MCQ

MCQ 511 Mark

The most frequently occurring data value in a data set is the __________.

A
median
B
arithmetic mean
C
population parameter
✓
mode

Answer

Correct option: D.

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.

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MCQ 521 Mark

Choose the correct answer. Let $\mathrm{x}_1, \mathrm{x}_2, \ldots, \mathrm{x}_n$ be n observations and $\overline{\mathrm{x}}$ be their arithmetic mean. The formula for the standard deviation is given by:

A
$\sum(\text{x}_\text{i}-\bar{\text{x}})^2$
B
$\frac{\sum(\text{x}_\text{i}-\bar{\text{x}})^2}{\text{x}}$
✓
$\sqrt{\frac{\sum(\text{x}_\text{i}-\bar{\text{x}})^2}{\text{n}}}$
D
$\frac{\sum\text{x}_\text{i}^2}{\text{n}}-(\bar{\text{x}})^{-2}$

Answer

Correct option: C.

$\sqrt{\frac{\sum(\text{x}_\text{i}-\bar{\text{x}})^2}{\text{n}}}$

$\sqrt{\frac{\sum(\text{x}_\text{i}-\bar{\text{x}})^2}{\text{n}}}$

Solution:
The formula for $\text{S.D}=\sigma=\sqrt{\frac{\sum(\text{x}_\text{i}-\bar{\text{x}})^2}{\text{n}}}$

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MCQ 531 Mark

Choose the correct answer. The mean deviation of the data 3, 10, 10, 4, 7, 10, 5 from the mean is:

A
2
✓
2.57
C
3
D
3.75

Answer

Correct option: B.

2.57

2.57

Solution:
Observations are fiven by 3, 10, 10, 4, 7, 10, and 5
$\therefore\ \bar{\text{x}}=\frac{3+10+10+4+7+10+5}{7}=\frac{49}{7}=7$
$\text{MD}=\frac{\sum\text{d}_\text{i}}{\text{n}}=\frac{18}{7}=2.57$

$x_i$	$\text{d}_\text{i}=\|\text{x}_\text{i}-\bar{\text{x}}\|$
3	4
10	3
10	3
4	3
7	0
10	3
5	2
Total	$\sum\text{d}_\text{i}=18$

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MCQ 541 Mark

The sum $ \displaystyle \sum _{\text{ r}=1 }^{ 10 }{ \left( {\text{ r} }^{ 2 }+1 \right) \times \left( \text{r}\ ! \right) }$ is equal to:

A
(11)!
✓
10 × (11)!
C
101 × (10)!
D
11 × (11)!

Answer

Correct option: B.

10 × (11)!

$ \sum(\text{r}^2+1)\text{r}!=\sum[\text{r}(\text{r }+1) -(\text{r}-1)\text{r}!$
$ =\sum\limits^{10}_\text{r=1}[\text{r}(\text{r }+1)!-(\text{r}-1)\text{r}!]$
$= (1×2!−0×1!)+(2×3!−1×2!)+......+(10×11!−9×10!)=10×11! $

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MCQ 551 Mark

If the first and the second letters of the word MISJUDGEMENTS are interchanged with the last and the second last letters respectively, and similarly the third and the fourth letters are interchanged with the third and the fourth letters from the last respectively , and so on,then what will be the fifth letter to the right of the third letter from the left end?

A
E
B
G
✓
D
D
T

Answer

Correct option: C.

M	I	S	J	U	D	G	E	M	E	N	T	S
1	2	3	4	5	6	7	8	9	10	11	12	13

The alphabet series is shown in the above diagram with positions.
Now, with the required interchange in positions,
as the question says, we get following letters interchanged1st → 13th
2nd → 12th3rd → 11th
4th → 10th
5th → 9th
6th → 8th
7th → 7thUsing above mentioned interchange we get complete reversal of the letters of this word. which is:
STNEMEGDUJSIM
Here Third letter from the left end is N. According to the question,
⇒ (5 + 3)th letter from the left
⇒ 8th letter from the left Hence, 8th letter from the left end is D.

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MCQ 561 Mark

The mean of x, x + 3, x + 4, x + 8 and x + 10:

A
x + 4
B
x + 8
C
x + 3
✓
x + 5

Answer

Correct option: D.

x + 5

By definition
$ \text{Average} =\cfrac{\text{x}+(\text{x}+3)+(\text{x}+4)+((\text{x}+8)+(\text{x}+10)}{5}$
$ =\cfrac{5\text{x}+25}{5}$
$ =(\text{x}+5)$

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MCQ 571 Mark

The average can be found only in __________ variables:

A
string
B
qualitative
✓
quantitative
D
none

Answer

Correct option: C.

quantitative

The average can be found only in quantitative variables.
Example: A quantitative variable is something that
can be measured and written out as a number.
So, we can find the average marks of 2020 students in 1212 class but,
we can not find the average of the
cleverness of students.

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MCQ 581 Mark

Choose the correct answer. Coefficient of variation of two distributions are 50 and 60, and their arithmetic means are 30 and 25 respectively. Difference of their standard deviation is:

✓
0
B
1
C
1.5
D
2.5

Answer

Correct option: A.

Here, $\text{CV}_1=50,\ \text{CV}_2=60,\ \bar{\text{x}}_1=30$ and $\bar{\text{x}}_2=25$
$\therefore\ \text{CV}_1=\frac{\sigma_1}{\bar{\text{x}}_1}\times100$
$\Rightarrow50=\frac{\sigma_1}{30}\times100$
$\therefore\ \sigma_1=\frac{30\times50}{100}=15$
and $\text{CV}_2=\frac{\sigma_2}{\bar{\text{x}}_2}\times100$
$\Rightarrow60=\frac{\sigma_2}{25}\times100$
$\therefore\ \sigma^2=\frac{60\times25}{100}=15$
Now, $\sigma_1-\sigma_2=15-15=0$

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MCQ 591 Mark

If the mode of five observations, in order, 0, 2, 3, m, 5 is 3 then m = _______.

A
5
B
2
✓
3
D
0

Answer

Correct option: C.

If the mode of five observation, in order, 0, 2, 3, m, 5 is 3, then m=3.

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MCQ 601 Mark

For dealing with qualitative data the best average is:

A
A.M.
B
G.M.
C
Mode
✓
Median

Answer

Correct option: D.

Median

Median is the middle most value.
Also for even number of observations, median is the average of to middle values.
Hence, it divides the whole series into two equal halv.
It gives the more accurate and best average for qualitative data.

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MCQ 611 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.

We have,
$= \frac{7+8+\text{x}+11+14}{5}$
$ = \text{x} = 40+\text{x} = 5\text{x} = \text{x} = 10$

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MCQ 621 Mark

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

A
50,000
B
250,000
✓
252500
D
255000

Answer

Correct option: C.

252500

Let $\overline{\text{x}}$ and $\sigma$ be the mean and standard deviation of 100 observations, respectively.
$\therefore\overline{\text{x}}=50,\ \sigma=5$ and n = 100
$\text{Mean},\ \overline{\text{x}}=50$
$\Rightarrow\frac{\sum\text{x}_\text{i}}{100}=50$
$\Rightarrow\sum\text{x}_\text{i}=5000\ ...(1)$
Now,
Standard deviation, $\sigma=5$
$\Rightarrow\sqrt{\frac{\sum\text{x}_\text{i}^2}{100}-\Big(\frac{\sum\text{x}_\text{i}}{100}\Big)^2}=5$
$\Rightarrow\frac{\sum\text{x}_\text{i}^2}{100}-\Big(\frac{5000}{100}\Big)^2=25$ [From (1)]
$\Rightarrow\frac{\sum\text{x}_\text{i}^2}{100}=25+2500=2525$
$\Rightarrow{\sum\text{x}_\text{i}^2}=252500$
Thus, the sum of all squares of all the observations is 252500.
Hence, the correct answer is option (c).

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MCQ 631 Mark

The following data has been arranged in ascending order. If their median is 63, find the value of x.34, 37, 53, 55, x, x + 2, 77, 83, 89 and 100.

A
65
B
68
✓
62
D
70

Answer

Correct option: C.

Solution:
The series in ascending order is: 34, 37, 53, 55, x, x + 2, 77, 83, 89 and
The series has 10 numbers, even numbers.
Hence, the median will be the mean of the two middle numbers:
median = mean of $5^{th}$ and $6^{th}$ terms
$63=\frac{\text{x}\ + \ \text{x}\ +\ 2}{2}$
$1260=2\text{x}+2$
$ 2\text{x}=124$
$ \text{x}=62$

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MCQ 641 Mark

If 150 is the mean of 200 observations and 100 is the mean of some 300 other observations, find the mean of the combination:

A
90
B
100
✓
120
D
150

Answer

Correct option: C.

120

Mean of 200 observations = 150
Sum of 200 observations = 200 × 150 = 30000
Mean of 300 observations = 100
Sum of 300 observations = 300 × 100 = 30000
Total Sum = 30000 + 30000 = 60000
Number of observations = 100 + 200 = 500
$\text{Mean} = \frac{\text{Sum}}{\text{Number of observations}}$
$ = \frac{60000}{500} = 120$

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MCQ 651 Mark

if $x_1, x_2, x_3, x_4, x_5$ are five consecutive odd numbers, then their average is:

A
$x_2$
✓
$x_3$
C
$x_4$
D
$x_5$

Answer

Correct option: B.

$x_3$

$x_3$

Solution:
The five consecutive odd numbers are $ \text{x}_1+ \text{x}_1+2, \text{x}_1 + 4,\text{x}_1 +6,\text{x}_1 +8$
$ \therefore \text{mean}=\frac{\text{x}_1 \ + \ \text{x}_1 \ +\ 2+\text{x}_1 \ +\ 4\ +\ \text{x}_1 \ +\ 6\text{x}_1 \ +\ 8}{6}$
$ =\frac{5\text{x}_1\ +\ 20}{5}$
$ =\text{x}_1\ +\ 4=\text{x}_3$

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MCQ 661 Mark

The standard deviation of the observations 6, 5, 9, 13, 12, 8, 10 is:

A
$6$
B
$\sqrt6$
C
$\frac{52}{7}$
✓
$\sqrt{\frac{52}{7}}$

Answer

Correct option: D.

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

The given observations are 6, 5, 9, 13, 12, 8, 10.
Now,
$\sum\text{x}_\text{i}=$ 6 + 5 + 9 + 13 + 12 + 8 + 10 = 63
$\sum\text{x}_\text{i}^2=$ 36 + 25 + 81 + 169 + 144 + 64 + 100 = 619
$\therefore$ Standard deviation of the observations, $\sigma$
$=\sqrt{\frac{1}{\text{N}}\sum\text{x}_\text{i}^2-\Big(\frac{1}{\text{N}}\sum\text{x}_\text{i}\Big)^2}$
$=\sqrt{\frac{1}{7}\times619-\Big(\frac{1}{7}\times63\Big)^2}$
$=\sqrt{\frac{619}{7}-81}$
$=\sqrt{\frac{619-567}{7}}$
$=\sqrt{\frac{52}{7}}$
Hence, the correct answer is option (d).

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MCQ 671 Mark

The average of a collection of 20 measurements was calculated to be 56 cm. But later it was found that a mistake occured in one of the measurements which was recorded as 64 cm but should have been 61 cm. What is the correct average?

A
39.55cm
B
29.55 cm
✓
55.85 cm
D
50.75cm

Answer

Correct option: C.

55.85 cm

Incorrect total of 20 measurement = 20 × 56 = 1120
Correct total = 1120 - 64 + 61 = 1117
$ ∴ \text{Correct average} = \displaystyle \frac{1117}{20} = 55.85\text{cm}$

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MCQ 681 Mark

If the mean of five observations x ,x + 2, x + 4, x + 6 and x + 8 is 11, then the mean of last three obsevations is:

A
11
✓
13
C
15
D
17

Answer

Correct option: B.

Given observations x, x + 2, x + 4, x + 6, x + 8
$\Rightarrow\frac{5\text{x}+ 20}{5}=11$
⇒ x = 7
So, the observations are 7, 9, 11, 13, 15
Req. mean $ =\frac{11+13+15}{3}=13$

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MCQ 691 Mark

The standard deviation of the data:

x	1	a	$a^2$	....	$a^n$
f	$^nC_0$	$^nC_1$	$^nC_2$	....	$^nC_2$

is,

A
$\Big(\frac{1+\text{a}^2}{2}\Big)^\text{n}-\Big(\frac{1+\text{a}}{2}\Big)^\text{n}$
B
$\Big(\frac{1+\text{a}^2}{2}\Big)^{2\text{n}}-\Big(\frac{1+\text{a}}{2}\Big)^\text{n}$
C
$\Big(\frac{1+\text{a}^2}{2}\Big)^{2\text{n}}-\Big(\frac{1+\text{a}^2}{2}\Big)^\text{n}$
✓
None of these

Answer

Correct option: D.

None of these

None of these

Solution:

x$_i$	$f_i$	$f_ix_i$	$x_i^2$	$f_ix_i^2$
1	$^nC_0$	$^nC_0$	1	1
a	$^nC_1$	$a^nC_1$	$a^2$	$a^2{}^nC_1$
a	$^nC_2$	$a^2{}^nC_2$	$a^4$	$a^{4 n}C_2$
a	$^nC_3$	$a^{3 n}C_3$	$a^6$	$a^{6 n}C_3$
: : :	: : :	: : :	: : :	: : :
$a^n$	$^nC_n$	$a^{n n}C_n$	$a^{2n}$	$a^{2n n}C_n$
	$\sum\limits_{\text{i}=1}^{\text{n}}\text{f}_\text{i}=2^\text{n}$	$\sum\limits_{\text{i}=1}^{\text{n}}\text{f}_\text{i}\text{x}_\text{i}=(1+\text{a})^\text{n}$		$\sum\limits_{\text{i}=1}^{\text{n}}\text{f}_\text{i}\text{x}_\text{i}^2=(1+\text{a}^2)^\text{n}$

Number of terms, $\text{N}=\sum\limits_{\text{i}=1}^\text{n}\text{f}_\text{i}=2^\text{n}$
$\sum\limits_{\text{i}=1}^\text{n}\text{f}_\text{i}\text{x}_\text{i}=^\text{n}\text{C}_0+\text{a}^\text{n}\text{C}_1+\text{a}^2{^\text{ n}\text{C}_2+...+\text{a}^\text{n}{^\text{ n}\text{C}_\text{n}}}=(1+\text{a})^\text{n}$
$\overline{\text{X}}=\frac{\sum\limits_{\text{i}=1}^\text{n}\text{f}_\text{i}\text{x}_\text{i}}{\text{N}}$
$=\frac{(1+\text{a})^\text{n}}{2^\text{n}}$
$\sum\limits_{\text{i}=1}^\text{n}\text{f}_\text{i}\text{x}_\text{i}^2=(1+\text{a}^2)^\text{n}$
$\sigma^2=\text{Variance}(\text{X})=\frac{1}{\text{N}}\sum\limits_{\text{i}=1}^\text{n}\text{f}_\text{i}\text{x}_\text{i}^2-\Bigg(\frac{\sum\limits_{\text{i}=1}^\text{n}\text{f}_\text{i}\text{x}_\text{i}}{\text{N}}\Bigg)^2$
$=\frac{(1+\text{a}^2)^\text{n}}{2^\text{n}}-\Big[\frac{(1+\text{a}^2)^\text{n}}{2^\text{n}}\Big]^2$
$=\Big[\frac{1+\text{a}^2}{2}\Big]^\text{n}-\Big[\frac{1+\text{a}}{2}\Big]^2\text{n}$
$\sigma=\sqrt{\text{Variance}(\text{X})}$
$=\sqrt{\Big[\frac{1+\text{a}^2}{2}\Big]^\text{n}-\Big[\frac{1+\text{a}}{2}\Big]^2\text{n}}$

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MCQ 701 Mark

Of the three numbers, the first is twice the second, and the second is twice the third. The average of the reciprocal of the numbers is $ \frac{7}{72}$. What are the three numbers?

✓
24, 12, 6
B
8, 4, 2
C
12, 6, 3
D
None of the above

Answer

Correct option: A.

24, 12, 6

Let the third number be x Then Second number = 2x
and first number = 4x
Sum of the reciprocals of these 3 numbers $ =\frac{1}{\text{4x}}+\frac{1}{2\text{x}}+\frac{1}{\text{x}}=\frac{1+2+4}{4\text{x}}$
$= \frac{7}{\text{4x}}$
Given, $\frac{7}{4\text{x}}=3\times \frac{7}{72}$
$=4\text{x}=24= \text{x}=6 $
Therefore, the three numbers are,
4 × 6, 2 × 6, 6 i.e. 24, 12, 6.

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MCQ 711 Mark

The weight of a body, calculated as the average of seven different experiments is 53.735g.The average of the first three experiments is 54.005g. The fourth was greater than the fifth by 0.0040.004 g and the average of sixth and seventh was 0.010g less than the average of the first three. Find the weight of the body in the fourth experiment.

A
52.071g
✓
53.072g
C
51.450g
D
53.005g

Answer

Correct option: B.

53.072g

53.072g

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MCQ 721 Mark

The mean of 100 numbers is 45. The mean of the last 99 numbers is 44. The first number is

A
143
B
141
✓
144
D
140

Answer

Correct option: C.

144

The mean of 100 numbers is 45
$ ∴$ Sum of all 100 numbers = 100 × 45 = 4500
The mean of last 99 numbers is 44
$ ∴$Sum of all last 99 numbers = 99 × 44 = 4356
⇒ The first number = 4500 - 4356 = 144

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MCQ 731 Mark

Which of the following is not changed for the observations 31, 48, 50, 60, 25, 8, 3x, 26, 32? (where x lies between 10 and 15):

A
A.M.
✓
Range
C
Median
D
Q.D.

Answer

Correct option: B.

Range

Range

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MCQ 741 Mark

Find the mean of first six natural numbers:

A
3.6
B
7
✓
3.5
D
None of these

Answer

Correct option: C.

3.5

First six natural numbers = 1, 2, 3, 4, 5, 6
$\text{ Mean} = \frac{\text{Sum}}{\text{Number of observations}} $
$\text{Mean} = \frac{1 + 2 + 3 + 4 + 5 +6}{6} = \dfrac{21}{6}$
$ =3.5$

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MCQ 751 Mark

On Thursday, 20 of the 25 students in a chemistry class took a test and their average (arithmetic mean) was 80. On Friday, the other 5 students took the test and their average (arithmetic mean) was 90. What was the average for the entire class?

✓
82
B
83
C
84
D
85

Answer

Correct option: A.

Average $ = \frac{20\left ( 80 \right )\ +\ 5\left ( 90 \right )}{25}$
$=\frac{1600\ +\ 450}{25}$
$=\frac{2050}{25}=82$

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MCQ 761 Mark

The mean deviation of the numbers 3, 4, 5, 6, 7 from the mean is:

A
25
B
5
✓
1.2
D
0

Answer

Correct option: C.

1.2

$\text{Mean}(\overline{\text{X}})=\frac{3+4+5+6+7}{5}$
$=\frac{25}{5}$
$=5$
Taking the absolute value of deviation of each term from the mean, we get:
$\text{MD}=\frac{|(3-5)|+|(4-5)|+|(5-5)|+|(6-5)|+|(7-5)|}{5}$
$=\frac{2+1+0+1+2}{5}$
$=\frac{6}{5}$
$=1.2$

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MCQ 771 Mark

When there are 2 observations in the middle, median is calculated by ______.

A
taking both the values as median
B
taking the mean of the two observations
C
$ \frac{(\text{N}+1)}{ 2}$
✓
both B and C

Answer

Correct option: D.

both B and C

Median is the middle most value of a series.
So when the series has odd number of elements then,
median can be calculated easily but when the series has even number of elements then,
The series has two middle values, so
median is calculated either by taking out the average of both the
value or the median is the$ \frac{(\text{N}+1)}{ 2}$ th element of the series.

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MCQ 781 Mark

The average age of a group of eight is same as it was 3 years ago when a young member is substituted for an old member the incoming member is younger to the outgoing member by:

A
11 years
✓
24 years
C
28 years
D
16 years

Answer

Correct option: B.

24 years

24 years

Solution:
let presently the member be $x_1, x_2, x_3 \ldots . x_8$
So the average age$ =\dfrac{\text{x}_1+\text{x}_2+\text{x}_3+.......+\text{x}_8}{8}$
......1Now the average age of all the members 3 years ago
$ =\dfrac{\text{x}_1+\text{x}_2+\text{x}_3+.......+\text{x}_8-24}{8}$
if $x_1$ the younger member is replaced by the older member $y_1$ then,
$ =\dfrac{\text{x}_1+\text{x}_2+\text{x}_3+.......+\text{x}_8}{8}$
$ =\dfrac{\text{x}_1+\text{x}_2+\text{x}_3+.......+\text{x}_8-24}{8}$
$ ⇒\text{x}1=\text{y}1+24\Rightarrow \text{x}_1-\text{y}_1=24 $

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MCQ 791 Mark

The mean of x, y, z is y, then x + z =:

A
y
✓
2y
C
3y
D
zy

Answer

Correct option: B.

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MCQ 801 Mark

Choose the correct answer. Standard deviations for first 10 natural numbers is:

A
5.5
B
3.87
C
2.97
✓
2.87

Answer

Correct option: D.

2.87

We know that SD of first n natural numbers $\sqrt{\frac{\text{n}^2-1}{12}}$
Here, $\text{n}=10$
$\therefore\ \text{SD}=\sqrt{\frac{(10)^2-1}{12}}=\sqrt{\frac{99}{12}} $
$=\sqrt{8.25}=2.87$

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MCQ 811 Mark

Given the following data set, what is the value of median (2 4 3 6 1 8 9 2 5 7).

A
2
B
4.7
✓
4.5
D
10

Answer

Correct option: C.

4.5

Median is the middle most value of a series.
So when the series has odd number of elements then median can be calculated easily but, when the series has even number of elements then the series has two middle values, so median is calculated by taking out the average of both the value.
The given series is first arranged into ascending; 1, 2, 2, 3, 4, 5, 6, 7, 8, 9
$\text{N} = 10$
$ \text{median}= \frac{(10+1)}{2}$
$ \text{th term} = \frac{11}{2}$
$ \text{th term} = 5.5 $
$ = \frac{( \text{value of 5th term }+ \text{value of 6th term)}}{2}$
$ = \frac{(4+5)}{2} = \frac{9}{2} = 4.5$

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MCQ 821 Mark

Size	2	3	4	5	6	7	8
Frequency	10	12	25	20	25	15	11

A
2
B
8
✓
Both 4 and 6
D
5

Answer

Correct option: C.

Both 4 and 6

Mode is that observation which have highest frequency. Since, both 4 and 6 have highest frequency
i.e. 25 and 25, they are the mode of the given distribution.
Hence, option (C) is correct.

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MCQ 831 Mark

Choose the correct answer. Let $\mathrm{x}_1, \mathrm{x}_2, \mathrm{x}_3, \mathrm{x}_4, \mathrm{x}_5$ be the observations with mean m and standard deviation s . The standard deviation of the observations $\mathrm{kx}_1, \mathrm{kx}_2, \mathrm{kx}_3, \mathrm{kx}_4, \mathrm{kx}_5$ is:

A
k + s
B
$\frac{\text{s}}{\text{k}}$
✓
ks
D
s

Answer

Correct option: C.

Solution:
Here, $\text{m}=\frac{\sum\text{x}_\text{i}}{\text{N}},$
$\text{S}=\sqrt{\frac{\sum\text{x}_\text{i}^2}{5}-\Big(\frac{\sum\text{x}_\text{i}}{5}\Big)^2}$
$\therefore\ \text{SD}=\sqrt{\frac{\text{K}^2\sum\text{x}_\text{i}^2}{5}-\Big(\frac{\text{K}\sum\text{x}_\text{i}}{5}\Big)^2}$
$=\sqrt{\frac{\text{K}^2\sum\text{x}_\text{i}^2}{5}-\text{K}^2\Big(\frac{\sum\text{x}_\text{i}}{5}\Big)^2}$
$=\text{K}=\sqrt{\frac{\sum\text{x}_\text{i}^2}{5}-\Big(\frac{\sum\text{x}_\text{i}}{5}\Big)^2}$
$=\text{K}.\text{S}$

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MCQ 841 Mark

Mean of a set of 23 values is 7, if each value is multiplied by 23 the new mean is:

A
529
✓
161
C
30
D
49

Answer

Correct option: B.

161

⇒ Sum of the observation = 23 × 7 = 161If each observation is multiplied by 23 then the sum is also multiplied by 23.
New sum = 23 × 161 = 3703
⇒ New mean $ =\frac{3703}{23}=161$

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MCQ 851 Mark

A batsman scores runs in 10 innings as 38, 70, 48, 34, 42, 55, 63, 46, 54 and 44. The mean deviation about mean is:

✓
8.6
B
6.4
C
10.6
D
7.6

Answer

Correct option: A.

8.6

Solution:
N = 10
$\overline{\text{X}}=\frac{38+70+48+34+42+55+63+46+54+44}{10}$
$=\frac{494}{10}$
$=49.4$

$x_i$	$\text{d}_\text{i}=\big\|\text{x}_\text{i}-49.4\big\|$
34	15.4
38	11.4
42	7.4
44	5.4
46	3.4
48	1.4
54	4.6
55	5.6
63	13.6
70	20.6
	$\sum\limits^{\text{n}}_{\text{i}=}\text{d}_\text{i}=88.8$

Mean deviation from the mean $=\frac{88.8}{10}$
$= 8.88$
Disclaimer: No option is matching the answer.

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MCQ 861 Mark

Median divides the total frequency into _____ equal parts:

✓
2
B
3
C
4
D
None of these

Answer

Correct option: A.

The 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.

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MCQ 871 Mark

The average of $ \displaystyle 1\frac{1}{6},2\frac{1}{3},6\frac{2}{3}161,231,632 \text{ and } \displaystyle 8\frac{5}{6}865 $ is:

A
$ \displaystyle 6\frac{3}{4}$
B
$ \displaystyle 5\frac{3}{4}$
✓
$ \displaystyle 4\frac{3}{4}$
D
$ \displaystyle 3\frac{3}{4}$

Answer

Correct option: C.

$ \displaystyle 4\frac{3}{4}$

$= \displaystyle 1\frac{1}{6}+2\frac{1}{3}+6\frac{2}{3}+ 8\frac{5}{6}$
$=\frac{7}{6}+ \frac{7}{3 }+ \frac{20}{3 }+\frac{53}{6}$
$ = \frac{6+7+14+40+53}{6} $
$= \frac{114}{6}$
$ 19\therefore \text{Average}=\frac{19}{4}=4\dfrac{3}{4}$

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MCQ 881 Mark

The average age of a teacher and three students is 20 years. If all students are of equal age and the difference between the age of the teacher and that of a student is 20 years, then the age of the teacher is:

A
25 years
B
30 years
✓
35 years
D
45 years

Answer

Correct option: C.

35 years

Let the age of each student be x years
Then, the age of teacher will be (x + 20) years
$\text{Mean age} =\frac{\left (\text{x}+20 \right )+3\text{x}}{4}$
$20=\frac{\text{4x}+20}{4}$
⇒ x = 15
Hence, age of the teacher = 35 years

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MCQ 891 Mark

The average of four consecutive even numbers is one fourth of the sum of these numbers. What is the difference between the first and last number?

A
4
✓
6
C
2
D
8

Answer

Correct option: B.

Let the numbers be 2x - 2, 2x, 2x + 2 and 2x + 4, where x is a natural number.
Then the difference between the first and last number = 2x + 4 - (2x - 2) = 6.

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MCQ 901 Mark

The most frequent value in a data set is?

A
Median
✓
Mode
C
Arithmetic mean
D
Geometric mean

Answer

Correct option: B.

Mode

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MCQ 911 Mark

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

The given numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.
If 1 is added to each number, then the new numbers obtained are
2, 3, 4, 5, 6, 7, 8, 9, 10, 11
Now,
$\sum\text{x}_\text{i}=$ 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 = 65
$\sum\text{x}_\text{i}^2=$ 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 + 100 + 121 = 505
$\therefore$ Variance of the numbers so obtained
$=\frac{\sum\text{x}_\text{i}^2}{10}-\Big(\frac{\sum\text{x}_\text{i}}{10}\Big)^2$
$=\frac{505}{10}-\Big(\frac{65}{10}\Big)^2$
$=50.5-42.25$
$=8.25$

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MCQ 921 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?

✓
The order of numbers.
B
14
C
18
D
None of these

Answer

Correct option: A.

The order of numbers.

To 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.

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MCQ 931 Mark

The captain of a cricket team of 11 members is 26 years old and the wicket keeper is 3 years older. If the ages of these two are excluded, the average age of the remaining players is one year less than the average age of the whole team. What is the average age of the team?

✓
23 years
B
24 years
C
25 years
D
None of these

Answer

Correct option: A.

23 years

Let the average age of the whole team by x years.
= 11 x - (26 + 29) = 9(x - 1)
= 11x - 9x = 46
= 2x = 46 ⇒ 2x = 46
= x = 23 ⇒ x = 23.
So, average age of the team is 23 years.

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MCQ 941 Mark

Two high school classes took the same test. One class of 20 students made an average grade of 80%; the other class of 30 students made an average grade of 70%. The average grade for all students in both classes is:

A
75%
✓
74%
C
77%
D
None of these

Answer

Correct option: B.

74%

$ \text{Average}=\frac{20.80+30.70}{20+30}=74$

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MCQ 951 Mark

A school has 20 teachers one of them retires at the age of 60 years and a new teacher replaces him this change reduces the average age of the staff by 2 years the age of new teacher is:

A
28 years
B
25 years
✓
20 years
D
18 years

Answer

Correct option: C.

20 years

Let the average age of the staff
= x Age of the new teacher
= y According to the questionNew age of the staff reduced by
$ 2 \text{years}\Rightarrow \dfrac{20\text{x}-60+\text{y}}{20}$
$\text{ x}-2⇒20\text{x}−60+\text{y}$
$ \Rightarrow 20\text{x}-60+\text{y}$
$ =20(\text{x}-2)⇒20\text{x}−60+\text{y}$
$ ⇒20\text{x}−60+\text{y}$
$ \Rightarrow\text{y}=60-40=20$
$ ⇒\text{y}=60−40$
= 20 Hence the age of the new teacher is 20 years.

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MCQ 961 Mark

For a frequency distribution standard deviation is computed by applying the formula:

✓
$\sigma=\sqrt{\frac{\sum\text{fd}^2}{\sum\text{f}}-\Big(\frac{\sum\text{fd}}{\sum\text{f}}\Big)^2}$
B
$\sigma=\sqrt{\Big(\frac{\sum\text{fd}}{\sum\text{f}}\Big)^2-\frac{\sum\text{fd}^2}{\sum\text{f}}}$
C
$\sigma=\sqrt{\frac{\sum\text{fd}^2}{\sum\text{f}}-\frac{\sum\text{fd}}{\sum\text{f}}}$
D
$\sqrt{\Big(\frac{\sum\text{fd}}{\sum\text{f}}\Big)^2-\frac{\sum\text{fd}^2}{\sum\text{f}}}$

Answer

Correct option: A.

$\sigma=\sqrt{\frac{\sum\text{fd}^2}{\sum\text{f}}-\Big(\frac{\sum\text{fd}}{\sum\text{f}}\Big)^2}$

$\sigma=\sqrt{\frac{\sum\text{fd}^2}{\sum\text{f}}-\Big(\frac{\sum\text{fd}}{\sum\text{f}}\Big)^2}$

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MCQ 971 Mark

Choose the correct answer. Following are the marks obtained by 9 students in a mathematics test: 50, 69, 20, 33, 53, 39, 40, 65, 59 The mean deviation from the median is:

A
9
B
10.5
✓
12.67
D
14.76

Answer

Correct option: C.

12.67

12.67

Solution:
$\therefore\ \text{Median}=5^{\text{th}}\text{ term}$
$\text{M}_\text{e}=50$

$x_i$	$d_i = \|x_i - M_e\|$
20	30
33	17
39	11
40	10
50	0
53	3
59	9
65	15
69	19
N = 2	$\sum\text{d}_\text{i}=114$

$\therefore\ \text{MD}=\frac{114}{9}=12.67$

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MCQ 981 Mark

Mode is:

✓
most frequent value
B
least frequent value
C
middle most value
D
none of thes

Answer

Correct option: A.

most frequent value

Mode is the value that occurs most often For example:
13, 13, 12, 14, 13 The Mode of the following is 13.

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MCQ 991 Mark

The mean of 20 observations is 15 On checking it was found that the two observations were wrongly copied as 3 and 6. The correct values are 8 and 4 , then correct mean will be given by:

✓
15.15
B
14.69
C
14.74
D
15.25

Answer

Correct option: A.

15.15

Mean of 20 observatios = 15
Sum of 20 observations = 15 × 20 = 300
Correct sum = 300 + 8 + 4 - 3 - 6 = 300
Correct mean $ = \dfrac{303}{20} = 15.15$

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MCQ 1001 Mark

The mean of 6 numbers is 42 If one number is excluded, the mean of remaining numbers is 45. Find the excluded number:

✓
27
B
25
C
30
D
32

Answer

Correct option: A.

mean of 6 numbers = 42
Sum of 6 numbers = 42 × 6 = 252
After excluding one number,
mean of 5 numbers = 45
Sum of 5 numbers = 45 × 5 = 225
Thus, the number excluded = 252 - 225 = 27

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MATHS MCQ