark

If the arithmetic mean of 7, 8, x, 11, 14 is x, then x =

A
9
B
9.5
✓
10
D
10.5

Answer

Correct option: C.

Arithmetic mean of 7, 8, x, 11, 14, is x
$\Rightarrow\frac{7+8+\text{x}+11+14}{5}=\text{x}$
$\Rightarrow\frac{40+\text{x}}{5}=\text{x}\Rightarrow40+\text{x}=5\text{x}$
$\Rightarrow5\text{x}-\text{x}=40\Rightarrow4\text{x}=40$
$\Rightarrow\text{x}=\frac{40}{4}=10$

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MCQ 281 Mark

Which of the following is not a measure of central tendency?

A
Mean.
B
Median.
C
Mode.
✓
Standard deviation.

Answer

Correct option: D.

Standard deviation.

Standard deviation is not a measure of central tendency. Only mean, median and mode are measures.

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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:

A
Mean.
✓
Median.
C
Mode.
D
All the three above.

Answer

Correct option: B.

Median.

The 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).

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MCQ 301 Mark

If the median of the data : 6, 7, x – 2, x, 17, 20, written in ascending order, is 16. Then x =

A
15
B
16
✓
17
D
18

Answer

Correct option: C.

Median of 6, 7, x - 2, x, 17, 20 is 16
Here n = 6
$\therefore\text{Median}=\frac{1}{2}\Big[\frac{\text{n}}{2}\text{th}+\Big(\frac{\text{n}}{2}+1\Big)\text{th}\Big]\text{term}$
$=\frac{1}{2}\Big[\frac{6}{2}+\Big(\frac{6}{2}+1\Big)\Big]\text{term}$
$=\frac{1}{2}(3\text{rd}+4\text{th})\text{term}$
$=\frac{1}{2}(\text{x}-2+\text{x})$
$=\frac{1}{2}(2\text{x}-2)=\text{x}-1$
$\therefore\text{x}-1=16$
$\Rightarrow\text{x}=16+1=17$

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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.

Answer

Correct 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).

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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
$\bar{\text{x}}+\frac{\text{m}}{\text{n}}$

Answer

Correct 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
$=\frac{\Big(\frac{\text{y}_1}{\text{m}}+\text{n}\Big)+\Big(\frac{\text{y}_2}{\text{m}}+\text{n}\Big)+.....+\Big(\frac{\text{y}_\text{k}}{\text{m}}+\text{n}\Big)}{\text{k}}$
$=\frac{\Big(\frac{\text{y}_1}{\text{m}}+\frac{\text{y}_2}{\text{m}}+.....+\frac{\text{y}_\text{k}}{\text{m}}+\Big)+(\text{n}+\text{n}+.....+\text{n})}{\text{k}}$
$=\frac{\text{y}_1+\text{y}_2+....+\text{y}_\text{k}}{\text{mk}}+\frac{\text{nk}}{\text{k}}$
$=\frac{\text{k}\bar{\text{x}}}{\text{mk}}+\frac{\text{nk}}{\text{k}}$
$=\frac{\bar{\text{x}}}{\text{m}}+\text{n}$

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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.

Answer

Correct option: C.

Mode = 3 Median – 2 Mean.

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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:

A
17
✓
17.5
C
18
D
18.5

Answer

Correct 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.

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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
$\text{n}^2$

Answer

Correct 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)
$=\frac{\text{n}}{2}[2+2\text{n}-2]=\frac{\text{n}}{2}\times2\text{n}=\text{n}^2$
$\therefore\text{Mean}=\frac{\text{Sum of n terms}}{\text{n}}=\frac{\text{n}^2}{\text{n}}=\text{n}$

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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 $\bar{\text{x}}$

A
23
B
24
✓
27
D
25

Answer

Correct option: C.

$\text{u}_\text{i}=\frac{\text{x}_\text{i}-25}{10},\ \sum\text{f}_\text{i}\text{u}_\text{i}=20,\ \sum\text{f}_\text{i}=100$
Here assumed mean = 25
and class interval (h) = 10
$\therefore\ \bar{\text{x}}=\text{A}+\frac{\sum\text{f}_\text{i}\text{u}_\text{i}}{\sum\text{f}_\text{i}}\times\text{h}$
$=25+\frac{20}{100}\times10$
$=25+2=27$

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MCQ 371 Mark

Which of the following cannot be determined graphically?

✓
Mean.
B
Median.
C
Mode.
D
None of these.

Answer

Correct option: A.

Mean.

Mean cannot be determind grafically,

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MCQ 381 Mark

The algebraic sum of the deviations of a frequency distribution from its mean is:

A
Always positive.
B
Always negative.
✓
$0.$
D
A non$-$zero number.

Answer

Correct option: C.

$0.$

The 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
$\therefore (\bar{\text{x}}-\text{x}_1)+(\bar{\text{x}}-\text{x}_2)+(\bar{\text{x}}-\text{x}_3)+ ....(\bar{\text{x}}-\text{x}_\text{n})$
$=\text{n}\bar{\text{x}}-(\text{x}_1+\text{x}_2+\text{x}_3+.....\text{x}_\text{n})$
$=\text{n}\bar{\text{x}}-\text{n}\bar{\text{x}}$
$=0$

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MCQ 391 Mark

The median of first 10 prime numbers is:

A
11
✓
12
C
13
D
14

Answer

Correct option: B.

First 10 prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29
Here n = 10
$\therefore\text{Median}=\frac{1}{2}\Big[\frac{\text{n}}{2}\text{th}+\Big(\frac{\text{n}}{2}+1\Big)\text{th}\Big]\text{term}$
$=\frac{1}{2}\Big[\frac{10}{2}\text{th}+\Big(\frac{10}{2}+1\Big)\text{th}\Big]\text{term}$
$=\frac{1}{2}[5\text{th}+6\text{th}]\text{term}$
$=\frac{1}{2}[11+13]=\frac{1}{2}\times24=12$

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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

A
3
✓
8
C
13
D
24

Answer

Correct option: B.

Mean = 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

$\therefore\text{Mean}=\frac{\sum\text{fx}}{\sum\text{f}}$
$\Rightarrow2.6=\frac{28+3\text{y}}{12+\text{y}}$
$\Rightarrow2.6(12+\text{y})=28+3\text{y}$
$\Rightarrow31.2+2.6\text{y}=28+3\text{y}$
$\Rightarrow3\text{y}-2.6\text{y}=31.2-28$
$\Rightarrow0.4\text{y}=3.2$
$\Rightarrow4\text{y}=32$
$\Rightarrow\text{y}=\frac{32}{4}=8$
$\text{y}=8$

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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
$\frac{\bar{\text{x}}}{\text{a}}$

Answer

Correct option: C.

$\bar{\text{x}}+\text{a}$

Meam of observations $x_1, x_2, \ldots \ldots, x_n$ is $\bar{x}$
$\frac{x_1+x_2+x_3 \ldots \ldots .+x_n}{n}=\bar{x}$
$x_1+a+x_2+a+x_3+a+\ldots \ldots x_n+a$
$=x_1+x_2+x_3+\ldots x_n+n a$
$\therefore \text { Mean of }\left(x_1+x_2+x_3 \ldots . .+x_n\right)+n a$
$=\bar{x}+\frac{n a}{n}$
$=\bar{x}+a$

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MCQ 421 Mark

If 35 is removed from the data : 30, 34, 35, 36, 37, 38, 39, 40, then the median increases by:

A
2
B
1.5
C
1
✓
0.5

Answer

Correct option: D.

0.5

Given 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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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
8th observation
B
7th observation
✓
average of 8th and 9th observation
D
average of 7th and 8th observation

Answer

Correct option: C.

average of 8th and 9th observation

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MCQ 441 Mark

The value of x for which the class mark of the class interval 30-x is 36 is

A
38
B
40
C
36
✓
42

Answer

Correct option: D.

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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

✓
27
B
37
C
17
D
25

Answer

Correct option: A.

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MCQ 461 Mark

If the mean of first in natural numbers is $\frac{5 n}{9}$, then the value of n is

A
5
B
4
✓
9
D
10

Answer

Correct option: C.

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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.

A
median
✓
mode
C
mean
D
range

Answer

Correct option: B.

mode

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MCQ 481 Mark

If value of each observation in a data is increased by 2, then the median of the new data

✓
increases by 2
B
increases by 2n
C
remains same
D
decreases by 2

Answer

Correct option: A.

increases by 2

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MCQ 491 Mark

The middle most observation of every ery data arranged in order is called

A
Mode
B
Median
✓
Mean
D
Deviation

Answer

Correct option: C.

Mean

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MCQ 501 Mark

If every term of the statistical data consisting of n terms is decreased by 2, then the mean of the data:

✓
decreased by 2
B
remains unchanged
C
decreases by 2n
D
decreases by 1

Answer

Correct option: A.

decreased by 2

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