Karnataka 1st PUC Mathematics Statistics MCQ with Answers

GOVERNMENT OF KARNATAKA
DEPARTMENT OF SCHOOL EDUCATION (PRE-UNIVERSITY)
18TH CROSS, MALLESHWARAM, BENGALURU – 560 012
CHAPTERWISE MULTIPLE CHOICE QUESTIONS FOR COMPETATIVE EXAM
SUBJECT: MATHEMATICS – I PUC
NAME OF THE CHAPTER: STATISTICS

Mean:
Mean for ungrouped data:
x +x2 +x3 +⋯+xn ∑x
Let x1 , x2 , x3 , … xn be n observations. Then their Mean, x‾ = 1 = n
n

Mean for grouped data:
Let x1 , x2 , x3 , … xn be n observations with frequencies f1 , f2 , f3 , … fn respectively .
1
Then their Mean 𝑖𝑠 x̄ = N ∑ni=1 fi xi

where N = ∑ni=1 fi
Combined Mean:
n1 x1 + n2 x2
x‾ = ,
n1 + n2
Where,
x1 is the mean of n1 number of observations
x2 is the mean of n2 number of observations
Properties of Arithmetic Mean
If arithmetic mean of x1 , x2 , x3 , … , xn is x‾, then
(i) arithmetic mean of x1 ± 𝑎, x2 ± 𝑎, x3 ± 𝑎, …, xn ± 𝑎 is x‾ ± 𝑎
(ii) arithmetic mean of ax1 , ax2 , ax3 , . . . , axn is ax‾
x1 x 2 x3 x x‾
(iii) arithmetic mean of , a , a , … , an is a
a

Median:
n + 1 th
( ) observation, if n is odd
2
Median =
n th n th
(2) + (2 + 1)
{ observation, if n is even
2

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Median for continuous frequency distribution
N
( −C)
2
M=l+ ×h
f
N
Here 2 gives median class interval, whose cumulative frequency is just greater than
N
or equal to 2 ,

N − is the sum of frequencies,
l- lower limit,
f – frequency of current class,
h - width of the median class,
C - the cumulative frequency of the class just preceding the median class.
Mode:
Mode for symmetrical distribution = Mean = Median
Mode for asymmetrical distribution = 3( Median ) − 2( Mean )
Mean deviation about mean [M.D. (𝒙̄ )]
For ungrouped data:
1 1
M.D (x̄ ) =n ∑ni=1|xi − x̄ | , where x̄ is the mean given by x̄ = n ∑ni=1 xi

For grouped data:
1 1
M.D (x̄ ) = N ∑ni=1 fi |xi − x̄ |, where x̄ is the mean given by x̄ = N ∑ni=1 fi xi and N = ∑ni=1 fi

Mean deviation about median [M.D. (M)]
1
 For ungrouped data: M.D.(M) = n ∑ni=1|xi − M|, Where M is the median
1
 For grouped data: M.D.(M) = N ∑ni=1 fi |xi − M|, where M is the median and N = ∑ni=1 fi

Variance & Standards Deviation:
Let x1 , x2 , x3 , … , xn be n observations with x‾ as mean.
1
(i) The variance denoted by σ2 is given by σ2 = n ∑(xi − x‾)2

1
(ii) The standard deviation denoted by σ is given by σ = √ ∑ (xi − x‾)2
n

(iii) Standard deviation for a discrete frequency
1
distribution is given by σ = √N ∑fi (x1 − x‾)2

1 ∑x2
(iv)σ2 = n ∑(xi − x‾)2 = n i − (x‾)2

(v) Combined variance of two data sets
1 n n
σ2 = n +n [n1 σ12 + n2 σ22 + n 1+n2 (x̅1 − x̅̅̅)
2 ]
2
1 2 1 2

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Some important Results on Variance:
n2 −1
 Variance of first ' n ' natural numbers: σ2 = 12

n2 −1
 Standard deviation of first ' n ' natural numbers: σ = √ 12
n2 −1
 Variance of first ' n ' odd natural numbers: σ2 = 3
n2 −1
 Variance of first ' n ' even natural numbers: σ2 = 3

Properties of Variance
Let x1 , x2 , x3 , … , xn be n observations with variance σ2 and standard deviation σ. Then
 variance of x1 ± k, x2 ± k, x3 ± k … , xn ± k is σ2
 Standard deviation of x1 ± k, x2 ± k, x3 ± k, … , xn ± k is σ.
 variance of kx1 , kx2 , kx3 , … , kxn is k 2 σ2
 Standard deviation of kx1 , kx2 , kx3 , … , kxn is kσ
 From above results Var(aX + b) = a2 Var(X)
Relationship between measure of dispersion : 9( Q.D. ) = 7.5 (M.D. ) = 6( S.D. )
5
(a) Q.D. = 6 (M.D.)
2
(b) QD = 3 ( S. D)
4
(c) M.D. = 5 ( S.D. )

Coefficient Of Variation:
σ
C. V = x‾ × 100, where σ is the standard deviation and x‾ is the arithmetic mean

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MEAN AND MEDIAN
1. Mean of 20 observations is 15.5. Later it was found that the observation 24 was
misread as 42. The corrected mean is (Average)
(A) 14.2 (B) 14.8 (C) 14.0 (D) 14.6
2. If the mean of four observations is 20 and when a constant 𝐜 is added to each
observation, the mean becomes 22 . The value of 𝐜 is (Easy)
(A) -2 (B) 2 (C) 4 (D) 6
3. Given (i) 85 observations which are not sorted and (ii) 150 observations which are
sorted and arranged in an increasing order.
The median values of (i) & (ii) respectively can be found as (Average)

(A) (i) 43rd observation (ii) A.M. of 75th and 76th observation

(B) (i) 43rd observation (ii) 76th observation
(C) (i) cannot be found (ii) cannot be found
(D) None of these
𝟒𝟔𝐧
4. If the mean of 𝐧 observations 𝟏𝟐 , 𝟐𝟐 , 𝟑𝟐 , … , 𝐧𝟐 is 𝟏𝟏 , then 𝐧 is equal to (Average)

(A) 11 (B) 12 (C) 23 (D) 22
5. In a batch of 15 students, if the marks of 10 students who passed are
𝟕𝟎, 𝟓𝟎, 𝟗𝟓, 𝟒𝟎, 𝟔𝟎, 𝟕𝟎, 𝟖𝟎, 𝟗𝟎, 𝟕𝟓, 𝟖𝟎 then the median marks of all the 15 students is
(Difficult)
(A) 40 (B) 50 (C) 60 (D) 70
6. The mean of six numbers is 30. If one number is excluded, the mean of the remaining
numbers is 29. The excluded number is (Easy)
(A) 29 (B) 30 (C) 35 (D) 45
7. The arithmetic mean of a set of observations is 𝐱‾. If each observation is divided by 𝛂 ,
then it is increased by 10, the mean of the new series is: (Average)
x‾ x‾+10 x‾+10α
(A) α (B) (C) (D) αx‾ + 10
α α

8. The observations 𝟐𝟗, 𝟑𝟐, 𝟒𝟖, 𝟓𝟎, 𝐱, 𝐱 + 𝟐, 𝟕𝟐, 𝟕𝟖, 𝟖𝟒, 𝟗𝟓 are arranged in ascending order.
What is the value of 𝐱 if the median of the data is 63? (Average)
(A) 61 (B) 62 (C) 62.5 (D) 63
9. The mean of 13 observations is 14. If the mean of the first 7 observations is 12 and
that of the last 7 observations is 16 , what is the value of the 𝟕th observation?(Average)
(A) 12 (B) 13 (C) 14 (D) 15

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10. The average of 5 quantities is 6, the average of three of them is 4, then the average of
remaining two numbers is (Easy)
(A) 9 (B) 6 (C) 10 (D) 5
11. The mean of a set of 20 observation is 19.3. The mean is reduced by 0.5 when a new
observation is added to the set. The new observation is (Average)
(A) 19.8 (B) 8.8 (C) 9.5 (D) 30.8
12. Consider the following data which represents the runs scored by two batsmen in their
last ten matches as
Batsman A: 𝟑𝟎, 𝟗𝟏, 𝟎, 𝟔𝟒, 𝟒𝟐, 𝟖𝟎, 𝟑𝟎, 𝟓, 𝟏𝟏𝟕, 𝟕𝟏
Batsman B:𝟓𝟑, 𝟒𝟔, 𝟒𝟖, 𝟓𝟎, 𝟓𝟑, 𝟓𝟑, 𝟓𝟖, 𝟔𝟎, 𝟓𝟕, 𝟓𝟐
Which of the following is/are true about the data?
I. Mean of batsman 𝐀 runs is 53.
II. Median of batsman 𝐀 runs is 42.
III. Mean of batsman 𝐁 runs is 53.
IV. Median of batsman 𝐁 runs is 53. (Difficult)
(A) Only I is true
(B) I and III are true
(C) I, III and IV are true
(D) All are true

MEAN DEVIATION
13. Find the mean deviation about the mean for the data 𝟒, 𝟕, 𝟖, 𝟗, 𝟏𝟎, 𝟏𝟐, 𝟏𝟑, 𝟏𝟕 (Average)
(A) 3 (B) 24 (C) 10 (D) 8
14. Assertion: The mean deviation of the data 2,9,9,3,6,9,4 from the mean is 2.57
∑|xi −x‾|
Reason: For individual observation, Mean deviation (X‾) = (Difficult)
n

(A) Assertion is correct, reason is correct; reason is a correct explanation for assertion.
(B) Assertion is correct, reason is correct; reason is not a correct explanation for assertion
(C) Assertion is correct, reason is incorrect
(D) Assertion is incorrect, reason is correct.
15. The mean deviation about the mean of the data 𝟑, 𝟏𝟎, 𝟏𝟎, 𝟒, 𝟕, 𝟏𝟎, 𝟓 : [2015]
(A)3 (B)2 (C)75 (D)2.57
16. The mean deviation about the mean for the data 𝟒, 𝟕, 𝟖, 𝟗, 𝟏𝟎, 𝟏𝟐, 𝟏𝟑, 𝟏𝟕 is [2025]
(A) 4.03 (B) 10 (C) 3 (D) 8.5

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17. Find the mean deviation about the mean for the data. (Average)
𝐱𝐢 5 10 15 20 25
𝐟𝐢 7 4 6 3 5
(A) 6 (B) 7.3 (C) 8 (D) 6.32
18. The mean deviation from the mean of the set of observations −𝟏, 𝟎 and 4 is(Average)
(A) 3 (B) 1 (C) -2 (D) 2
19. Consider the following data
𝐱𝐢 15 21 27 30 35
𝐟𝐢 3 5 6 7 8

Then, the mean deviation about the median for the data is (Average)
(A) 5 (B) 5.3 (C) 5.1 (D) 5.2
20. Find the mean deviation about the median for the data given below:
𝟏𝟏, 𝟑, 𝟖, 𝟕, 𝟓, 𝟏𝟒, 𝟏𝟎, 𝟐, 𝟗 (Average)
(A) 3 (B) 2 (C) 5 (D) 4
STANDARD DEVIATION
21. If for a distribution 𝚺(𝐱 − 𝟓) = 𝟑,
𝚺(𝐱 − 𝟓)𝟐 = 𝟒𝟑 and the total number of items is 18. Find the standard deviation.
(Average)
(A) 1.53 (B) 1.43 (C) 1.55 (D) None of these
22. The standard deviation of a distribution is 30 and each item is raised by 3 , then new
𝐒. 𝐃.is (Easy)
(A) 32 (B) 28 (C) 27 (D) None of these
23. From a frequency distribution consisting of 18 observations, the mean and the
standard deviation were found to be 7 and 4 respectively. But on comparison with the
original data, it was found that a figure 12 was miscopied as 21 in calculations. Find
the correct standard deviation. (Difficult)
(A) 3.5 (B) 4.5 (C) 5.5 (D) 2.5
24. The mean and variance for the data 𝟔, 𝟕, 𝟏𝟎, 𝟏𝟐, 𝟏𝟑, 𝟒, 𝟖, 𝟏𝟐 respectively are(Average)
(A) 8, √26.25 (B) 9, √9.25 (C) 8,26.25 (D) 9,9.25
25. The mean and standard deviation of 100 observations were calculated as 40 and 5.1,
respectively by a student who took by mistake 50 instead of 40 for one observation.
Find the correct mean. (Average)
(A) 38.9 (B) 37.9 (C) 39.9 (D) 36.9
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26. If each of the observations 𝐱 𝟏 , 𝐱 𝟐 , … , 𝐱 𝐧 is increased by ' 𝐚′ , where 𝐚 is negative or
positive number, then the variance (Easy)
(A) same (B) increased by a2 times
(C) decreased by a2 times (D) None of these
𝟒𝟗
27. The variance of the numbers 𝟐, 𝟑, 𝟏𝟏 and 𝐱 is 𝟒 . Find the value of 𝐱. (Average)
14 14 16 13
(A) 6, 3 (B) 6, 5 (C) 6, 3 (D) 4, 5

28. Standard deviation of the first 𝟐𝐧 + 𝟏 natural numbers is equal to (Average)
n(n+1) n(n+1)(2n+1) n(n+1) n(n−1)
(A) √ (B) √ (C) √ (D) √
2 3 3 2

29. The variance of first 50 even natural numbers is (Easy)
833 437
(A) (B) 833 (C) 437 (D)
4 4

30. Find the 𝐒. D. of first 10 multiples of 3. (Average)
(A) 8.61 (B) 8.21 (C) 8.51 (D) 8.31
31. The mean of 5 observations is 4.4 and their variance is 8.24. If three of the
observations are 1,2 and 6 , find the other two observations. (Easy)
(A) 4,9 (B) 3,9 (C) 4,4 (D) 9,9
32. The standard deviations of two sets containing 10 and 20 members are 2 and 3
respectively measured from their common mean 5. The S.D. for the whole set of 30
members is (Average)
2 22
(A)√3 (B)√6 (C) √ 3 (D)√3

𝟐
33. If 𝐱 𝟏 , 𝐱 𝟐 , … … 𝐱 𝟏𝟖 are observations such, that ∑𝟏𝟖 𝟏𝟖
𝐣=𝟏 (𝐱 𝐣 − 𝟖) = 𝟗 and ∑𝐣=𝟏 (𝐱 𝐣 − 𝟖) = 𝟒𝟓, then

the standard deviation of these observations is (Average)
81 3
(A) √34 (B) 5 (C) √5 (D) 2

34. If the coefficient of variation and standard deviation are 60 and 21 respectively, the
arithmetic mean of distribution is [2014]
(A)60 (B)30 (C)35 (D)21
35. Mean and standard deviation of 100 items are 50 and 4 respectively. The sum of all
squares of the items is [2019 ,2026]
(A)261600 (B)266000 (C)256100 (D)251600
36. The Standard Deviation of the data 6, 7, 8, 9, 10 is [2020]
(A) √10 (B) 2 (C) 10 (D) √2

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37. The Standard Deviation of the numbers 31,32, 𝟑𝟑 … … . 𝟒𝟔, 𝟒𝟕 is [2021]
17 472 −1
(A) √12 (B) √ 12 (C) 2√6 (D) 4√3

38. If the standard deviation of the numbers −𝟏, 𝟎, 𝟏, 𝐤 is √𝟓 where 𝐤 > 𝟎 then 𝐤 is equal to
[2022]
10 5
(A)2√ 3 (B)2√6 (C)4√3 (D)√6

39. The mean of 100 observations is 50 and their standard deviation is 5. Then the sum
of squares of all observations is [2013]
(A) 250000 (B) 255000 (C) 50000 (D) 252500
40. Let 𝒂, 𝒃, 𝒄, 𝒅 and e be the observations with mean m and standard deviation S . The
standard deviation of the observations 𝒂 + 𝒌, 𝒃 + 𝒌, 𝒄 + 𝒌, 𝒅 + 𝒌 and 𝒆 + 𝒌 is [2024]
S
(A) Ks (B) S + k (C) r (D) S

STATISTICS
1 2 3 4 5 6 7 8 9 10
D B D A C C C B C A
11 12 13 14 15 16 17 18 19 20
B C A A B C D D C A
21 22 23 24 25 26 27 28 29 30
A D D D C A A C B A
31 32 33 34 35 36 37 38 39 40
A C D C D D C B D D

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