2020
JEE Mains
MCQ
JEE Main 2020 (Online) 2nd September Morning Slot
Let X = {x
$ \in $ N : 1
$ \le $ x
$ \le $ 17} and
Y = {ax + b: x $ \in $ X and a, b $ \in $ R, a > 0}. If mean
and variance of elements of Y are 17 and 216
respectively then a + b is equal to :
Y = {ax + b: x $ \in $ X and a, b $ \in $ R, a > 0}. If mean
and variance of elements of Y are 17 and 216
respectively then a + b is equal to :
A.
7
B.
9
C.
-7
D.
-27
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 9th January Morning Slot
Let the observations xi (1 $ \le $ i $ \le $ 10) satisfy the
equations, $\sum\limits_{i = 1}^{10} {\left( {{x_1} - 5} \right)} $ = 10 and $\sum\limits_{i = 1}^{10} {{{\left( {{x_1} - 5} \right)}^2}} $ = 40.
If $\mu $ and $\lambda $ are the mean and the variance of the
observations, x1 – 3, x2 – 3, ...., x10 – 3, then
the ordered pair ($\mu $, $\lambda $) is equal to :
equations, $\sum\limits_{i = 1}^{10} {\left( {{x_1} - 5} \right)} $ = 10 and $\sum\limits_{i = 1}^{10} {{{\left( {{x_1} - 5} \right)}^2}} $ = 40.
If $\mu $ and $\lambda $ are the mean and the variance of the
observations, x1 – 3, x2 – 3, ...., x10 – 3, then
the ordered pair ($\mu $, $\lambda $) is equal to :
A.
(6, 6)
B.
(3, 3)
C.
(3, 6)
D.
(6, 3)
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 8th January Evening Slot
The mean and variance of 20 observations are
found to be 10 and 4, respectively. On
rechecking, it was found that an observation 9
was incorrect and the correct observation was
11. Then the correct variance is
A.
3.98
B.
3.99
C.
4.01
D.
4.02
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 8th January Morning Slot
The mean and the standard deviation (s.d.) of
10 observations are 20 and 2 resepectively.
Each of these 10 observations is multiplied by
p and then reduced by q, where p $ \ne $ 0 and
q $ \ne $ 0. If the new mean and new s.d. become
half of their original values, then q is equal to
A.
10
B.
-20
C.
-10
D.
-5
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 6th September Evening Slot
Consider the data on x taking the values
0, 2, 4, 8,....., 2n with frequencies
nC0 , nC1 , nC2 ,...., nCn respectively. If the
mean of this data is ${{728} \over {{2^n}}}$, then n is equal to _________ .
0, 2, 4, 8,....., 2n with frequencies
nC0 , nC1 , nC2 ,...., nCn respectively. If the
mean of this data is ${{728} \over {{2^n}}}$, then n is equal to _________ .
Correct Answer: 6
Explanation:
Mean = ${{\sum {{x_1}.{f_1}} } \over {\sum {{f_1}} }}$
= ${{0.{}^n{C_0} + 2.{}^n{C_1} + {2^2}.{}^n{C_2} + ... + {2^n}.{}^n{C_n}} \over {{}^n{C_0} + {}^n{C_1} + ... + {}^n{C_n}}}$
We know,
(1 + x)n = ${{}^n{C_0} + {}^n{C_1}x + {}^n{C_2}{x^2} + ... + {}^n{C_n}{x^n}}$ ...(1)
Put x = 2, at (1) we get
$ \Rightarrow $ 3n - 1 = ${2.{}^n{C_1} + {2^2}.{}^n{C_2} + ... + {2^n}.{}^n{C_n}}$
And Putting x = 1 in (1), we get
2n = ${{}^n{C_0} + {}^n{C_1} + ... + {}^n{C_n}}$
$ \therefore $ Mean = ${{{3^n} - 1} \over {{2^n}}}$
According to question,
${{{3^n} - 1} \over {{2^n}}}$ = ${{728} \over {{2^n}}}$
$ \Rightarrow $ 3n = 729
$ \Rightarrow $ n = 6
= ${{0.{}^n{C_0} + 2.{}^n{C_1} + {2^2}.{}^n{C_2} + ... + {2^n}.{}^n{C_n}} \over {{}^n{C_0} + {}^n{C_1} + ... + {}^n{C_n}}}$
We know,
(1 + x)n = ${{}^n{C_0} + {}^n{C_1}x + {}^n{C_2}{x^2} + ... + {}^n{C_n}{x^n}}$ ...(1)
Put x = 2, at (1) we get
$ \Rightarrow $ 3n - 1 = ${2.{}^n{C_1} + {2^2}.{}^n{C_2} + ... + {2^n}.{}^n{C_n}}$
And Putting x = 1 in (1), we get
2n = ${{}^n{C_0} + {}^n{C_1} + ... + {}^n{C_n}}$
$ \therefore $ Mean = ${{{3^n} - 1} \over {{2^n}}}$
According to question,
${{{3^n} - 1} \over {{2^n}}}$ = ${{728} \over {{2^n}}}$
$ \Rightarrow $ 3n = 729
$ \Rightarrow $ n = 6
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 4th September Evening Slot
If the variance of the following frequency
distribution :
Class : 10–20 20–30 30–40
Frequency : 2 x 2
is 50, then x is equal to____
Class : 10–20 20–30 30–40
Frequency : 2 x 2
is 50, then x is equal to____
Correct Answer: 4
Explanation:
xi = midpoint of class interval
Variance($\sigma^{2})$$ =\frac{\sum f_{i}\left( x_{i}-\bar{x} \right)^{2} }{\sum f_{i}} $
Also, $\bar{x} =\frac{\sum f_{i}x_{i}}{\sum f_{i}} $
= $\frac{30+25x+70}{2+2+x} $ = 25
Given, Variance = 50
$ \therefore $ 50 = $\frac{200+0+200}{2+2+x} $
$ \Rightarrow $ x = 4
Variance($\sigma^{2})$$ =\frac{\sum f_{i}\left( x_{i}-\bar{x} \right)^{2} }{\sum f_{i}} $
Also, $\bar{x} =\frac{\sum f_{i}x_{i}}{\sum f_{i}} $
= $\frac{30+25x+70}{2+2+x} $ = 25
Given, Variance = 50
$ \therefore $ 50 = $\frac{200+0+200}{2+2+x} $
$ \Rightarrow $ x = 4
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 2nd September Evening Slot
If the variance of the terms in an increasing A.P.,
b1 , b2 , b3 ,....,b11 is 90, then the common difference of this A.P. is_______.
b1 , b2 , b3 ,....,b11 is 90, then the common difference of this A.P. is_______.
Correct Answer: 3
Explanation:
Let the common difference = d
and ${b_1} = a$
${b_2} = a + d$
${b_3} = a + 2d$
... ${b_{11}} = a + 10d$
Variance = ${{\sum {a_i^2} } \over {11}} - {\left( {{{\sum {{a_i}} } \over {11}}} \right)^2} = 90$
$ \Rightarrow {{{a^2} + {{\left( {a + d} \right)}^2} + ... + {{\left( {a + 10d} \right)}^2}} \over {11}} - {\left( {{{a + \left( {a + d} \right) + ... + \left( {a + 10d} \right)} \over {11}}} \right)^2} = 90$
$ \Rightarrow 11\left[ {11{a^2} + 385{d^2} + 110ad} \right] - {\left[ {11a + 55d} \right]^2} = 10890$
$ \Rightarrow 1210{d^2} = 10890$
$ \Rightarrow {d^2} = 9$
$ \Rightarrow d = \pm 3$
As A.P is increasing so d should be positive
$ \therefore $ d = 3
and ${b_1} = a$
${b_2} = a + d$
${b_3} = a + 2d$
... ${b_{11}} = a + 10d$
Variance = ${{\sum {a_i^2} } \over {11}} - {\left( {{{\sum {{a_i}} } \over {11}}} \right)^2} = 90$
$ \Rightarrow {{{a^2} + {{\left( {a + d} \right)}^2} + ... + {{\left( {a + 10d} \right)}^2}} \over {11}} - {\left( {{{a + \left( {a + d} \right) + ... + \left( {a + 10d} \right)} \over {11}}} \right)^2} = 90$
$ \Rightarrow 11\left[ {11{a^2} + 385{d^2} + 110ad} \right] - {\left[ {11a + 55d} \right]^2} = 10890$
$ \Rightarrow 1210{d^2} = 10890$
$ \Rightarrow {d^2} = 9$
$ \Rightarrow d = \pm 3$
As A.P is increasing so d should be positive
$ \therefore $ d = 3
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 7th January Evening Slot
If the mean and variance of eight numbers 3, 7, 9, 12, 13, 20, x and y be 10 and 25 respectively,
then x.y is equal to _______.
Correct Answer: 54
Explanation:
Mean = ${{3 + 7 + 9 + 12 + 13 + 20 + x + y} \over 8}$ = 10
16 = x + y ....(1)
Variance (${\sigma ^2}$) = 25
$ \Rightarrow $ ${{{3^2} + {7^2} + {9^2} + {{12}^2} + {{13}^2} + {{20}^2} + {x^2} + {y^2}} \over 8}$ - 100 = 25
$ \Rightarrow $ 125 × 8 = 9 + 49 + 81 + 144 + 169 + 400 + x2 + y2 - 800
$ \Rightarrow $ x2 + y2 = 148
We know, (x + y)2 = x2 + y2 + 2xy
$ \Rightarrow $ 256 = 148 + 2xy
$ \Rightarrow $ x.y = 54
16 = x + y ....(1)
Variance (${\sigma ^2}$) = 25
$ \Rightarrow $ ${{{3^2} + {7^2} + {9^2} + {{12}^2} + {{13}^2} + {{20}^2} + {x^2} + {y^2}} \over 8}$ - 100 = 25
$ \Rightarrow $ 125 × 8 = 9 + 49 + 81 + 144 + 169 + 400 + x2 + y2 - 800
$ \Rightarrow $ x2 + y2 = 148
We know, (x + y)2 = x2 + y2 + 2xy
$ \Rightarrow $ 256 = 148 + 2xy
$ \Rightarrow $ x.y = 54
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 7th January Morning Slot
If the variance of the first n natural numbers is 10 and the variance of the first m even natural
numbers is 16, then m + n is equal to_____.
Correct Answer: 18
Explanation:
Variance ${\sigma ^2} = {{\sum {x_i^2} } \over N} - {\mu ^2}$
variance of (1, 2, ….. n)
10 = ${{{1^2} + {2^2} + .... + {n^2}} \over n} - {\left( {{{1 + 2 + 3 + .... + n} \over n}} \right)^2}$
on solving we get n = 11
variance of 2, 4, 6…….2m = 16
$ \Rightarrow $ ${{{2^2} + {4^2} + .... + {{\left( {2m} \right)}^2}} \over m} - {\left( {m + 1} \right)^2}$ = 16
$ \Rightarrow $ m2 = 49
$ \Rightarrow $ m = 7
$ \therefore $ m + n = 18
variance of (1, 2, ….. n)
10 = ${{{1^2} + {2^2} + .... + {n^2}} \over n} - {\left( {{{1 + 2 + 3 + .... + n} \over n}} \right)^2}$
on solving we get n = 11
variance of 2, 4, 6…….2m = 16
$ \Rightarrow $ ${{{2^2} + {4^2} + .... + {{\left( {2m} \right)}^2}} \over m} - {\left( {m + 1} \right)^2}$ = 16
$ \Rightarrow $ m2 = 49
$ \Rightarrow $ m = 7
$ \therefore $ m + n = 18
2020
TS-EAMCET
MCQ
TS EAMCET 2020 (Online) 14th September Evening Shift
If $S_1$ and $S_2$ are the variances of the first $2 k$ and $k(k>1)$ natural numbers respectively, then ( $S_1 / S_2$ ) lies in the interval
A.
$[4, \infty)$
B.
$(1,4]$
C.
$(4,5]$
D.
$[7, \infty)$
2020
TS-EAMCET
MCQ
TS EAMCET 2020 (Online) 14th September Evening Shift
The standard deviations of two sets of observations $X=\left\{x_i\right\}$ and $Y=\left\{y_i\right\}(i=1,2, \ldots, 100)$ are respectively 5 and 6 . If $\bar{x}, \bar{y}$ are their means and $\sum_{i=1}^{100}\left(x_i-\bar{x}\right)\left(y_i-\bar{y}\right)=600$, then the standard deviation of $Z=\left\{z_i / z_i=x_i-y_i\right)$ is
A.
12
B.
6
C.
7
D.
10
2020
TS-EAMCET
MCQ
TS EAMCET 2020 (Online) 14th September Evening Shift
In a discrete data $\frac{1 \text { th }}{4}$ of the observations are equal to $a$, another $\frac{1 \text { th }}{4}$ of the observations are equal to $-a$. Out of the remaining, half of them are equal to $b$ and the rest are equal to $-b$. If the variance of all the observations is $(a b)$, then
A.
$a^2=4 b^2$
B.
$a=-2 b$
C.
$a=b$
D.
$a=-3 b$
2020
TS-EAMCET
MCQ
TS EAMCET 2020 (Online) 14th September Evening Shift
For the following distribution, the mean deviation about the median is
$ \begin{array}{cccccccc} \hline x_i & 6 & 12 & 18 & 24 & 30 & 36 & 42 \\ \hline f_i & 4 & 7 & 9 & 18 & 15 & 10 & 5 \\ \hline \end{array} $
A.
8.0
B.
7.5
C.
7.2
D.
7.0
2020
TS-EAMCET
MCQ
TS EAMCET 2020 (Online) 10th September Evening Shift
Assertion (A) Variance of $4 x_1, 4 x_2, \ldots, 4 x_n$ is 16 times the variance of $x_1, x_2, x_3, \ldots, x_n$
Reason (R) If $y=a x+b$, then variance of $y$ is a $($ variance of $x)+b$
The correct option among the following is
A.
(A) is true, (R) is true and (R) is the correct explanation for (A).
B.
(A) is true, (R) is true but (R) is not the correct explanation for (A).
C.
(A) is true but (R) is false.
D.
(A) is false but (R) is true.
2020
TS-EAMCET
MCQ
TS EAMCET 2020 (Online) 10th September Evening Shift
If $\alpha, \beta$ are respectively the mean deviation about the mean and variance of the first five prime numbers, then the ordered pair ( $\alpha, \beta$ )
A.
$(2.27,10.42)$
B.
$(2.27,10.24)$
C.
$(2.72,10.24)$
D.
$(2.72,10.42)$
2020
TS-EAMCET
MCQ
TS EAMCET 2020 (Online) 10th September Morning Shift
For the following frequency distribution, the variance is approximately equal to
$ \begin{array}{cccccc} \hline \begin{array}{c} \text { Class } \\ \text { Interval } \end{array} & 0-5 & 5-10 & 10-15 & 15-20 & 20-25 \\ \hline \text { Frequency } & 4 & 1 & 10 & 3 & 2 \\ \hline \end{array} $
A.
33.1
B.
30.55
C.
34.75
D.
37.50
2020
TS-EAMCET
MCQ
TS EAMCET 2020 (Online) 10th September Morning Shift
If the mean of the discrete distribution $8,9,6,5, x, 4$, 6, 5 is 6 , then its standard deviation (nearest to two decimal places) is
A.
2.50
B.
1.58
C.
0.51
D.
1.41
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th April Morning Slot
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 :
A.
$\sqrt 2 $
B.
2
C.
2$\sqrt 2 $
D.
4
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th April Evening Slot
If both the mean and the standard deviation of 50 observations x1, x2,..., x50 are equal to 16, then the mean of (x1 – 4)2
, (x2 – 4)2
,....., (x50 – 4)2
is :
A.
400
B.
480
C.
380
D.
525
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th April Morning Slot
If for some x $ \in $ R, the frequency distribution of the marks obtained by 20 students in a test is :
then the mean of the marks is
| Marks | 2 | 3 | 5 | 7 |
|---|---|---|---|---|
| Frequency | (x + 1)2 | 2x - 5 | x2 - 3x | x |
then the mean of the marks is
A.
3.0
B.
2.8
C.
2.5
D.
3.2
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th April Evening Slot
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 ${y \over x}$ is equal to
A.
${7 \over 2}$
B.
${8 \over 3}$
C.
${9 \over 4}$
D.
${7 \over 3}$
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th April Morning Slot
If the standard deviation of the numbers
–1, 0, 1, k is $\sqrt 5$ where k > 0, then k is equal to
A.
2$\sqrt 6 $
B.
$\sqrt 6 $
C.
$2\sqrt {{{5} \over 6}} $
D.
$2\sqrt {{{10} \over 3}} $
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 8th April Evening Slot
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
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
A.
$100 \over {\sqrt 3}$
B.
$10 \over {\sqrt 3}$
C.
$10 \over3$
D.
$100 \over3$
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 8th April Morning Slot
The mean and variance of seven observations are
8 and 16, respectively. If 5 of the observations are
2, 4, 10, 12, 14, then the product of the remaining
two observations is :
A.
40
B.
48
C.
49
D.
45
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th January Evening Slot
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 :
A.
1
B.
7
C.
3
D.
5
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th January Morning Slot
If the sum of the deviations of 50 observations from 30 is 50, then the mean of these observations is :
A.
31
B.
50
C.
51
D.
30
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 11th January Morning Slot
The outcome of each of 30 items was observed; 10 items gave an outcome ${1 \over 2}$ – d each, 10 items gave outcome ${1 \over 2}$ each and the remaining 10 items gave outcome ${1 \over 2}$+ d each. If the variance of this outcome data is ${4 \over 3}$ then |d| equals :
A.
${2 \over 3}$
B.
${{\sqrt 5 } \over 2}$
C.
${\sqrt 2 }$
D.
2
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th January Evening Slot
If mean and standard deviation of 5 observations x1, x2, x3, x4, x5 are 10 and 3, respectively, then the variance of 6 observations x1, x2, ….., x5 and –50 is equal to
A.
582.5
B.
507.5
C.
586.5
D.
509.5
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th January Morning Slot
The mean of five observations is 5 and their variance is 9.20. If three of the given five observations are 1, 3 and 8, then a ratio of other two observations is -
A.
6 : 7
B.
10 : 3
C.
4 : 9
D.
5 : 8
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th January Evening Slot
A data consists of n observations : x1, x2, . . . . . . ., xn.
If $\sum\limits_{i = 1}^n {{{\left( {{x_i} + 1} \right)}^2}} = 9n$ and
$\sum\limits_{i = 1}^n {{{\left( {{x_i} - 1} \right)}^2}} = 5n,$
then the standard deviation of this data is :
If $\sum\limits_{i = 1}^n {{{\left( {{x_i} + 1} \right)}^2}} = 9n$ and
$\sum\limits_{i = 1}^n {{{\left( {{x_i} - 1} \right)}^2}} = 5n,$
then the standard deviation of this data is :
A.
2
B.
$\sqrt 5 $
C.
5
D.
$\sqrt 7 $
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th January Morning Slot
5 students of a class have an average height 150 cm and variance 18 cm2. A new student, whose height is 156 cm, joined them. The variance (in cm2) of the height of these six students is :
A.
16
B.
22
C.
20
D.
18
2018
JEE Mains
MCQ
JEE Main 2018 (Online) 16th April Morning Slot
The mean and the standard deviation(s.d.) of five observations are9 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 :
A.
0
B.
1
C.
2
D.
4
2018
JEE Mains
MCQ
JEE Main 2018 (Offline)
If $\sum\limits_{i = 1}^9 {\left( {{x_i} - 5} \right)} = 9$ and
$\sum\limits_{i = 1}^9 {{{\left( {{x_i} - 5} \right)}^2}} = 45$, then the standard deviation of the 9 items
${x_1},{x_2},.......,{x_9}$ is
$\sum\limits_{i = 1}^9 {{{\left( {{x_i} - 5} \right)}^2}} = 45$, then the standard deviation of the 9 items
${x_1},{x_2},.......,{x_9}$ is
A.
3
B.
9
C.
4
D.
2
2018
JEE Mains
MCQ
JEE Main 2018 (Online) 15th April Evening Slot
If the mean of the data : 7, 8, 9, 7, 8, 7, $\lambda $, 8 is 8, then the variance of this data is :
A.
${7 \over 8}$
B.
1
C.
${9 \over 8}$
D.
2
2018
JEE Mains
MCQ
JEE Main 2018 (Online) 15th April Morning Slot
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 :
A.
${1 \over 3}$
B.
${2 \over 3}$
C.
${4 \over 3}$
D.
${10 \over 3}$
2017
JEE Mains
MCQ
JEE Main 2017 (Online) 9th April Morning Slot
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 :
A.
8.25
B.
8.50
C.
8.00
D.
9.00
2017
JEE Mains
MCQ
JEE Main 2017 (Online) 8th April Morning Slot
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 :
A.
25
B.
30
C.
35
D.
40
2016
JEE Mains
MCQ
JEE Main 2016 (Online) 10th April Morning Slot
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 :
A.
2.4
B.
2.8
C.
2.5
D.
2.6
2016
JEE Mains
MCQ
JEE Main 2016 (Online) 9th April Morning Slot
If the mean deviation of the numbers 1, 1 + d, ..., 1 +100d from their mean is 255, then a value of d is :
A.
10.1
B.
20.2
C.
10
D.
5.05
2016
JEE Mains
MCQ
JEE Main 2016 (Offline)
If the standard deviation of the numbers 2, 3, a and 11 is 3.5, then which of the following is true?
A.
3$a$2 - 26$a$ + 55 = 0
B.
3$a$2 - 32$a$ + 84 = 0
C.
3$a$2 - 34$a$ + 91 = 0
D.
3$a$2 - 23$a$ + 44 = 0
2015
JEE Mains
MCQ
JEE Main 2015 (Offline)
The mean of the data set comprising of 16 observations is 16. If one of the observation 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 :
A.
15.8
B.
14.0
C.
16.8
D.
16.0
2014
JEE Mains
MCQ
JEE Main 2014 (Offline)
The variance of first 50 even natural numbers is
A.
833
B.
437
C.
${{437} \over 4}$
D.
${{833} \over 4}$
2013
JEE Mains
MCQ
JEE Main 2013 (Offline)
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?
A.
median
B.
mode
C.
variance
D.
mean
2012
JEE Mains
MCQ
AIEEE 2012
Let x1, x2,........., xn be n observations, and let $\overline x $ be their arithematic mean and ${\sigma ^2}$ be their variance.
Statement 1 : Variance of 2x1, 2x2,......., 2xn is 4${\sigma ^2}$.
Statement 2 : : Arithmetic mean of 2x1, 2x2,......, 2xn is 4$\overline x $.
Statement 1 : Variance of 2x1, 2x2,......., 2xn is 4${\sigma ^2}$.
Statement 2 : : Arithmetic mean of 2x1, 2x2,......, 2xn is 4$\overline x $.
A.
Statement 1 is false, statement 2 is true
B.
Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1
C.
Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1
D.
Statement 1 is true, statement 2 is false
2011
JEE Mains
MCQ
AIEEE 2011
If the mean deviation about the median of the numbers a, 2a,........., 50a is 50, then |a| equals
A.
4
B.
5
C.
2
D.
3
2010
JEE Mains
MCQ
AIEEE 2010
For two data sets, each of size 5, the variances are given to be 4 and 5 and the corresponding
means are given to be 2 and 4, respectively. The variance of the combined data set is
A.
${5 \over 2}$
B.
${11 \over 2}$
C.
6
D.
${13 \over 2}$
2009
JEE Mains
MCQ
AIEEE 2009
If the mean deviation of number 1, 1 + d, 1 + 2d,........, 1 + 100d from their mean is 255, then the d is
equal to
A.
20.0
B.
10.1
C.
20.2
D.
10.0
2009
JEE Mains
MCQ
AIEEE 2009
Statement - 1 : The variance of first n even natural numbers is ${{{n^2} - 1} \over 4}$
Statement - 2 : The sum of first n natural numbers is ${{n\left( {n + 1} \right)} \over 2}$ and the sum of squares of first n natural numbers is ${{n\left( {n + 1} \right)\left( {2n + 1} \right)} \over 6}$
Statement - 2 : The sum of first n natural numbers is ${{n\left( {n + 1} \right)} \over 2}$ and the sum of squares of first n natural numbers is ${{n\left( {n + 1} \right)\left( {2n + 1} \right)} \over 6}$
A.
Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1
B.
Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1
C.
Statement-1 is true, Statement-2 is false
D.
Statement-1 is false, Statement-2 is true
2008
JEE Mains
MCQ
AIEEE 2008
The mean of the numbers a, b, 8, 5, 10 is 6 and the variance is 6.80. Then which one of the following
gives possible values of a and b?
A.
a = 0, b = 7
B.
a = 5, b = 2
C.
a = 1, b = 6
D.
a = 3, b = 4
2007
JEE Mains
MCQ
AIEEE 2007
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
A.
80
B.
60
C.
40
D.
20