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

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JEE MAIN Previous Year Questions (PYQs)

JEE MAIN Statistics Measures Of Central Tendency PYQ

Qus : 1 JEE MAIN PYQ

The mean and variance of 5 observations are 5 and 8 respectively. If 3 observations are 1, 3, 5, then the sum of cubes of the remaining two observations is :

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2023 (1 February Morning Shift) PYQ

Solution

Qus : 2 JEE MAIN PYQ

Let the mean and the standard deviation of the observation $2,3,3,3,4,5,7,a,b$ be $4$ and $\sqrt{2}$ respectively. Then the mean deviation about the mode of these observations is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2025 (4 April Evening Shift) PYQ

Solution

Qus : 3 JEE MAIN PYQ

If $\sum\limits_{i = 1}^n {\left( {{x_i} - a} \right)} = n$ and $\sum\limits_{i = 1}^n {{{\left( {{x_i} - a} \right)}^2}} = na$(n, a > 1) then the standard deviation of n observations x₁, x₂, ..., x_n is :

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 6 September 2020 (Morning) PYQ

Solution

Qus : 4 JEE MAIN PYQ

If mean and standard deviation of 5 observations $x_1,x_2,x_3,x_4,x_5$ are $10$ and $3$ respectively, then the variance of 6 observations $x_1,x_2,\ldots,x_5$ and $-50$ is equal to :

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2019 (10 January Evening Shift) PYQ

Solution

Qus : 5 JEE MAIN PYQ

All the students of a class performed poorly in Mathematics. The teacher decided to give grace marks of $10$ to each of the students. Which of the following statistical measures will not change even after the grace marks were given?

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2013 (Offline) PYQ

Solution

Qus : 6 JEE MAIN PYQ

The outcome of each of 30 items was observed; 10 items gave an outcome $\dfrac{1}{2}-d$ each, 10 items gave outcome $\dfrac{1}{2}$ each and the remaining 10 items gave outcome $\dfrac{1}{2}+d$ each. If the variance of this outcome data is $\dfrac{4}{3}$ then $|d|$ equals :

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2019 (11 January Morning Shift) PYQ

Solution

Qus : 7 JEE MAIN PYQ

The mean and variance of a set of $15$ numbers are $12$ and $14$ respectively. The mean and variance of another set of $15$ numbers are $14$ and $\sigma^{2}$ respectively. If the variance of all the $30$ numbers in the two sets is $13$, then $\sigma^{2}$ is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2023 (6 April Morning Shift) PYQ

Solution

Qus : 8 JEE MAIN PYQ

The mean and standard deviation of $100$ observations are $40$ and $5.1$, respectively. By mistake one observation is taken as $50$ instead of $40$. If the correct mean and the correct standard deviation are $\mu$ and $\sigma$ respectively, then $10(\mu+\sigma)$ is equal to

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2025 (7 April Morning Shift) PYQ

Solution

Qus : 9 JEE MAIN PYQ

The mean of set of $30$ observations is $75$. If each observation is multiplied by a non-zero number $\lambda$ and then each of them is decreased by $25$, their mean remains the same. Then $\lambda$ is equal to :

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2018 (15 April Morning Shift) PYQ

Solution

Qus : 10 JEE MAIN PYQ

The mean and standard deviation of 20 observations were calculated as 10 and 2.5 respectively. It was found that by mistake one data value was taken as 25 instead of 35. if $\alpha$ and $\sqrt \beta $ are the mean and standard deviation respectively for correct data, then ($\alpha$, $\beta$) is :

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2021 (26 August Morning Shift) PYQ

Solution

Qus : 11 JEE MAIN PYQ

For a statistical data $x_1, x_2, \ldots, x_{10}$ of $10$ values, a student obtained the mean as $5.5$ and $\sum_{i=1}^{10} x_i^2 = 371$. He later found that he had noted two values in the data incorrectly as $4$ and $5$, instead of the correct values $6$ and $8$, respectively. The variance of the corrected data is

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2025 (24 January Morning Shift) PYQ

Solution

Qus : 12 JEE MAIN PYQ

Let $a_1,a_2,\ldots,a_{10}$ be $10$ observations such that $\displaystyle \sum_{k=1}^{10} a_k = 50$ and $\displaystyle \sum_{i
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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2024 (27 January Morning Shift) PYQ

Solution

Qus : 13 JEE MAIN PYQ

If the mean of the data $7, 8, 9, 7, 8, 7, \lambda, 8$ is $8$, then the variance of this data is :

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2018 (15 April Evening Shift) PYQ

Solution

Qus : 14 JEE MAIN PYQ

If the sum of the deviations of 50 observations from 30 is 50, then the mean of these observations is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2019 (12 January Morning Shift) PYQ

Solution

Qus : 15 JEE MAIN PYQ

The mean and variance of $12$ observations are $\dfrac{9}{2}$ and $4$ respectively. Later, it was observed that two observations were considered as $9$ and $10$ instead of $7$ and $14$ respectively. If the correct variance is $\dfrac{m}{n}$, where $m$ and $n$ are coprime, then $m+n$ is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2023 (8 April Evening Shift) PYQ

Solution

Qus : 16 JEE MAIN PYQ

If $\displaystyle \sum_{i=1}^{9}(x_i-5)=9$ and $\displaystyle \sum_{i=1}^{9}(x_i-5)^{2}=45$, then the standard deviation of the $9$ items $x_1,x_2,\ldots,x_9$ is :

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2018 (Offline) PYQ

Solution

Qus : 17 JEE MAIN PYQ

The mean and standard deviation of $20$ observations are found to be $10$ and $2$, respectively. On rechecking, one observation recorded as $8$ was actually $12$. The corrected standard deviation is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2024 (6 April Morning Shift) PYQ

Solution

Qus : 18 JEE MAIN PYQ

The mean and the variance of five observations are $4$ and $5.20$, respectively. If three of the observations are $3, 4$ and $4$, then the absolute value of the difference of the other two observations is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2019 (12 January Evening Shift) PYQ

Solution

Qus : 19 JEE MAIN PYQ

The mean and the standard deviation (s.d.) of five observations are $9$ and $0$, respectively. If one of the observations is changed such that the mean of the new set of five observations becomes $10$, then their s.d. is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2018 (16 April Morning Shift) PYQ

Solution

Qus : 20 JEE MAIN PYQ

Let $\mu$ be the mean and $\sigma$ be the standard deviation of the distribution where $\displaystyle \sum f_i = 62$. If $[x]$ denotes the greatest integer $\le x$, then $[\mu^2 + \sigma^2]$ is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2023 (10 April Evening Shift) PYQ

Solution

Qus : 21 JEE MAIN PYQ

If the mean and variance of five observations are $\dfrac{24}{5}$ and $\dfrac{104}{25}$ respectively, and the mean of the first four observations is $\dfrac{7}{2}$, then the variance of the first four observations is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2024 (29 January Evening Shift) PYQ

Solution

Qus : 22 JEE MAIN PYQ

The mean of the numbers a, b, 8, 5, 10 is 6 and their variance is 6.8. If M is the mean deviation of the numbers about the mean, then 25 M is equal to :

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2022 (26 June Morning Shift) PYQ

Solution

Qus : 23 JEE MAIN PYQ

Let $X = \{x \in \mathbb{N} : 1 \leq x \leq 17\}$ and $Y = \{ax + b : x \in X,\; a \in \mathbb{R},\; b \in \mathbb{R},\; a > 0\}$. If mean and variance of elements of $Y$ are $17$ and $216$ respectively, then $a + b$ is equal to :

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2 September 2020 (Morning) PYQ

Solution

Qus : 24 JEE MAIN PYQ

Consider three observations a, b, and c such that b = a + c. If the standard deviation of a + 2, b + 2, c + 2 is d, then which of the following is true?

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2021 (16 March Morning Shift) PYQ

Solution

Qus : 25 JEE MAIN PYQ

Let sets A and B have 5 elements each. Let the mean of the elements in sets A and B be 5 and 8 respectively and the variance of the elements in sets A and B be 12 and 20 respectively. A new set C of 10 elements is formed by subtracting 3 from each element of

and adding 2 to each element of

. Then the sum of the mean and variance of the elements of

is ___________.

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2023 (11 April Morning Shift) PYQ

Solution

Qus : 26 JEE MAIN PYQ

The mean and standard deviation of 50 observations are 15 and 2 respectively. It was found that one incorrect observation was taken such that the sum of correct and incorrect observations is 70. If the correct mean is 16, then the correct variance is equal to :

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2022 (26 June Evening Shift) PYQ

Solution

Qus : 27 JEE MAIN PYQ

The mean and variance of seven observations are $8$ and $16$, respectively. If five of the observations are $2,4,10,12,14$, then the product of the remaining two observations is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2019 (8 April Morning Shift) PYQ

Solution

Qus : 28 JEE MAIN PYQ

Let M denote the median of the following frequency distribution
Then 20M is equal to :

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2024 (30 January Morning Shift) PYQ

Solution

Qus : 29 JEE MAIN PYQ

Let the six numbers $a_{1},a_{2},a_{3},a_{4},a_{5},a_{6}$ be in A.P. and $a_{1}+a_{3}=10$. If the mean of these six numbers is $\dfrac{19}{2}$ and their variance is $\sigma^{2}$, then $8\sigma^{2}$ is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2023 (24 January Evening Shift) PYQ

Solution

Qus : 30 JEE MAIN PYQ

Let the mean of 6 observations $1, 2, 4, 5, x, y$ be $5$ and their variance be $10$. Then their mean deviation about the mean is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2023 (11 April Evening Shift) PYQ

Solution

Qus : 31 JEE MAIN PYQ

For the frequency distribution :
Variate (x) : x₁ x₂ x₃ .... x₁₅
Frequency (f) : f₁ f₂ f₃...... f₁₅
where 0 < x₁ < x₂ < x₃ < ... < x₁₅ = 10 and $\sum\limits_{i = 1}^{15} {{f_i}} $ > 0, the standard deviation cannot be :

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 3 September 2020 (Morning) PYQ

Solution

Qus : 32 JEE MAIN PYQ

The mean and variance of the marks obtained by the students in a test are 10 and 4 respectively. Later, the marks of one of the students is increased from 8 to 12. If the new mean of the marks is 10.2, then their new variance is equal to :

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2023 (25 January Morning Shift) PYQ

Solution

Qus : 33 JEE MAIN PYQ

The mean age of 25 teachers in a school is 40 years. A teacher retires at the age of 60 years and a new teacher is appointed in his place. If now the mean age of the teachers in this school is 39 years, then the age (in years) of the newly appointed teacher is :

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2017 (8 April Morning Shift) PYQ

Solution

Qus : 34 JEE MAIN PYQ

In an increasing geometric progression of positive terms, the sum of the second and sixth terms is $\dfrac{70}{3}$ and the product of the third and fifth terms is $49$. Then the sum of the $4^{\text{th}},6^{\text{th}}$ and $8^{\text{th}}$ terms is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2024 (8 April Evening Shift) PYQ

Solution

Qus : 35 JEE MAIN PYQ

The mean and variance of the data 4, 5, 6, 6, 7, 8, x, y, where x < y, are 6 and ${9 \over 4}$ respectively. Then ${x^4} + {y^2}$ is equal to :

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2022 (27 June Evening Shift) PYQ

Solution

Qus : 36 JEE MAIN PYQ

A student scores the following marks in five tests: $45,,54,,41,,57,,43$. His score is not known for the sixth test. If the mean score is $48$ in the six tests, then the standard deviation of the marks in six tests is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2019 (8 April Evening Shift) PYQ

Solution

Qus : 37 JEE MAIN PYQ

The mean and variance of five observations are $\dfrac{24}{5}$ and $\dfrac{194}{25}$ respectively. If the mean of the first four observations is $\dfrac{7}{2}$, then the variance of the first four observations is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2025 (29 January Evening Shift) PYQ

Solution

Qus : 38 JEE MAIN PYQ

If the standard deviation of the numbers $-1,0,1,k$ is $\sqrt{5}$ where $k>0$, then $k$ is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2019 (9 April Morning Shift) PYQ

Solution

Qus : 39 JEE MAIN PYQ

The sum of $100$ observations and the sum of their squares are $400$ and $2475$, respectively. Later on, three observations, $3,4$ and $5$, were found to be incorrect. If the incorrect observations are omitted, then the variance of the remaining observations is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2017 (9 April Morning Shift) PYQ

Solution

Qus : 40 JEE MAIN PYQ

The frequency distribution of the age of students in a class of 40 students is given below.

If the mean deviation about the median is $1.25$, then $4x + 5y$ is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2024 (9 April Morning Shift) PYQ

Solution

Qus : 41 JEE MAIN PYQ

If the standard deviation of the numbers $2, 3, a,$ and $11$ is $3.5$, then which of the following is true?

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2016 (Offline) PYQ

Solution

Qus : 42 JEE MAIN PYQ

The mean and the median of the following ten numbers in increasing order $10,22,26,29,34,x,42,67,70,y$ are $42$ and $35$ respectively, then $\dfrac{y}{x}$ is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2019 (9 April Evening Shift) PYQ

Solution

Qus : 43 JEE MAIN PYQ

If the mean and the variance of $6,4, a, 8, b, 12,10,13$ are 9 and 9.25 respectively, then $a+b+a b$ is equal to :

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2025 (2 April Evening Shift) PYQ

Solution

Qus : 44 JEE MAIN PYQ

The mean and standard deviation of 10 observations are 20 and 8 respectively. Later on, it was observed that one observation was recorded as 50 instead of 40. Then the correct variance is :

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2023 (15 April Morning Shift) PYQ

Solution

Qus : 45 JEE MAIN PYQ

If the variance of the frequency distribution

is $160$, then the value of $c\in\mathbb{N}$ is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2024 (9 April Evening Shift) PYQ

Solution

Qus : 46 JEE MAIN PYQ

Let x_i (1 $ \le $ i $ \le $ 10) be ten observations of arandom variable X. If
$\sum\limits_{i = 1}^{10} {\left( {{x_i} - p} \right)} = 3$ and $\sum\limits_{i = 1}^{10} {{{\left( {{x_i} - p} \right)}^2}} = 9$ where 0 $ \ne $ p $ \in $ R, then the standard deviation of these observations is :

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 3 September 2020 (Evening) PYQ

Solution

Qus : 47 JEE MAIN PYQ

Let the mean and the variance of 5 observations $x_1, x_2, x_3, x_4, x_5$ be $\dfrac{24}{5}$ and $\dfrac{194}{25}$ respectively. If the mean and variance of the first 4 observations are $\dfrac{7}{2}$ and $a$ respectively, then $(4a + x_5)$ is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2022 (29 June Morning Shift) PYQ

Solution

Qus : 48 JEE MAIN PYQ

The mean and variance of 8 observations are 10 and 13.5, respectively. If 6 of these observations are 5, 7, 10, 12, 14, 15, then the absolute difference of the remaining two observations is :

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 4 September 2020 (Morning) PYQ

Solution

Qus : 49 JEE MAIN PYQ

The mean and variance of 7 observations are 8 and 16 respectively. If two observations are 6 and 8, then the variance of the remaining 5 observations is :

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2021 (31 August Evening Shift) PYQ

Solution

Qus : 50 JEE MAIN PYQ

If ${\Delta _1} = \left| {\matrix{ x & {\sin \theta } & {\cos \theta } \cr { - \sin \theta } & { - x} & 1 \cr {\cos \theta } & 1 & x \cr } } \right|$ and
${\Delta _2} = \left| {\matrix{ x & {\sin 2\theta } & {\cos 2\theta } \cr { - \sin 2\theta } & { - x} & 1 \cr {\cos 2\theta } & 1 & x \cr } } \right|$, $x \ne 0$ ;

then for all $\theta \in \left( {0,{\pi \over 2}} \right)$ :

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2019 (10 April Morning Shift) PYQ

Solution

Qus : 51 JEE MAIN PYQ

Let the mean and the variance of $6$ observations $a, b, 68, 44, 48, 60$ be $55$ and $194$, respectively. If $a>b$, then $a+3b$ is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2024 (31 January Evening Shift) PYQ

Solution

Qus : 52 JEE MAIN PYQ

Five students of a class have an average height $150\ \mathrm{cm}$ and variance $18\ \mathrm{cm}^2$. A new student, whose height is $156\ \mathrm{cm}$, joins them. The variance (in $\mathrm{cm}^2$) of the heights of these six students is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2019 (9 January Morning Shift) PYQ

Solution

Qus : 53 JEE MAIN PYQ

If the mean deviation of the numbers $1,, 1 + d,, \ldots,, 1 + 100d$ from their mean is $255$, then a value of $d$ is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2016 (9 April Morning Shift) PYQ

Solution

Qus : 54 JEE MAIN PYQ

The number of values of a $\in$ N such that the variance of 3, 7, 12, a, 43 $-$ a is a natural number is :

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2022 (29 June Evening Shift) PYQ

Solution

Qus : 55 JEE MAIN PYQ

If for some $x\in\mathbb{R}$, the frequency distribution of the marks obtained by $20$ students in a test is:

then the mean of the marks is

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2019 (10 April Morning Shift) PYQ

Solution

Qus : 56 JEE MAIN PYQ

Let $S$ be the set of all values of $a_1$ for which the mean deviation about the mean of $100$ consecutive positive integers $a_1,a_2,a_3,\ldots,a_{100}$ is $25$. Then $S$ is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2023 (30 January Evening Shift) PYQ

Solution

Qus : 57 JEE MAIN PYQ

A data consists of $n$ observations: $x_1,x_2,\ldots,x_n$. If $\displaystyle \sum_{i=1}^{n}(x_i+1)^2=9n$ and $\displaystyle \sum_{i=1}^{n}(x_i-1)^2=5n$, then the standard deviation of this data is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2019 (9 January Evening Shift) PYQ

Solution

Qus : 58 JEE MAIN PYQ

The mean of $5$ observations 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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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2016 (10 April Morning Shift) PYQ

Solution

Qus : 59 JEE MAIN PYQ

The mean and variance of 7 observations are 8 and 16, respectively. If five observations are 2, 4, 10, 12, 14, then the absolute difference of the remaining two observations is :

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 5 September 2020 (Morning) PYQ

Solution

Qus : 60 JEE MAIN PYQ

If both the mean and the standard deviation of $50$ observations $x_{1},x_{2},\ldots,x_{50}$ are equal to $16$, then the mean of $(x_{1}-4)^{2},(x_{2}-4)^{2},\ldots,(x_{50}-4)^{2}$ is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2019 (10 April Evening Shift) PYQ

Solution

Qus : 61 JEE MAIN PYQ

Consider $10$ observations $x_{1},x_{2},\ldots,x_{10}$ such that $\displaystyle \sum_{i=1}^{10}(x_{i}-\alpha)=2$ and $\displaystyle \sum_{i=1}^{10}(x_{i}-\beta)^{2}=40$, where $\alpha,\beta$ are positive integers. Let the mean and the variance of the observations be $\dfrac{6}{5}$ and $\dfrac{84}{25}$ respectively. Then $\dfrac{\beta}{\alpha}$ is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2024 (1 February Evening Shift) PYQ

Solution

Qus : 62 JEE MAIN PYQ

Let the mean and variance of five observations $x_1=1,\ x_2=3,\ x_3=a,\ x_4=7,\ x_5=b,\ a>b$ be $5$ and $10$ respectively. Then the variance of the observations $n+x_n,\ n=1,2,\ldots,5$ is

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2025 (3 April Evening Shift) PYQ

Solution

Qus : 63 JEE MAIN PYQ

The mean of the data set comprising of $16$ observations is $16$. If one of the observations valued $16$ is deleted and three new observations valued $3,4$ and $5$ are added to the data, then the mean of the resultant data is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2015 (Offline) PYQ

Solution

Qus : 64 JEE MAIN PYQ

Let the mean and standard deviation of marks of class $A$ of $100$ students be respectively $40$ and $\alpha\ (>\,0)$, and the mean and standard deviation of marks of class $B$ of $n$ students be respectively $55$ and $30-\alpha$. If the mean and variance of the marks of the combined class of $100+n$ students are respectively $50$ and $350$, then the sum of variances of classes $A$ and $B$ is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2023 (31 January Evening Shift) PYQ

Solution

Qus : 65 JEE MAIN PYQ

If the mean and the standard deviation of thedata 3, 5, 7, a, b are 5 and 2 respectively, then a and b are the roots of the equation :

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 5 September 2020 (Evening) PYQ

Solution

Qus : 66 JEE MAIN PYQ

If the mean deviation about median for the numbers 3, 5, 7, 2k, 12, 16, 21, 24, arranged in ascending order, is 6, then the median is :

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2022 (25 July Evening Shift) PYQ

Solution

Qus : 67 JEE MAIN PYQ

Let $\alpha, \beta \in \mathbb{R}$. Let the mean and the variance of 6 observations $-3,\, 4,\, 7,\,-6,\, \alpha,\, \beta$ be $2$ and $23$, respectively. The mean deviation about the mean of these 6 observations is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2024 (4 April Morning Shift) PYQ

Solution

Qus : 68 JEE MAIN PYQ

If the data x₁, x₂,......., x₁₀ is such that the mean of first four of these is 11, the mean of the remaining six is 16 and the sum of squares of all of these is 2,000 ; then the standard deviation of this data is :

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2019 (12 April Morning Shift) PYQ

Solution

Qus : 69 JEE MAIN PYQ

Marks obtained by all the students of class 12 are presented in a frequency distribution with classes of equal width. The median of this grouped data is $14$ with median class interval $12$–$18$ and median class frequency $12$. If the number of students whose marks are less than $12$ is $18$, then the total number of students is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2025 (23 January Morning Shift) PYQ

Solution

Qus : 70 JEE MAIN PYQ

If the center and radius of the circle $\left|\dfrac{z-2}{z-3}\right|=2$ are respectively $(\alpha,\beta)$ and $\gamma$, then $3(\alpha+\beta+\gamma)$ is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2023 (1 February Morning Shift) PYQ

Solution

Qus : 71 JEE MAIN PYQ

The variance of first $50$ even natural numbers is :

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2014 (Offline) PYQ

Solution

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