Qus : 1 3 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 2.5 2 2.6 3 2.8 4 2.4 Go to Discussion Qus : 2 1 If the mean deviation 1, 1+d, 1+2d, … , 1+100d from their mean is 255, then d is equal to
1 10.1 2 10.2 3 10.02 4 10.4 Go to Discussion
Solution Qus : 3 4 If
and
, then a possible value of n is among the following is
1 9 2 12 3 16 4 18 Go to Discussion Qus : 5 1 The 10th and 50th percentiles of the observation 32, 49, 23, 29, 118 respectively are
1 21, 32 2 23, 32 3 23, 33 4 22, 31 Go to Discussion
Solution Qus : 6 3 The first three moments of a distribution about 2 are 1, 16, -40 respectively. The mean and variance of the distribution are
1 (2,16) 2 (2,15) 3 (3,15) 4 (1,16) Go to Discussion Qus : 7 4 If {{a}}_1,{{a}}_2,\ldots,{{a}}_n are any real numbers and n is
any positive integer, then
1 n\sum ^n_{i=1}{{{a}}_i}^2{\lt}{\Bigg{(}\sum ^n_{i=1}{{{a}}_i}\Bigg{)}}^2 2 n\sum ^n_{i=1}{{{a}}_i}^2{\geq}{\Bigg{(}\sum ^n_{i=1}{{{a}}_i}\Bigg{)}}^2 3 \sum ^n_{i=1}{{{a}}_i}^2{\geq}{\Bigg{(}\sum ^n_{i=1}{{{a}}_i}\Bigg{)}}^2 4 None of the above Go to Discussion
Solution Qus : 8 2 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 49.5 2 25 3 50
4 99 Go to Discussion Solution
Statistics Puzzle: Variance under Linear Transformation
Given:
Y = 10.5 + 2X
Var(Y) = 100
Formula: If Y = a + bX, then Var(Y) = b² × Var(X)
Apply:
100 = 2² × Var(X)
100 = 4 × Var(X)
Var(X) = 100 / 4 = 25
✅ Final Answer: The variance of the marks given by the first judge is 25 .
Qus : 9 3 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 32 2 36 3 38 4 42 Go to Discussion
Solution Qus : 10 4 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 at most 1 2 at most 2 3 at least 1 4 at least 2 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 1 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 1,4 2 1,9 3 3,9 4 9,8 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 3 If the mean of the squares of first n natural numbers be 11, then n is equal to?
1 -13/2 2 11 3 5 4 4 Go to Discussion Solution Qus : 13 3 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 24 2 34 3 30 4 20 Go to Discussion
Solution Qus : 14 2 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 50 2 25 3 \frac{{2}^{50}}{50} 4 51 Go to Discussion
Solution Qus : 15 3 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 40.965 2 39.035 3 39.965 4 40.035 Go to Discussion
Solution Qus : 16 4 The mean deviation from the mean of the AP a, a + d, a + 2d, ..., a + 2nd, is:
1 \frac{n}{n+1} \, d 2 \frac{n}{2n+1} \, d 3 \frac{n+1}{2n+1} \, d 4 \frac{n(n+1)}{2n+1} \, d Go to Discussion
Solution Qus : 17 2 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 81, 25 2 82, 24 3 83, 23 4 84, 22 Go to Discussion
Solution Qus : 18 2 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 20.0 2 10.1 3 20.2 4 10.0 Go to Discussion
Solution Qus : 19 4 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 Mean of A is equal to Mode of C 2 Mean of C is equal to Median of B 3 Median of B is equal to Mode of A 4 Mean, Median and Mode of A are same 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 1 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 1.607 2 9.0 3 5.0 4 1.88 Go to Discussion Solution Total number of items in the distribution = Σ fi = 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}
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}
∴ 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 3 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 2.5 2 2.6 3 2.8 4 2.4 Go to Discussion
Solution Qus : 22 3 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 39.9, 14.97 2 39.5, 14 3 39.55, 14.97 4 40.19, 15.1 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 4 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 80, 25 2 83, 22 3 79, 26 4 82, 23 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 1 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 80% 2 60% 3 40% 4 20% Go to Discussion
Solution [{"qus_id":"3754","year":"2018"},{"qus_id":"3895","year":"2019"},{"qus_id":"3936","year":"2019"},{"qus_id":"3928","year":"2019"},{"qus_id":"3927","year":"2019"},{"qus_id":"9439","year":"2020"},{"qus_id":"9440","year":"2020"},{"qus_id":"10676","year":"2021"},{"qus_id":"10694","year":"2021"},{"qus_id":"10711","year":"2021"},{"qus_id":"11113","year":"2022"},{"qus_id":"11126","year":"2022"},{"qus_id":"11132","year":"2022"},{"qus_id":"11139","year":"2022"},{"qus_id":"11503","year":"2023"},{"qus_id":"11506","year":"2023"},{"qus_id":"11507","year":"2023"},{"qus_id":"11652","year":"2024"},{"qus_id":"11670","year":"2024"},{"qus_id":"10200","year":"2015"},{"qus_id":"10448","year":"2014"},{"qus_id":"10450","year":"2014"},{"qus_id":"10486","year":"2014"},{"qus_id":"11900","year":"2024"}]
and 
, then a possible value of n is among the following is 
