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

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nimcet Statistics Measures Of Central Tendency PYQ

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:

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nimcet Previous Year PYQnimcet NIMCET 2019 PYQ

Solution

Qus : 2nimcet PYQ

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

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nimcet Previous Year PYQnimcet NIMCET 2019 PYQ

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Qus : 3nimcet PYQ

and

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

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nimcet Previous Year PYQnimcet NIMCET 2019 PYQ

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Qus : 4nimcet PYQ

Not Available

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nimcet Previous Year PYQnimcet NIMCET 2019 PYQ

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Qus : 5nimcet PYQ

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

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nimcet Previous Year PYQnimcet NIMCET 2022 PYQ

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Qus : 6nimcet PYQ

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

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nimcet Previous Year PYQnimcet NIMCET 2022 PYQ

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Qus : 7nimcet PYQ

If ${{a}}_1,{{a}}_2,\ldots,{{a}}_n$ are any real numbers and $n$ is any positive integer, then

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nimcet Previous Year PYQnimcet NIMCET 2022 PYQ

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Qus : 8nimcet PYQ

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:

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nimcet Previous Year PYQnimcet NIMCET 2024 PYQ

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Qus : 9nimcet PYQ

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

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nimcet Previous Year PYQnimcet NIMCET 2022 PYQ

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Qus : 10nimcet PYQ

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?

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nimcet Previous Year PYQnimcet NIMCET 2024 PYQ

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 : 11nimcet PYQ

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

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nimcet Previous Year PYQnimcet NIMCET 2024 PYQ

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 : 12nimcet PYQ

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

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nimcet Previous Year PYQnimcet NIMCET 2018 PYQ

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Qus : 13nimcet PYQ

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

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nimcet Previous Year PYQnimcet NIMCET 2021 PYQ

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Qus : 14nimcet PYQ

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

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nimcet Previous Year PYQnimcet NIMCET 2021 PYQ

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Qus : 15nimcet PYQ

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:

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nimcet Previous Year PYQnimcet NIMCET 2014 PYQ

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Qus : 16nimcet PYQ

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

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nimcet Previous Year PYQnimcet NIMCET 2014 PYQ

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Qus : 17nimcet PYQ

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

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nimcet Previous Year PYQnimcet NIMCET 2021 PYQ

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Qus : 18nimcet PYQ

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

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nimcet Previous Year PYQnimcet NIMCET 2015 PYQ

Solution

Qus : 19nimcet PYQ

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:

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nimcet Previous Year PYQnimcet NIMCET 2020 PYQ

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 : 20nimcet PYQ

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

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nimcet Previous Year PYQnimcet NIMCET 2020 PYQ

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 : 21nimcet PYQ

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:

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nimcet Previous Year PYQnimcet NIMCET 2023 PYQ

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Qus : 22nimcet PYQ

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

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nimcet Previous Year PYQnimcet NIMCET 2023 PYQ

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 : 23nimcet PYQ

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

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nimcet Previous Year PYQnimcet NIMCET 2023 PYQ

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 : 24nimcet PYQ

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

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nimcet Previous Year PYQnimcet NIMCET 2014 PYQ

Solution

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nimcet Statistics Measures Of Central Tendency PYQ

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Given two sets:

✅ Answer:

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Mean, Median, and Mode Relation

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

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