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JEE 2023 Class 11th
7. ● Introduction
● Measures of central tendency
● Relation between mean, median and mode
● Measures of dispersion
Statistics
8. Statistics is all about studying the behavior of big data. In statistics we
define some parameters which help us to do that. So, here in this chapter
we will be studying about two such parameters namely “Central
tendency” & “Dispersion”.
Introduction
9. Measures of central
tendency
We are already acquainted with these measures. We normally study 3
measures of central tendency:
(a) Mean
(b) Median
(c) Mode
10. Mean:
As such finding mean is very simple
(a)
(b)
For class wise data (i.e. continuous data), we take mid points of
classes as xi
Note
11. (a) Mean of 1, 2, 4, 6, 8 is
(b) Mean of is
(c) Mean of is
So, finding mean is never a challenge
13. Question!
The mean of runs of Kedar Jadhav in 15 matches is 4. If
mean of his runs in first 3 matches is 7, then find mean of
his runs in remaining matches.
Q
18. Mean of 100 items is 49. It was discovered that three items
which should have been 60, 70 and 80 were wrongly read
as 40, 20 and 50, respectively. The correct mean is
Q
A
B
C
D
48
50
80
Question!
19. Mean of 100 items is 49. It was discovered that three items
which should have been 60, 70 and 80 were wrongly read
as 40, 20 and 50, respectively. The correct mean is
Q
A
B
C
D
48
50
80
Question!
C
20. Solution :
Sum of 100 items = 49 × 100 = 4900
Sum of items added = 60 + 70 + 80 = 210
Sum of items replaced = 40 + 20 + 50 = 110
New sum = 4900 − 110 + 210 = 5000
21. The mean of data set comprising of 16 observations is 16. If one
of the observation valued 16 is deleted and 3 observations 3, 4
& 5 are added then mean of resultant data is:
Q
A
B
C
D
16.8
16.0
15.8
14
JEE Main 2015
Question!
22. The mean of data set comprising of 16 observations is 16. If one
of the observation valued 16 is deleted and 3 observations 3, 4
& 5 are added then mean of resultant data is:
Q
A
B
C
D
16.8
16.0
15.8
14
JEE Main 2015
Question!
D
24. The mean age of a combined group of men and women is 25
years. If the mean age of the group of men is 26 years and
that of the group of women is 21 years, then the percentage
of men and women in the group are respectively.
Q
A
B
C
D
60, 40
80, 20
20, 80
40, 60
Question!
25. Q
A
B
C
D
60, 40
80, 20
20, 80
40, 60
Question!
B
The mean age of a combined group of men and women is 25
years. If the mean age of the group of men is 26 years and
that of the group of women is 21 years, then the percentage
of men and women in the group are respectively.
37. Median:
(i) Median of an individual series:
Step 1: Arrange the data in ascending or descending order.
Step 2: (i) If n is odd, then
Median = value of the middle term
(ii) If n is even, then
Median = mean of the two middle terms
38. The median of distribution
83, 54, 78, 64, 90, 59, 67, 72, 70, 73 is
Q
A
B
C
D
71
70
72
None of these
Question!
39. The median of distribution
83, 54, 78, 64, 90, 59, 67, 72, 70, 73 is
Q
A
B
C
D
71
70
72
None of these
Question!
A
41. (ii) Median of a discrete series:
Step 1: Arrange the values of the variate in ascending or descending
order
Step 2: Prepare a cumulative frequency table
Step 3: Median is the observation whose cumulative frequency is equal
to or just greater than N/2, where N is sum of frequencies.
42. Step 1: Prepare the cumulative frequency table
Step 2: Find the median class, i.e. the class whose cumulative
frequency is equal or just greater than (N/2)
Step 3: The median value is given by the formula
(iii) Median of a continuous series:
43. Where
l = lower limit of the median class
N = total frequency
f = frequency of the median class
h = width of the median class
Cf = cumulative frequency of the class preceding the median class
44. Let’s do an example on finding median of continuous data, that is the only
case where finding median is sometimes challenging
45. Median, for the following distribution is
Q
A B
C D
24.46 14.80
34.21 None of these
Question!
Wages (₹) Number of labour
0-10 22
10-20 38
20-30 46
30-40 35
40-50 20
46. Median, for the following distribution is
Q
A B
C D
24.46 14.80
34.21 None of these
Question!
Wages (₹) Number of labour
0-10 22
10-20 38
20-30 46
30-40 35
40-50 20
A
48. Question!
Consider the following frequency distribution:
If mean and median = 14, then the value (a − b)2 is
equal to______.
Q
JEE Main 22 July, 2021 - Shift 1
Class 0-6 6-12 12-18 18-24 24-30
Frequency a b 12 9 5
49. Question!
Q
JEE Main 22 July, 2021 - Shift 1
Class 0-6 6-12 12-18 18-24 24-30
Frequency a b 12 9 5
Ans: 4
Consider the following frequency distribution:
If mean and median = 14, then the value (a − b)2 is
equal to______.
50. Solution :
Class Frequency xi fi xi
0-6 a 3 3a
6-12 b 9 9b
12-18 12 15 180
18-24 9 21 189
24-30 5 27 135
N = (26+ a +b) (504 + 3a + 9b)
52. Mode:
The mode is that value in a series of observations which occurs with
greatest frequency. In case of individual series, the mode is the value
which occurs more frequently.
54. In a moderately skewed distribution the values of mean and
median are 5 and 6, respectively. The value of mode in such
a situation is approximately equal to
Q
A
B
C
D
8
11
16
None of these
Question!
55. In a moderately skewed distribution the values of mean and
median are 5 and 6, respectively. The value of mode in such
a situation is approximately equal to
Q
A
B
C
D
8
11
16
None of these
Question!
A
56. Solution :
Given that, mean = 5, median = 6
For a moderately skewed distribution, we have
Mode = 3 median − 2 mean
⇒ Mode = 3(6) − 2(5) = 8
58. Consider scores of two batsmen in three matches
Batsman A : 0, 150, 0
Batsman B : 48, 50, 52
Try to observe average score of both is same i.e. 50, but it’s very evident
that they have completely different playing styles. That thing which is
making you think like that is “dispersion”. So, to select a batsman from
these two batsmen, mean alone is not giving complete indication of
nature of player, “dispersion” together with “mean” tells you more about
the batting style & hence facilitates better selection.
60. Measures of dispersion
We primarily have 5 measures of dispersion, namely
(1) Mean deviation about
(a) Mean
(b) Median
(c) Any given number
(2) Standard deviation
(3) Variance
(4) Range
(5) Coefficient of variation
63. Mean deviation:
Mean deviation about any number ‘A’ is:
For mean deviation about “Mean” we take in place of A
For mean deviation about “Median” we take median ‘M’ in place of A
(a)
(b)
64. Variable (x) 10 30 50 70 90
Frequency (f) 4 24 28 16 8
Question!
Calculate the mean deviation from mean for the following
frequency distribution:
Q
65. Let us make the table of the given data and append other columns after
calculations
xi fi fi xi
10 4 40 40 160
30 24 720 20 480
50 28 1400 0 0
70 16 1120 20 320
90 8 720 40 320
80 4000 1280
Solution :
68. Question!
Find the minimum possible number of the mean deviations
from the following data
Q
Class interval 0 - 6 6 - 12 12 - 18 18 - 24 24 - 30
Frequency 4 5 3 6 2
69. Question!
Find the minimum possible number of the mean deviations
from the following data
Q
Class interval 0 - 6 6 - 12 12 - 18 18 - 24 24 - 30
Frequency 4 5 3 6 2
Ans: 7
73. Standard deviation:
Given a data, its S.D. (denoted by σ) is found by using following formula
Also,
Variance = 𝜎2
Note
74. The mean of 100 observations is 50 and their standard
deviation is 5. The sum of all squares of all the observations is
Q
A
B
C
D
50000
250000
252500
255000
Question!
75. The mean of 100 observations is 50 and their standard
deviation is 5. The sum of all squares of all the observations is
Q
A
B
C
D
50000
250000
252500
255000
Question!
C
77. 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 :
Q
A
B
C
D
JEE Main 2019
Question!
78. Q
A
B
C
D
JEE Main 2019
Question!
A
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 :
80. 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:
Q
A
B
C
D
509.5
586.5
582.5
507.5
JEE Main 2019
Question!
81. 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:
Q
A
B
C
D
509.5
586.5
582.5
507.5
JEE Main 2019
Question!
D
83. The mean and variance of 10 observations are found to be
10 and 4 respectively. On rechecking it was found that an
observation 8 was incorrect. If it is replaced by 18, then the
correct variance is
Q
A
B
C
D
7
8
9
Question!
84. The mean and variance of 10 observations are found to be
10 and 4 respectively. On rechecking it was found that an
observation 8 was incorrect. If it is replaced by 18, then the
correct variance is
Q
A
B
C
D
7
8
9
Question!
C
86. We need to know the dependence of 𝜎 on change of origin & scale, as it is
going to provide step deviation method to find “𝜎” which makes involved
calculations easier.
87. Dependence of 𝛔 on change of origin & scale:
Consider data x1, x2, …, xn with SD as 𝝈x
(a) If yi = xi = A then 𝝈y = 𝝈x
i.e. SD is independent of origin
(b) If then
What about Variance?
Note
88. 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?
Q
A
B
C
D
Mean
Median
Mode
Variance
JEE Main 2013
Question!
89. 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?
Q
A
B
C
D
Mean
Median
Mode
Variance
JEE Main 2013
Question!
D
98. 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______
Q
JEE Main 2020
Question!
100. Let the observations xi(1 ≤ I ≤ 10) satisfy the equations,
. . If μ and λ are the
mean and the variance of the observations, x1 – 3, x2 – 3,
…… x10 – 3, then the ordered pair (μ, λ) is equal to
Q
A
B
C
D
(3, 3)
(3, 6)
(6, 6)
(6, 3)
JEE Main 2014
Question!
101. Q
A
B
C
D
(3, 3)
(3, 6)
(6, 6)
(6, 3)
JEE Main 2014
Question!
A
Let the observations xi(1 ≤ I ≤ 10) satisfy the equations,
. . If μ and λ are the
mean and the variance of the observations, x1 – 3, x2 – 3,
…… x10 – 3, then the ordered pair (μ, λ) is equal to
105. Benefits of Pro
Subscription
Pro Lite Pro Classic Pro Plus
₹1,350/- ₹1,800/- ₹2,700/-
LIVE Interactive Online Classes
Test series and analysis
Assignments and notes
Doubt solving during class
Doubt Solving on mobile app
Personal mentor
COUPON CODE: AVKPRO
JEE 2023 Class 11th
