Phrases Statistics Measures of Central Tendency Previous Year Question Paper and Solution with PDF

Aspire's Library

A Place for Latest Exam wise Questions, Videos, Previous Year Papers,
Study Stuff for MCA Examinations

Phrases
Previous Year Questions

Phrases
Previous Year Questions
Phrases

Phrases Previous Year Questions (PYQs)

Phrases Statistics Measures Of Central Tendency PYQ

Qus : 1

The mean of 5 observation is 5 and their variance is 12.4. If three of the observations are 1,2 and 6; then the mean deviation from the mean of the data is:

1
2
3
4
Go to Discussion

Solution

Qus : 2

If the mean deviation 1, 1+d, 1+2d, … , 1+100d from their mean is 255, then d is equal to

1
2
3
4
Go to Discussion

Solution

Qus : 3

and

, then a possible value of n is among the following is

1
2
3
4
Go to Discussion

Solution

Qus : 4

Not Available

1
2
3
4
Go to Discussion

Solution

Qus : 5

The 10th and 50th percentiles of the observation 32, 49, 23, 29, 118 respectively are

1
2
3
4
Go to Discussion

Solution

Qus : 6

The first three moments of a distribution about 2 are 1, 16, -40 respectively. The mean and variance of the distribution are

1
2
3
4
Go to Discussion

Solution

Qus : 7

${{a}}_1,{{a}}_2,\ldots,{{a}}_n$ are any real numbers and

$n$ is any positive integer, then

1
2
3
4
Go to Discussion

Solution

Qus : 8

In a reality show, two judges independently provided marks base do the performance of the participants. If the marks provided by the second judge are given by Y = 10.5 + 2x, where X is the marks provided by the first judge. If the variance of the marks provided by the second judge is 100, then the variance of the marks provided by the first judge is:

1
2
3
4
Go to Discussion

Solution

Qus : 9

The mean of 25 observations was found to be 38. It was later discovered that 23 and 38 were misread as 25 and 36, then the mean is

1
2
3
4
Go to Discussion

Solution

Qus : 10

Given a set A with median

$m_1 = 2$ and set B with median

$m_2 = 4$

What can we say about the median of the combined set?

1
2
3
4
Go to Discussion

Solution

Given two sets:

Set $A$ has median $m_1 = 2$
Set $B$ has median $m_2 = 4$

What can we say about the median of the combined set $A \cup B$ ?

✅ Answer:

The combined median depends on the size and values of both sets.

Without that information, we only know that:

$\text{Combined Median} \in [2, 4]$

So, the exact median cannot be determined with the given data.

Qus : 11

It is given that the mean, median and mode of a data set is

$1, 3^x$ and

$9^x$ respectively. The possible values of the mode is

1
2
3
4
Go to Discussion

Solution

Mean, Median, and Mode Relation

Given:

Mean = 1
Median = $3^x$
Mode = $9^x$

Use empirical formula:

$\text{Mode} = 3 \cdot \text{Median} - 2 \cdot \text{Mean}$

$9^x = 3 \cdot 3^x - 2 \Rightarrow (3^x)^2 = 3 \cdot 3^x - 2$

Let $y = 3^x$ , then:

$y^2 = 3y - 2 \Rightarrow y^2 - 3y + 2 = 0 \Rightarrow (y - 1)(y - 2) = 0$

So, $y = 1 \text{ or } 2 \Rightarrow 9^x = y^2 = 1 \text{ or } 4$

✅ Final Answer: $\boxed{1 \text{ or } 4}$

Qus : 12

If the mean of the squares of first n natural numbers be 11, then n is equal to?

1
2
3
4
Go to Discussion

Solution

Qus : 13

The standard deviation of 20 numbers is 30. If each of the numbers is increased by 4, then the new standard deviation will be

1
2
3
4
Go to Discussion

Solution

Qus : 14

If a variable takes values 0, 1, 2,…, 50 with frequencies

$1,\, {{50}}_{{{C}}_1},{{50}}_{{{C}}_2},\ldots..,{{50}}_{{{C}}_{50}}$ , then the AM is

1
2
3
4
Go to Discussion

Solution

Qus : 15

In a group of 200 students, the mean and the standard deviation of scores were found to be 40 and 15, respectively. Later on it was found that the two scores 43 and 35 were misread as 34 and 53, respectively. The corrected mean of scores is:

1
2
3
4
Go to Discussion

Solution

Qus : 16

The mean deviation from the mean of the AP a, a + d, a + 2d, ..., a + 2nd, is:

1
2
3
4
Go to Discussion

Solution

Qus : 17

Consider the following frequency distribution table.

Class Interval	10-20	20-30	30-40	40-50	50-60	60-70	70-80
Frequency	180	$f_1$	34	180	136	$f_2$	50

If the total frequency is 686 and the median is 42.6, then the value of

$f_1$ ;and

$f_2$ are

1
2
3
4
Go to Discussion

Solution

Qus : 18

If the mean deviation of the numbers 1, 1 + d, 1 + 2d, ....., 1 + 100d from their mean is 255, then the value of d is

1
2
3
4
Go to Discussion

Solution

Qus : 19

A, B, C are three sets of values of x:

A: 2,3,7,1,3,2,3

B: 7,5,9,12,5,3,8

C: 4,4,11,7,2,3,4

Select the correct statement among the following:

1
2
3
4
Go to Discussion

Solution

A: 2, 3, 7, 1, 3, 2, 3

Increasing Order : A: 1, 2, 2, 3, 3, 3, 7

Mode = 3 (occurs maximum number of times)

Median = 3 (the middle term)

Mean =

$\frac{(1+2+2+3+3+3+7)}{7}$

$=\frac{21}{7} = 3$

Hence. Mean=Median=Mode

Qus : 20

Standard deviation for the following distribution is

Size of item	6	7	8	9	10	11	12
Frequency	3	6	9	13	8	5	4

1
2
3
4
Go to Discussion

Solution

Total number of items in the distribution = Σ f_i = 3 + 6 + 9 + 13 + 8 + 5 + 4 = 48.

The Mean (x̅) of the given set = $\rm \dfrac{\sum f_i x_i}{\sum f_i}$ .

⇒ x̅ = $\rm \dfrac{6\times3+7\times6+8\times9+9\times13+10\times8+11\times5+12\times4}{48}=\dfrac{432}{48}$ = 9.

Let's calculate the variance using the formula: $\rm \sigma^2 =\dfrac{\sum x_i^2}{n}-\bar x^2$ .

$\rm \dfrac{\sum {x_i}^2}{n}=\dfrac{6^2\times3+7^2\times6+8^2\times9+9^2\times13+10^2\times8+11^2\times5+12^2\times4}{48}=\dfrac{4012}{48}$ = 83.58.

∴ σ² = 83.58 - 9² = 83.58 - 81 = 2.58.

And, Standard Deviation (σ) = $\rm \sqrt{\sigma^2}=\sqrt{Variance}=\sqrt{2.58}$ ≈ 1.607.

Qus : 21

The mean of 5 observation is 5 and their variance is 124. If three of the observations are 1,2 and 6; then the mean deviation from the mean of the data is:

1
2
3
4
Go to Discussion

Solution

Qus : 22

For a group of 100 candidates, the mean and standard deviation of scores were found to be 40 and 15 respectively. Later on, it was found that the scores 25 and 35 were misread as 52 and 53 respectively. Then the corrected mean and standard deviation corresponding to the corrected figures are

1
2
3
4
Go to Discussion

Solution

Corrected Mean and Standard Deviation

Original Mean: 40, Standard Deviation: 15

Two scores were misread: 25 → 52 and 35 → 53

Corrected Mean:

$\mu' = \frac{3955}{100} = \boxed{39.55}$

Corrected Standard Deviation:

$\sigma' = \sqrt{\frac{178837}{100} - (39.55)^2} \approx \boxed{14.96}$

✅ Final Answer: Mean = 39.55, Standard Deviation ≈ 14.96

Qus : 23

Consider the following frequency distribution table.

Class interval	10-20	20-30	30-40	40-50	50-60	60-70	70-80
Frequency	180	$f_1$	34	180	136	$f_2$	50

If the total frequency is 685 & median is 42.6 then the values of

$f_1$ and

$f_2$ are

1
2
3
4
Go to Discussion

Solution

Median & Frequency Table

Given: Median = 42.6, Total Frequency = 685

Using Median Formula:

$\text{Median} = L + \left( \frac{N/2 - F}{f} \right) \cdot h$

Median Class: 40–50
Lower boundary $L = 40$
Frequency $f = 180$
Class width $h = 10$
Cumulative freq before median class $F = 214 + f_1$

Substituting values:

$42.6 = 40 + \left( \frac{128.5 - f_1}{180} \right) \cdot 10 \Rightarrow f_1 = \boxed{82}$

Using total frequency:

$662 + f_2 = 685 \Rightarrow f_2 = \boxed{23}$

✅ Final Answer: $f_1 = 82,\quad f_2 = 23$

Qus : 24

The average marks of boys in a class is 52 and that of girls is 42. The average marks of boys and girls combined is 50. The percentage of boys in the class is

1
2
3
4
Go to Discussion

Solution

Phrases

Online Test Series,
Information About Examination,
Syllabus, Notification
and More.

Click Here to
View More

Phrases

Online Test Series,
Information About Examination,
Syllabus, Notification
and More.

Click Here to
View More

Ask Your Question or Put Your Review.

Feedback

Info.

Information

Aspire's Library

Phrases Previous Year Questions (PYQs)

Phrases Statistics Measures Of Central Tendency PYQ

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Given two sets:

✅ Answer:

Solution

Mean, Median, and Mode Relation

✅ Final Answer: 1 or 4\boxed{1 \text{ or } 4}

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Corrected Mean and Standard Deviation

Solution

Median & Frequency Table

Solution

Phrases

Phrases

Info.

Confirm Delete

Edit

✅ Final Answer: $\boxed{1 \text{ or } 4}$