MCQ

MCQ 1011 Mark

For a frequency distribution mean deviation from mean is computed by:

A
$\text{M.D.}=\frac{\sum\text{f}}{\sum\text{f}\ |\text{d}|}$
B
$\text{M.D.}=\frac{\sum\text{d}}{\sum\text{f}}$
C
$\text{M.D.}=\frac{\sum\text{fd}}{\sum\text{f}}$
✓
$\text{M.D.}=\frac{\sum\text{f}\ |\text{d}|}{\sum\text{f}}$

Answer

Correct option: D.

$\text{M.D.}=\frac{\sum\text{f}\ |\text{d}|}{\sum\text{f}}$

$\text{M.D.}=\frac{\sum\text{f}\ |\text{d}|}{\sum\text{f}}$

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MCQ 1021 Mark

The weights in kilogram of 9 members in a school boxing team are 54, 59, x, 53, 73, 49, 50, 58, 45 If the average is 56 then x is:

A
61Kg
B
62Kg
C
64Kg
✓
63Kg

Answer

Correct option: D.

63Kg

$ \displaystyle \frac{54+59+\text{x}+53+73+49+50+58+45}{9}=56$
On simplification $ \text{x} = 63$

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MCQ 1031 Mark

Choose the correct answer. Consider the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. If 1 is added to each number, the variance of the numbers so obtained is:

A
6.5
B
2.87
C
3.87
✓
8.25

Answer

Correct option: D.

8.25

Given numbers are 1, 2, 3,4, 5, 6, 7, 8, 9 and 10
If 1 is added to each number, then observations will be 2, 3,4, 5, 6,7, 8, 9, 10 and 11
$\therefore\ \sum\text{x}_\text{i}=2+3+4+\ ....\ +11$
$=\frac{10}{2}\big[2\times2+9\times1\big]=5[4+9]=65$
and $\sum\text{x}^2_\text{i}=2^2+3^2+4^2+5^2+\ .....\ +11^2=(1^2+2^2+3^2+\ .....\ +11^2)-(1^2)$
$=\frac{11\times12\times23}{6}-1=505$
$\therefore\ \text{s}^2=\frac{\sum\text{x}^2_\text{i}}{\text{n}}-\Big(\frac{\sum\text{x}_\text{i}}{10}\Big)^2$
$=\frac{505}{10}-\Big(\frac{65}{10}\Big)^2$
$=50.5-(6.5)^2$
$=50.5-42.35$
$=8.25$

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MCQ 1041 Mark

Which average shows the most common variable in the data set?

A
Mean
✓
Mode
C
Media
D
All of the above

Answer

Correct option: B.

Mode

Mode is the highest occurring figure in a series.
It is the value in a series of observation that repeats maximum number of times and
which represents the whole series as most of the values in the series revolves around this value.
Therefore, the most common variable in the series of observations is known as mode.

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MCQ 1051 Mark

The mean of five numbers is 18. If one number is excluded, then their mean is 16, the excluded number is ___________.

A
24
✓
26
C
28
D
25

Answer

Correct option: B.

Mean of 55 numbers = 18
Sum of these 55 numbers = 18 × 5 = 90
Let number that has been excluded be x New mean = $ \dfrac{90-\text{x}}{4} = 16$
Solving this, we get 90 - x = 64
x = 26

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MCQ 1061 Mark

In a factory, the average salary of the employees is Rs. 70. If the average salary of 12 officers is Rs. 400 and that of the remaining employees is Rs. 60, then the number of employees are ...........

A
396
B
400
✓
408
D
404

Answer

Correct option: C.

408

⇒ Let total number of employees be
x ⇒ Average salary of total employee
= Rs. 70 = Average of 12 employees = Rs. 400 = Rs. 400 ⇒ Average of remaining employees
$ \text{Rs}.60∴ 70=\frac{400\times 12+(\text{x}-12)\times 60}{\text{x}}$
$ ∴ 70\text{x}=4800+60\text{x}-720$
$ ∴ 70\text{x}=4080+60\text{x}$
$ ∴ 10\text{x}=4080\therefore\text{x}=408$
Total number of employees are 408.

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MCQ 1071 Mark

Choose the correct answer. The standard deviation of the data 6, 5, 9, 13, 12, 8, 10 is:

✓
$\sqrt{\frac{52}{7}}$
B
$\frac{52}{7}$
C
$\sqrt{6}$
D
$6$

Answer

Correct option: A.

$\sqrt{\frac{52}{7}}$

$\sqrt{\frac{52}{7}}$

Solution:
Given data are 6, 5, 9, 13, 12, 8 and 10

$x_i$	$x_i{}^2$
6	36
5	25
9	81
13	169
12	144
8	64
10	100
$\sum\text{x}_\text{i}=63$	$\sum\text{x}_\text{i}^2=619$

$\therefore\ \text{SD}=\sigma=\sqrt{\frac{\sum\text{x}_\text{i}^2}{\text{N}}-\Big(\frac{\sum\text{x}_\text{i}}{\text{N}}\Big)^2}$
$=\sqrt{\frac{619}{7}-\Big(\frac{63}{7}\Big)^2}=\sqrt{\frac{4333-396}{49}}$
$=\sqrt{\frac{396}{49}}=\sqrt{\frac{52}{7}}$

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MCQ 1081 Mark

A grocer has a sale of Rs. 6435, Rs. 6927, Rs. 6855, Rs. 7230 and Rs. 6562Rs. for 5 consecutive months. How much sale must he have in the sixth month so that he gets an average sale of Rs. 6500?

✓
Rs. 4991
B
Rs. 5991
C
Rs. 6001
D
Rs. 6991

Answer

Correct option: A.

Rs. 4991

Total sale of 5 months = Rs. (6435 + 6927 + 7230 + 6562) = Rs. 34009.
Required sale = Rs. [(6500 × 6) - 34009]
= Rs. (39000 - 34009)
= Rs. 4991.

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MCQ 1091 Mark

If the mean of first n natural numbers is equal to $ \dfrac{\text{n}+7}{3}$ then nn is equal to:

✓
11
B
13
C
25
D
None of these

Answer

Correct option: A.

$ 1\ +\ 2+\ 3\ +....\text{n}=\frac{\text{n}\times(\text{n}\ +\ 1)}{2} $
$ \text{Then u}=\frac{\text{n}\times(\text{n}+1)}{2\times \text{n}}$
$ ∴\text{u}=\frac{(\text{n}+1)}{2}$
But Given: $ \text{u}=\dfrac{(\text{n}+7)}{3}$
Thus, $ \text{u}=\frac{(\text{n}+1)}{2}=\frac{(\text{n}+7)}{3}$
Solving above we get,
n=11

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MCQ 1101 Mark

The median of the following data 46, 64, 87, 41, 58, 77, 35, 90, 55, 33, 92 is:

A
87
B
77
✓
58
D
60.2

Answer

Correct option: C.

Arrange the given data in ascending order.
We have, 33,35,41,46,55,58,64,77,87,90 and 92.
The sixth entry is 58.
Median is 58.

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MCQ 1111 Mark

The mean of all possible factor of 10 is:

A
4
B
3
C
5
✓
4.5

Answer

Correct option: D.

4.5

Factors of 10 are, 1, 2, 5, 10Hence required mean $ =\frac{1+2+5+10}{4}$
$ =\frac{18}{4}=4.5$

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MCQ 1121 Mark

What is the modal value for the numbers 5, 8, 6, 4, 10, 15, 18, 10?

A
18
✓
10
C
14
D
None of above

Answer

Correct option: B.

Mode is the highest occurring figure in a series.
It is the value in a series of observation that repeats maximum number of times and, which represents the whole series as most of the values in the series revolves around this value.
Since in the given series, 10 is occurring the highest number of times.
Therefore, 10 is the mode of the series of given observations.

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MCQ 1131 Mark

Find the mean of all the positive factors of 72:

✓
16.25
B
17.25
C
18.25
D
None of these

Answer

Correct option: A.

16.25

Factors of 72 are: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72
$\text{mean}=\frac{\text{sum}}{\text{count}}$
$\text{Mean}=\frac{1+2+3+4+6+8+9+12+18+24+36+72}{12}$
$ \text{mean}=\frac{195}{12}$
$ \text{Mean}=16.25$

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MCQ 1141 Mark

In the first 10 overs of a cricket game, the run rate was only 3.2 What should be the run rate in the remaining 40 overs to reach the target of 282 runs?

✓
6.25
B
6.5
C
6.75
D
7

Answer

Correct option: A.

6.25

For first 10 overs, run rate = 3.2
⇒ Runs scored = 3.2 × 10 = 32
Total runs to be scored = 282
Runs Left to be scored in 4040 overs $ = \frac{282-32}{40}$
$ =\dfrac{250}{40}$
Required Run-Rate in remaining overs = 6.25

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MCQ 1151 Mark

The mean of 30, 32, 24, 34, 26, 28, 30, 35, 33, 25 is 29.7
If true then enter 1 and if false then enter 0:

A
0
✓
1
C
None of these
D
Can not be determined

Answer

Correct option: B.

Mean of the series: 30, 32, 24, 34, 26, 28, 30, 35, 33, 25
$ \text{mean} = \frac{\text{Sum}}{\text{Number of observations}}$
$\text{mean} = \frac{30 + 32 + 24 + 34 + 26 + 28 + 30 + 35 + 33 + 25}{10}$
$ \text{mean} = \frac{297}{10} = 29.7$

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MCQ 1161 Mark

The average age of two brothers is 9 years It is increased by 9 years when their mothers age is also included then the age of mother is:

A
35 years
✓
36 years
C
37 years
D
38 years

Answer

Correct option: B.

36 years

Average age of the two brother = 9 years
$ \therefore$ Age of two brother = 9 × 2 = 18 years
If their mother age is included then the average age is increased by 9
$ \therefore$ Average age of 3 = 9 + 9 = 18 years
Now Total age of three = 3 × 18 = 54 Years
$ \therefore$ Mothers age = 54 - 18 = 36 years.

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MCQ 1171 Mark

x	10	15	20	25	35
f	6	p	26	10	8

✓
10
B
12
C
24
D
26

Answer

Correct option: A.

$ ∑ \text{fx} = 15\text{p} + 1110; ∑ \text{f} = 50 + \text{p}$
$ \text{x} = 21$
$ 21=\frac{15_p\ +\ 1110}{50 \ +\ p}$
$ ⇒ 21\text{p} − 15\text{p} = 1110 − 1050$
$ \Rightarrow\text{p}=\frac{60}{6}=10$

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MCQ 1181 Mark

Choose the correct answer. Let $\mathrm{x}_1, \mathrm{x}_2, \ldots \mathrm{x}_{\mathrm{n}}$ be n observations. Let $\mathrm{w}_{\mathrm{i}}=\mathrm{x}_{\mathrm{i}}+\mathrm{k}$ for $\mathrm{i}=1,2, \ldots \mathrm{n}$, where I and k are constants. If the mean of $x_i^{\prime} s$ is 48 and their standard deviation is 12 , the mean of $w_i^{\prime} s$ is 55 and standard deviation of $w_i^{\prime} s$ is 15 , the values of I and k should be:

✓
l = 1.25, k = -5
B
l = -1.25, k = 5
C
l = 2.5, k = -5
D
l = 2.5, k = 5

Answer

Correct option: A.

l = 1.25, k = -5

l = 1.25, k = -5

Solution:
Given, $\text{w}_\text{i}=\text{lx}_\text{i}+\text{k},\ \bar{\text{x}}_\text{i}=48,\text{ sx}_\text{i}=12,\text{ w}_\text{i}=55$ and $\text{sw}_\text{i}=15$
Then, $\bar{\text{w}}_\text{i}=\text{l}\bar{\text{x}}_\text{i}+\text{k}$ $\big[\text{where }\bar{\text{w}}_\text{i}\text{ is mean w}_\text{i}{'\text{s}}\text{ and }\bar{\text{x}}_\text{i}\text{ is mean of x}_\text{i}{'\text{s}}\big]$
$\Rightarrow55=\text{l}\times48+\text{k}\ ...(\text{i})$
Now, $\text{SD of w}_\text{i}=\text{l}(\text{SD of x}_\text{i})$
$\Rightarrow15=\text{l}\times12$
$\Rightarrow\text{l}=\frac{15}{12}=12.5$
From Eq. (i) we get k = 55 - 1.25 × 48 = -5

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MCQ 1191 Mark

Which one of the following statements is correct?

A
The Standard deviation for a given distribution is the square of the variance.
✓
The standard deviation for a given distribution is the square root of the variance.
C
The standard deviation for a given distribution is equal to the variance.
D
The standard deviation for a given distribution is half of the variance.

Answer

Correct option: B.

The standard deviation for a given distribution is the square root of the variance.

The standard deviation for a given distribution is the square root of the variance.

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MCQ 1201 Mark

Choose the correct answer. When tested, the lives (in hours) of 5 bulbs were noted as follows: 1357, 1090, 1666, 1494, 1623 The mean deviations (in hours) from their mean is:

✓
178
B
179
C
220
D
356

Answer

Correct option: A.

178

Solution:
The lines of 5 bulbs are given by
1357, 1090, 1666, 1494, 1623
$\therefore\ \text{Mean}=\frac{1357+1090+1666+1494+1623}{5}$
$\Rightarrow\bar{\text{x}}=\frac{7230}{5}=1446$

$x_i$	$\text{d}_\text{i}=\|\text{x}_{\text{i}}-\bar{\text{x}}\|$
1357	89
1090	356
1666	220
1494	48
1623	177
Total	$\sum\text{d}_\text{i}=890$

$\therefore\ \text{MD}=\frac{\sum\text{d}_\text{i}}{\text{n}}=\frac{890}{5}=178$

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MCQ 1211 Mark

The mean of 20 observations is 12.5 By error, one observation was noted as -15 instead of 15. Then the correct mean is __________:

A
11.75
B
11
✓
14
D
13

Answer

Correct option: C.

Mean of 20 observations = 12.5 Sum of 20 observations = 12.5 × 20 = 250
Since 15 was misread as,
15 New sum = 250 - (-15) + 15 = 280
The correct $\text{mean} = \dfrac{280}{20} = 14$

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MCQ 1221 Mark

The mean of the following data is :
45, 35, 20, 30, 15, 25, 40:

A
15
B
25
C
35
✓
30

Answer

Correct option: D.

Mean is given by $ \text{mean}=\dfrac{\text{sum of the elements}}{\text{total number of elements}}$
$=\text{mean}=\frac{45+35+20+30+15+25+40}{7}$
$=\frac{210}{7}$
$ =30$

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MCQ 1231 Mark

The median 31, 16,19, 25, 14, 13,12, 4, 28, 45 is.

A
14
B
20
✓
17.5
D
none of thes

Answer

Correct option: C.

17.5

Arranging the given data in ascending order 4, 12, 13, 16, 19, 28, 31,
The middle terms are 16, 19 Hence median
$ =\frac{16 \ +\ 19}{2}$
$ =17.5$

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MCQ 1241 Mark

Value of the middle-most observation (s) is called:

A
Mean
✓
Median
C
Mode
D
None of thes

Answer

Correct option: B.

Median

To 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.

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MCQ 1251 Mark

If the median of $ \frac{\text{x}}{5}\text{x} \frac{\text{x}}{4} \frac{\text{x}}{2}\ \text{and}\ \frac{\text{x}}{3}$ (where x > 0) is 8 then the value of x would be:

✓
24
B
32
C
8
D
16

Answer

Correct option: A.

Arranging is ascending order the values are
$\frac{\text{x}}{5},\frac{\text{x}}{4},\frac{\text{x}}{3},\frac{\text{x}}{2},\text{x}$
Middle value $ =\frac{\text{x}}{3}\Rightarrow\frac{\text{x}}{3}=\text{x}=24$

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MCQ 1261 Mark

Let $\mathrm{x}_1, \mathrm{x}_2, \ldots, \mathrm{x}_n$ be values taken by a variable X and $\mathrm{y}_1, \mathrm{y}_2, \ldots, \mathrm{y}_n$ be the values taken by a variable Y such that $\mathrm{y}_{\mathrm{i}}=a \mathrm{x}_{\mathrm{i}}+$ $\mathrm{b}, \mathrm{i}=1,2, \ldots, \mathrm{n}$. Then,

✓
$\operatorname{Var}(Y)=a^2 \operatorname{Var}(X)$
B
$\operatorname{Var}(X)=a^2 \operatorname{Var}(Y)$
C
$\operatorname{Var}(\mathrm{X})=\operatorname{Var}(\mathrm{X})+\mathrm{b}$
D
None of these

Answer

Correct option: A.

$\operatorname{Var}(Y)=a^2 \operatorname{Var}(X)$

$\operatorname{Var}(Y)=a^2 \operatorname{Var}(X)$

Solution:
$\text{Var}(\text{x})=\frac{\sum\limits^\text{n}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)^2}{\text{n}}$ where Mean $\Big(\overline{\text{X}}\Big)=\frac{\sum\limits^\text{n}_{\text{i}=1}\text{x}_\text{i}}{\text{n}}$
$\text{Var}(\text{Y})=\frac{\sum\limits^\text{n}_{\text{i}=1}\Big(\text{y}_\text{i}-\overline{\text{Y}}\Big)^2}{\text{n}}$ and $\overline{\text{Y}}=\frac{\sum\limits^\text{n}_{\text{i}=1}\text{y}_\text{i}}{\text{n}}$
We have,
$\text{y}_\text{i}=\text{ax}_\text{i}+\text{b}$
$\overline{\text{Y}}=\frac{\sum\limits^\text{n}_{\text{i}=1}\text{y}_\text{i}}{\text{n}}$
$\overline{\text{Y}}=\frac{\sum\limits^\text{n}_{\text{i}=1}\text{ax}_\text{i}+\text{b}}{\text{n}}$
$=\frac{\sum\limits^\text{n}_{\text{i}=1}\text{x}_\text{i}}{\text{n}}+\frac{\text{nd}}{\text{n}}$
$=\text{a}\overline{\text{X}}+\text{b}$
$\text{Var}(\text{Y})=\frac{\sum\limits^\text{n}_{\text{i}=1}\Big(\text{y}_\text{i}-\overline{\text{Y}}\Big)^2}{\text{n}}$
$=\frac{\sum\limits^\text{n}_{\text{i}=1}\Big\{\text{ax}_\text{i}+\text{b}-\big(\text{a}\overline{\text{X}}+\text{b}\big)\Big\}^2}{\text{n}}$
$=\frac{\sum\limits^\text{n}_{\text{i}=1}\big(\text{ax}_\text{i}-\text{a}\overline{\text{X}}\big)^2}{\text{n}}$
$=\text{a}^2\frac{\sum\limits^\text{n}_{\text{i}=1}\big(\text{x}_\text{i}-\overline{\text{X}}\big)^2}{\text{n}}$
$=\text{a}^2\text{Var}(\text{X})$

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MCQ 1271 Mark

The average of 15 numbers is 18 The average of first 8 is 19 and that last 8 is 17 then the 8th number is:

A
15
B
16
✓
18
D
20

Answer

Correct option: C.

Average of 15 numbers 15 × 18
= 270 Average of first 8 number is 19
$\therefore$ Sum of first 8 number=19 × 8 = 152 = 19 × 8 = 152
Average of first 8 number = 17
$\therefore$ Sum of 8 number = 8 × 17 = 136 = 8 × 17 = 136
∴ 8th number is = (152 + 136) - 270 ⇒ 288 - 270 - 8

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MCQ 1281 Mark

Aman is his 12th innings makes a score of 63 runs and increases his average score to 2.What is his average after the 12th innings ?

A
15
B
29
C
69
✓
41

Answer

Correct option: D.

Let the average score till 11th innings be x according to question $ ⇒ \frac{11\text{x}+63}{12}$
= x + 2 ⇒ 11x + 63 = 12x + 24 ⇒ x = 39
12th inning average ⇒ 39 + 2 = 41

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MCQ 1291 Mark

The mean of $x_1, x_2 \ldots x_{50} M$, if every $x_{i,}=1,2 \ldots 50$ is replaced by $\frac{\text{x}_i}{50}$ then the mean is:

A
$\text{m}$
B
$\text{ M}+\frac{1}{50}$
C
$ \displaystyle \frac{50}{\text{M}}$
✓
$ \displaystyle \frac{\text{M}}{50}$

Answer

Correct option: D.

$ \displaystyle \frac{\text{M}}{50}$

$ \displaystyle \frac{\text{M}}{50}$

Solution:
Given $ \text{mean}= \frac{\text{x}_{1}+\text{x}_{2}+\text{x}_{3}...........\text{x}_{50}}{50}$
$ \text{mean} =\frac{\frac{\text{x}_{1}}{50}+\frac{\text{x}_{2}}{50}+...........\frac{\text{x}_{50}}{50}}{50}$
$ =\frac{\text{x}_1+\text{x}_2+....\text{x}_{50}}{50\times50}$
$ \displaystyle \frac{\text{M}}{50}$

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MCQ 1301 Mark

The modal value is the value of the variate which divides the total frequency into two equal parts:

A
True
✓
False
C
Neither
D
Either

Answer

Correct option: B.

False

False. Modal value is the value which occurs maximum number of times in the data.

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MCQ 1311 Mark

Find the mean of 23, 28, 13, 16, 20:

✓
20
B
25
C
23
D
none of these

Answer

Correct option: A.

Given observations 23, 28, 13, 16, 20 No. of observations are 5 Mean of
observations $ \dfrac{23+28+13+16+20}{5}$
$ =\dfrac{100}{5}$
$ =20$

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MCQ 1321 Mark

The _____ of a set of data is the middlemost number in the set.

A
mean
B
mode
C
range
✓
media

Answer

Correct option: D.

media

The median of a set of data is the middlemost number in the set.
Example: 3, 4, 5, 1, 1, 8, 10
So, first arrange the data in order.
So, 1, 1, 3, 4, 5, 8, 10
The middle number is median = 4

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MCQ 1331 Mark

The combined mean of three groups is 12 and the combined mean of first two groups is 3. If the first, second and third group have their mean as 2, 3 and 5 times respectively, then the mean of third group is:

A
10
✓
21
C
12
D
13

Answer

Correct option: B.

Let in common no. in each group is X
Then Member of each group is 2X, 3X and 5X
Total of three group = (2X + 3X + 5X) 12 = 120x
And total of Two group = (2X + 3X)3 = 15X (given mean of two group is 3)
Then total of third group = 120X - 15X = 115X
Mean of third group $ =\frac{115\text{X}}{5\text{X}}=21$

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MCQ 1341 Mark

In the formula for mode of a grouped data, $\text{ mode} =\text{l}+\left \{\frac {\text{f}_1-\text{f}_0}{2\text{f}_2-\text{f}_0-\text{f}_2}\right \}\times\text{h}$ where symbols have their usual meaning $\mathrm{f}_0=$ represents:

A
Frequency of modal class
B
Frequency of median class
✓
Frequency of the class preceding the modal class
D
Frequency of the class succeeding the modal clas

Answer

Correct option: C.

Frequency of the class preceding the modal class

Frequency of the class preceding the modal class

Solution:
In the formula for mode of a grouped data,
$\text{ mode} =\text{l}+\left \{\frac {\text{f}_1-\text{f}_0}{2\text{f}_2-\text{f}_0-\text{f}_2}\right \}\times\text{h}$ where symbols have their usual meaning
fo represents Frequency of the class preceding the modal classwhere
$\mathrm{f}=$ Frequency,
1 = Lowest value of the modal range,
$\mathrm{f}_1=$ Frequency of modal class,
$\mathrm{f}_2$ Frequency of class succeeding the modal class and
$\mathrm{f}_0=$ Frequency of class preceding the modal class. Hence, option C is correct.

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MCQ 1351 Mark

The average of 50 numbers is 38. If two numbers namely 45 and 55 are discarded, the average of the remaining numbers is:

✓
37.5
B
38.5
C
36.5
D
37

Answer

Correct option: A.

37.5

Average of remaining 48 numbers
$= \frac{(50 \times38 )- 55 - 45}{48} = 37.5$

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MCQ 1361 Mark

The ________ is the difference between the greatest and the least value of the variate:

✓
Range
B
Data
C
Average
D
Variance

Answer

Correct option: A.

Range

Range as the name indicates gives us all the area available under light and hence statement is true.

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MCQ 1371 Mark

State true or false: The mode is the most frequently occurring observation:

✓
True
B
False
C
Can't determine
D
None of these

Answer

Correct option: A.

True

The observation occurring the most number of times or which has highest frequency is called the mode.
Thus, the given statement is true.

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MCQ 1381 Mark

The mean of the following natural numbers 1, 2, 3,...10 is:

A
6.5
B
4.5
✓
5.5
D
5.4

Answer

Correct option: C.

5.5

Numbers are 1, 2, 3,..10
$\text{Sum of the numbers} = {\frac{\text{n}(\text{n}+1)}{2}}= \frac{10\times11}{2} = 55$
$\text{Mean} = \frac{55}{10} =5.5$

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MCQ 1391 Mark

The mean of 200 items was 50. Later on, it was discovered that two items were misread as 92 and 8 instead of 192 and 8. The correct mean is:

A
50
B
1
✓
50.9
D
None of these

Answer

Correct option: C.

50.9

Mean of 200 observations = 50
Sum of 200 observations = 50 × 200 = 10000
After replacing the misread observation 92 to 192 and 8 to 88
Sum of 200200 observations = 10000 - 92 + 192 - 8 + 88 = 10180
$ \text{New mean} = \frac{10180}{200} = 50.9$

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MCQ 1401 Mark

Choose the correct answer. The mean of 100 observations is 50 and their standard deviation is 5. The sum of all squares of all the observations is:

A
50000
B
250000
✓
252500
D
255000

Answer

Correct option: C.

252500

Here, $\bar{\text{x}}=\frac{\sum\text{x}_\text{i}}{\text{n}}$
$\Rightarrow50=\frac{\sum\text{x}_\text{i}}{100}$
$\Rightarrow{\sum\text{x}_\text{i}}=5000$
$\therefore\ \text{SD}=\sqrt{\frac{\sum\text{x}_\text{i}^2}{\text{n}}-\Big(\frac{\sum\text{x}_\text{i}}{\text{n}}\Big)^2}$
$\Rightarrow\ 5=\sqrt{\frac{\sum\text{x}_\text{i}^2}{\text{n}}-\Big(\frac{5000}{100}\Big)^2}$
$\Rightarrow25=\frac{\sum\text{x}_\text{i}^2}{100}=2525$
$\therefore\ {\sum\text{x}_\text{i}^2}=2525\times100=252500$

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MCQ 1411 Mark

If two variates X and Y are connected by the relation $\text{Y}=\frac{\text{aX}+\text{b}}{\text{c}},$ where a, b, c are constants such that ac < 0, then

A
$\sigma\text{Y}=\frac{\text{a}}{\text{c}}\sigma\text{X}$
✓
$\sigma\text{Y}=-\frac{\text{a}}{\text{c}}\sigma\text{X}$
C
$\sigma\text{Y}=-\frac{\text{a}}{\text{c}}\sigma\text{X}+\text{b}$
D
None of these

Answer

Correct option: B.

$\sigma\text{Y}=-\frac{\text{a}}{\text{c}}\sigma\text{X}$

$\text{Y}=\frac{\text{aX}+\text{b}}{\text{c}}$
$\overline{\text{Y}}=\frac{\sum\limits_{\text{i}=1}^\text{n}\frac{\text{aX}+\text{b}}{\text{c}}}{\text{n}}$
$=\frac{\frac{\text{a}\sum\limits_{\text{i}=1}^\text{n}\text{X}+\text{nb}}{\text{c}}}{\text{n}}$
$=\frac{\frac{\text{a}}{\text{c}}\sum\limits_{\text{i}=1}^\text{n}\text{X}}{\text{n}}+\frac{\text{b}}{\text{c}}$
$=\frac{\text{a}\overline{\text{X}}}{\text{c}}+\frac{\text{b}}{\text{c}}$
We know:
$\text{Var}(\text{X})=\frac{\sum\limits_{\text{i}=1}^\text{n}\big(\text{x}_\text{i}-\overline{\text{X}}\big)^2}{\text{n}}$
$=\sigma^2$
$\text{Var}(\text{Y})=\frac{\sum\limits_{\text{i}=1}^\text{n}\big(\text{y}_\text{i}-\overline{\text{Y}}\big)^2}{\text{n}}$
$=\frac{\sum\limits_{\text{i}=1}^\text{n}\big(\frac{\text{aX}}{\text{c}}+\frac{\text{b}}{\text{c}}-\frac{\text{a}}{\text{c}}\overline{\text{X}}-\frac{\text{b}}{\text{c}}\big)^2}{\text{n}}$
$=\frac{\sum\limits_{\text{i}=1}^\text{n}\big(\frac{\text{aX}}{\text{c}}-\frac{\text{a}}{\text{c}}\overline{\text{X}}\big)^2}{\text{n}}$
$=\Big(\frac{\text{a}}{\text{c}}\Big)^2\frac{\sum\limits_{\text{i}=1}^{\text{n}}\big(\text{x}_\text{i}-\overline{\text{X}}\big)^2}{\text{n}}$
$=\Big(\frac{\text{a}}{\text{c}}\Big)^2\sigma^2$
$\text{SD of Y}\big(\sigma_\text{y}\big)=\sqrt{\Big(\frac{\text{a}}{\text{c}}\Big)^2\sigma^2}$
$=\Big|\frac{\text{a}}{\text{c}}\Big|\sigma$
$\text{ac}<0$
$\Rightarrow\text{a}<0\text{ or }\text{c}<0$
$\therefore\Big|\frac{\text{a}}{\text{c}}\Big|=-\frac{\text{a}}{\text{c}}$
$\Rightarrow\sigma\text{Y}=-\frac{\text{a}}{\text{c}}\sigma\text{X}$

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MCQ 1421 Mark

If 16 observations are arranged in ascending order, then median is:

A
8th observation
B
9th observation
✓
$ \frac{8\text{th}\ \text{observation}\ +\ 9\text{th}\ \text{observation}}{2}$
D
$ \frac{7\text{th}\ \text{observation}\ +\ 8\text{th}\ \text{observation}}{2}$

Answer

Correct option: C.

$ \frac{8\text{th}\ \text{observation}\ +\ 9\text{th}\ \text{observation}}{2}$

For even number of observations median is the mean of $ \frac{\text{n}}{2}$
th observation and $ \Big(\frac{\text{n}}{2}+1\Big)$th observation:
So, median of 16 observation$= \frac{8\text{th}\ \text{observation}\ +\ 9\text{th}\ \text{observation}}{2}$

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MCQ 1431 Mark

If for a sample of size 60, we have the following information $\sum\text{x}_\text{i}^2=18000$ and $\sum\text{x}_\text{i}=960$ then the variance is:

A
6.63
B
16
C
22
✓
44

Answer

Correct option: D.

Given $\sum\text{x}_\text{i}^2=18000,\ \sum\text{x}_\text{i}=960$ and n = 60
$\therefore$ Variance
$=\frac{\sum\text{x}_\text{i}^2}{\text{n}}-\bigg(\frac{\sum\text{x}_\text{i}}{\text{n}}\bigg)^2$
$=\frac{18000}{60}-\Big(\frac{960}{60}\Big)^2$
$=300-256$
$=44$
Hence, the correct answer is option (d).

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MCQ 1441 Mark

The mean of 10 observation is 25. If one observation namely 25, is deleted, the new mean is:

✓
25
B
20
C
28
D
22

Answer

Correct option: A.

Mean of 1010 observations = 25
Sum of 1010 observations = 25 × 10 = 250
After removing an observation with a value = 25
New sum = 250 - 25 = 225
New $ \text{mean} = \frac{225}{9} = 25$

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MCQ 1451 Mark

Emmy did a survey of how many games each of 2020 friends owned, and got the following data: 5, 7, 12, 13, 4, 6, 8, 12, 9, 16, 13, 12, 5, 13, 7, 17, 3, 9, 12, 14. Find the mean:

A
8.55
B
7.59
C
5.49
✓
9.85

Answer

Correct option: D.

9.85

9.85

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MCQ 1461 Mark

The variance is the _______ of the standard deviation:

✓
Square
B
Cube
C
Square root
D
Cube root

Answer

Correct option: A.

Square

Square

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MCQ 1471 Mark

Seven of the eight numbers in a distribution are $11, 16,6, 10, 13, 11, 13$. Given that the mean of the distribution is $12,$ if $12$ will be included then find the new mean of the distribution.

A
$12$
✓
$11.5$
C
$16$
D
$12.2$

Answer

Correct option: B.

$11.5$

$ \text{mean}=\frac{6+10+11+11+13+13+13+16+12}{8}=11.6$

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MCQ 1481 Mark

The average age of 15 students of a class is 15 years. Out of these, the average age of 5 students is 14 years and that of the other nine students is 16 years. What is the age of the 15th student?

A
17 years
B
13 years
✓
11 years
D
18 years

Answer

Correct option: C.

11 years

Total age of 15 students = (15 × 15)years = 225years
Total age of 5 students = (5 × 14)years = 70years
Total age of other 9 students = (9 × 16)years = 144years
$ \therefore$ Age of the 15th student = 225 - (70 + 144) = 225 - 214 = 11 years.

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MCQ 1491 Mark

The mean deviation of the data 3, 10, 10, 4, 7, 10, 5 from the mean is:

A
2
✓
2.57
C
3
D
3.57

Answer

Correct option: B.

2.57

The given observations are 3, 10, 10, 4, 7, 10, 5.
$\therefore\text{Mean},\ \overline{\text{x}}=\frac{3+10+10+4+7+10+5}{7}=\frac{49}{7}=7$
Now,
Mean deviation from mean, MD
$=\frac{\sum|\text{x}_\text{i}-7|}{7}$
$=\frac{|3-7|+|10-7|+|10-7|+|4-7|+|7-7|+|10-7|+|5-7|}{7}$
$=\frac{4+3+3+3+0+3+2}{7}$
$=\frac{18}{7}$
$=2.57$
Hence, the correct answer is (b).

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MCQ 1501 Mark

If different values of variable x are 9.8, 5.4, 3.7, 1.7, 1.8, 2.6, 2.8, 8.6, 10.5 and 11.1; find the mean:

✓
5.8
B
7.8
C
9.8
D
None of these

Answer

Correct option: A.

5.8

Values of x are: 9.8, 5.4, 3.7, 1.7, 1.8, 2.6, 2.8, 8.6, 10.5and11.1
$\text{Mean}=\frac{\ \text{Sum}}{\text{Count}}$
$\text{Mean}=\frac{9.8+5.4+3.7+1.7+1.8+2.6+2.8+8.6+10.5+11.1}{10}$
$ \text{Mean}=\frac{58}{10}$
$\text{Mean}=5.8$

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MATHS MCQ