JEE Main Mathematics - Statistics
Exam Duration: 60 Mins Total Questions : 30
The measure of central tendency which takes into account all the data items is
- (a)
mean
- (b)
median
- (c)
mode
- (d)
quartiles
The mean of squares of first n natural numbers, is
- (a)
\(\frac { \left( n+1 \right) }{ 2 } \)
- (b)
\(\frac { n }{ 2 } \)
- (c)
\(\frac { 2n+1 }{ 6 } \)
- (d)
\(\frac { \left( n+1 \right) \left( 2n+1 \right) }{ 6 } \)
If a variable takes values 0,1,2,3,......,n with frequencies as nC0, nC1, nC2,........nCn respectively, then mean of the given data is,
- (a)
\(\frac { n }{ 2 } \)
- (b)
\(\frac { n\left( n+1 \right) }{ 2 } \)
- (c)
\(\frac { n }{ 4 } \)
- (d)
NONE OF THESE
The mode of the following data is
Class-interval | Frequency |
0-10 | 2 |
10-20 | 4 |
20-30 | 10 |
30-40 | 6 |
40-50 | 3 |
- (a)
22
- (b)
23
- (c)
24
- (d)
26
To determine the intelligence quotient of students, the most suitable measure of the central location, is
- (a)
mean
- (b)
median
- (c)
mode
- (d)
quartiles
The mean and S.D.of the marks of 200 candidates were found to be 40 and 15 respectively. Later, it was discovered that a score 40 was read wrong as 50. The correct mean and S.D.respectively are
- (a)
14.98,39.95
- (b)
39.95,14.98
- (c)
39.95,224.5
- (d)
NONE OF THESE
In a data distribution
- (a)
\(S.D.\ge M.D.\)
- (b)
\(S.D.
- (c)
\(S.D.\le M.D.\)
- (d)
\(S.D.=M.D.\)
In an english learning classes, there are 8 men, 7 women and 5 children whose mean ages separately are respectively 24,20 and 6 yr. The mean age of English learning classes candidate is
- (a)
18.0
- (b)
18.1
- (c)
18.2
- (d)
18.3
The SD of a variate x is \(\sigma\). The SD of the variate \(ax+b\over c\), where a,b and c are constants is
- (a)
\(({a\over c})\sigma\)
- (b)
\(\begin{vmatrix} \frac { a }{ c } \end{vmatrix}\sigma \)
- (c)
\(({a^2\over C^2})\sigma\)
- (d)
None of these
The sum of squares of deviations taken from mean 50 is 250. The coeficient of variation is
- (a)
10%
- (b)
40%
- (c)
50%
- (d)
None of these
The mean of the given data is 30.
Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
Frequency | 8 | 10 | f1 | 15 | f2 |
If total data is 70, then missing numbers are
- (a)
14, 23
- (b)
25, 21
- (c)
24, 13
- (d)
40, 31
The average weight of 25 boys was calculated to be 78.4 kg. It was later discovered that one weight was misread as 69 kg instead of 96 kg. The correct average is
- (a)
79 kg
- (b)
79.48 kg
- (c)
81.32 kg
- (d)
80.04 kg
A batsman in his 16th innings makes a score of 70 runs, and thereby increases his average by 2 runs. If he had never been 'not out', then his average after 16th innings is
- (a)
36
- (b)
38
- (c)
40
- (d)
42
If each of the observations x1,x2, .....;xn is increased by 'a', where a is a negative or positive number, then the variance
- (a)
same
- (b)
increased by a2 times
- (c)
decreased by a2 times
- (d)
None of these
The standard deviation of a distribution is 30 and each item is raised by 3, then new S.D. is
- (a)
32
- (b)
28
- (c)
27
- (d)
None of these
The standard deviation of the numbers 31, 32, 33, ......, 46, 47 is
- (a)
\(\sqrt { \frac { 17 }{ 12 } } \)
- (b)
\(\sqrt { \frac { { 47 }^{ 2 }-1 }{ 12 } } \)
- (c)
\(2\sqrt { 6 } \)
- (d)
\(4\sqrt { 3 } \)
Given below are the diameters of circles (in mm) drawn in a design.
Diameter | 33-36 | 37-40 | 41-44 | 45-48 | 49-52 |
Number of circles | 15 | 17 | 21 | 22 | 25 |
Find the variance.
- (a)
30.84
- (b)
33.84
- (c)
40.84
- (d)
42.84
In a survey of 950 families in a village, the following distribution of number of children was obtained.
Number of children | 0-2 | 2-4 | 4-6 | 6-8 | 8-10 | 10-12 |
Number of families | 272 | 328 | 205 | 120 | 15 | 10 |
Find the S.D.
- (a)
1.23
- (b)
2.23
- (c)
4.23
- (d)
3.23
The frequency distribution table is given here.
xi | 140 | 145 | 150 | 155 | 160 | 165 | 170 | 175 |
fi | 4 | 6 | 15 | 30 | 36 | 24 | 8 | 2 |
Find the variance.
- (a)
51.7336
- (b)
52.7136
- (c)
50.7336
- (d)
53.7236
Mean deviation for n observations x1,x2, ....., xn from their mean \(\bar { x } \) is given by
- (a)
\(\sum _{ i=1 }^{ n }{ ({ x }_{ i }-\bar { x } ) } \)
- (b)
\(\frac { 1 }{ n } \sum _{ i=1 }^{ n }{ |{ x }_{ i }-\bar { x } | } \)
- (c)
\(\sum _{ i=1 }^{ n }{ { ({ x }_{ i }-\bar { x } ) }^{ 2 } } \)
- (d)
\(\frac { 1 }{ n } \sum _{ i=1 }^{ n }{ { ({ x }_{ i }-\bar { x } ) }^{ 2 } } \)
When tested, the lives (in hours) of 5 bulbs were noted as follows: 1357, 1090, 1666, 1494, 1623 The mean deviations (in hours) from their mean is
- (a)
178
- (b)
179
- (c)
220
- (d)
356
Consider the following data
xi | 6 | 10 | 14 | 18 | 24 | 28 | 30 |
fi | 2 | 4 | 7 | 12 | 8 | 4 | 3 |
Statement-I: The mean of the data is 19.
Statement-II: The variance of the data is 43.4.
- (a)
If both Statement-I and Statement-II are true and Statement-II is the correct explanation of Statement -1.
- (b)
If both Statement -I and Statement-II are true but Statement-II is not the correct explanation of Statement -1.
- (c)
If Statement-I is true but Statement-II is false
- (d)
If Statement-I is false and Statement-II is true.
Statement-I: The variance of first n natural numbers is \(\frac { { n }^{ 2 }-1 }{ 12 } \)
Statement-II: The sum of first n natural numbers is \(\frac { n(n+1) }{ 2 } \) and the sum of squares of first n natural numbers is \(\frac { n(n+1)(2n+1) }{ 6 } \)
- (a)
If both Statement-I and Statement-II are true and Statement-II is the correct explanation of Statement -1.
- (b)
If both Statement -I and Statement-II are true but Statement-II is not the correct explanation of Statement -1.
- (c)
If Statement-I is true but Statement-II is false
- (d)
If Statement-I is false and Statement-II is true.
If for a distribution \(\\ \sum { (x-5) } \) = 3, \(\sum { { (x-5) }^{ 2 } } \) = 43 and the total number of items is 18. Find the standard deviation.
- (a)
1.53
- (b)
1.43
- (c)
1.55
- (d)
None of these
The mean and variance of n values of a variable x are 0 and \({ \sigma }^{ 2 }\),respectively. If the variable y = x2, then mean of y is
- (a)
\(\sigma \)
- (b)
\({ \sigma }^{ 2 }\)
- (c)
1
- (d)
\(\sigma /2\)
Coefficient of variation of two distributions are 60% and 75%, and their standard deviations are 18 and 15 respectively. Find their arithmetic means respectively.
- (a)
30, 30
- (b)
30,20
- (c)
20, 30
- (d)
20, 20
Fill in the blanks.
(i) The standard deviation of a data is ___P____ of any change in origin, Put is _____Q_____ on the change of scale.
(ii) The sum of squares of the deviations of the values of the variable is _____R____ when taken about their arithmetic mean.
(iii) The mean deviation of the data is ______S______ when measured from the median.
(iv) The standard deviation is _____T_____ to the mean deviation taken from the arithmetic mean.
- (a)
P Q R S T independent dependent minimum least greater than or equal - (b)
P Q R S T dependent independent minimum least greate - (c)
P Q R S T independent dependent minimum least equal - (d)
P Q R S T independent dependent maximum least equal
Which of the following statements is/are true?
Statement-I: Mean and standard deviation of 100 observations were found to be 40 and 10, respectively. If at the time of calculation two observations were wrongly taken as 30 and 70 in place of 3 and 27 respectively, then the correct standard deviation is 10.24.
Statement-II: While calculating the mean and variance of 10 readings, a student wrongly used the reading 52 for the correct reading 25. He obtained the mean and variance as 45 and 16 respectively. The correct mean and the variance is 50 and 43.81.
- (a)
Only Statement-I
- (b)
Only Statement-II
- (c)
Both Statement-I and Statement-II
- (d)
Neither Statement-I nor Statement-II
The mean and standard deviation of some data for the time taken to complete a test are calculated with the following results:
Number of observations = 25, mean = 18.2 seconds, standard deviation = 3.25 seconds
Further, another set of 15 observations x1,x2,.....,x15, also in seconds, is now available and we have \(\sum _{ i=1 }^{ 15 }{ { x }_{ i }=279 } \) and \(\sum _{ i=1 }^{ 15 }{ { x }_{ i }^{ 2 } } =5524\). Then the standard deviation based on all 40 observations is
- (a)
8.47
- (b)
4.87
- (c)
3.87
- (d)
5.87