Maths MCQ Class 11 Ch-15 | Statistics

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  Mathematics

MCQ | Class 11 | Chapter 15
STATISTICS

Multiple Choice Questions (MCQ)

  • MCQ Based on the  Mean Deviation about Mean.

  • MCQ  Based on the Mean Deviation about Median.

  • MCQ Based on the Variance and Standard Deviation of the given data.

Features

  • In this pdf given below you find the important MCQ which are strictly according to the CBSE syllabus and are very useful for the CBSE Examinations. 
  • Solution Hints are also given to some difficult problems. 
  • Each MCQ contains four options from which one option is correct. 
  • On the right hand side column of the pdf Answer option is given.

Action Plan

  • First of all students should Learn and write all basic points and Formulas related to the Statistics.
  • Start solving  the NCERT Problems with examples.
  • Solve the important assignments on the Statistics Chapter 15 Class XI.
  • Then start solving the following MCQ.

MCQ | CHAPTER 15 | CLASS 11
STATISTICS

Q 1)  Find the mean deviation
about the median of the scores of a batsman given below

Trending
Maths MCQ Class 12 Ch-2 | Inverse Trigonometric Functions

Innings

1

2

3

4

5

6

Scores

20

5

0

84

11

120

a) 10
b)10.5
c) 11
d) 9.
Ans: b

Q 2) If mode of a series exceeds its mean by 12, then mode exceeds the median by:
a. 4
b. 8
c. 6
d. 10

 Ans: b

Q 3) If the variance of the data is 121 then the standard deviation of the data is
(a) 121
(b) 11
(c) 12
(d) 21 
Ans: b

Q 4)  What
is the mean deviation from the mean for the following data?

117

156

206

198

223

a) 0
b) 3
c) 1
d) 5 

Ans: a

Q 5) Range of the data 4, 7, 8, 9, 10, 12, 13 and 17 is
(a) 4
(b) 17
(c) 13
(d) 21 
Ans: c

Q 6) The mean of 1, 3, 4, 5, 7, 4 is m. The numbers 3, 2, 2, 4, 3, 3, p have mean m – 1 and median q. Then, p + q =
a. 5
b. 7
c. 4
d. 6 
Ans: b

Q 7) The mean deviation of an ungrouped data is 150. If each observation is increased by 3.5%, then what is the new mean deviation?
a) 153.5
b) 3.5
c) 155.25
d) 150 
Ans: c

Q 8)  Find
the mean deviation about mean from the following data:

xi

3

5

20

25

27

fi

5

12

20

8

15

a) 7.7
b) 15
c) 8.7
d) 6.2 

Ans: a
Q 9)  If the mean of the following
data is 20.6, then the value of p is

X

10

15

p

25

35

F

3

10

25

7

5

(a) 30
(b) 20
(c) 25
(d) 10 

Ans: b

Q 10) In a class there are 20 juniors, 15 seniors and 5 graduate students. If the junior averaged 65 in the midterm exam, the senior averaged 70 and the graduate students averaged 91, then what is the mean of the centre class approximately?
a) 71
b) 74
c) 70
d) 72 
Ans: c

Q 11) The most frequently occurring number in a set of values is called
a. Median
b. Range
c. Mean
d. Mode 
Ans: d

Q 12)  Find
the variance of the observation values taken in the lab.

4.2

4.3

4

4.1

a) 0.27
b) 0.28
c) 0.3
d) 0.31
Ans: b

Q 13) If the standard deviation of a data is 0.012. Find the variance.
a) 0.144
b) 0.00144
c) 0.000144
d) 0.0000144 

Ans: c

Q 14) When tested the lives (in hours) of 5 bulbs were noted as follows: 1357, 1090, 1666, 1494, 1623. The mean of the lives of 5 bulbs is

(a) 1445
(b) 1446
(c) 1447
(d) 1448
Ans: b

Q 15) Find the variance of the first 10 natural numbers.
a) 7.25
b) 7
c) 8.25
d) 8
Ans: c

Q 16) The standard deviation of some temperature data in °C is 5. If the data were converted into °F, the variance would be

a. 81
b. 36
c. 25
d. 57
Ans: a

Q 17) The algebraic sum of the deviation of 20 observations measured from 30 is 2. So, the mean of observations is

(a) 30.0
(b) 30.1
(c) 30.2
(d) 30.3
Ans: b
Solution Hint
Given, algebraic sum of of the deviation of 20 observations measured from 30 is 2
⇒ ∑(xi – 30) = 2 {1 ≤ i ≤ 20} ⇒ ∑xi – 30 × 20 = 2
⇒ (∑xi)/20 – (30 × 20)/20 = 2/20 ⇒ (∑xi)/20 – 30 = 0.1
⇒ Mean – 30 = 0.1 ⇒ Mean = 30 + 0.1 ⇒ Mean = 30.1

Q 18) Assuming the variance of four numbers w, x, y, and z as 9. Find the variance of 5w, 5x, 5y and 5z.
a) 225
b) 5/9
c) 9/5
d) 54
Ans: a

Solution Hint (σx)2 = h2(σu)2, if u = (x –
h)/a
Now, h = (1/5). ⇒ Variance, (σu)2 = 9 × 25 = 225
Q 19) The median and SD of a distributed are 20 and 4 respectively. If each item is increased by 2, the new median and SD are
(a) 20, 4
(b) 22, 6
(c) 22, 4
(d) 20, 6
Ans: c
Solution Hint 
Since each value is increased by 2, therefore the median value is also increased by 2. So, new median = 22
Again, the variance is independent of the change of origin. So it remains the same. Hence standard deviation also remain same.



Q 20) What is the median and standard deviation of a distribution are 50 and 5 respectively, if each item is increased by 4.
a) Median will increase and S.D. will increase
b) Both will remain same
c) median will go up by 2 but S.D. will remain same
d) median will increase and S.D. will decrease
Ans: c
Solution Hint 
Median will change if the observations are changes but standard deviation is remains unaffected by the change in origin. So, here the median will go up by 4 and S.D. will remain same.

Q 21) The difference between the upper and the lower class limits is called
a. mean
b. class size
c. frequency
d. mid-points 

Ans: b

Q 22) If the S.D. is a set of observations is 4 and if each observation is divided by 4, find the S.D. of the new observations.
a) 4
b) 3
c) 2
d) 1 
Ans: d

Q 23) The change in which of following terms does not affect the standard deviation?
a) Origin
b) Scale
c) Origin and scale
d) Neither origin nor scale
Ans: a
Solution Hint Change in origin does not affect the standard deviation, whereas standard deviation is affected by scale.

Q 24) The mean and Standard deviation of a sample were found to be 9.5 and 2.5, respectively. Later, an additional observation 15 was added to the original data. Find the S.D. of the 11 observation.
a) 2.6
b) 2.8
c) 2.86
d) 3.24 
Ans: c

Q 25) The following marks were obtained by the students in a test: 81, 72, 90, 90, 86, 85,92, 70, 71, 83, 89, 95,85, 79, 62 The range of the marks is
a. 9
b. 33
c. 27
d. 17 
Ans: b

Q 26) The mean of 5 observations is 3 and variance is 2. If three of the five observations are 1, 3, 5, find the other two.
a) 2, 6
b) 3, 3
c) 1, 5
d) 2, 4 
Ans: d

Q 27) The average of four numbers is 60. If first number is one-fourth of the sum of the last three, the first number is:
a. 48
b. 15
c. 42
d. 45 
Ans: a

Q 28) The mean of a group of 100 observations was found to be 20. Later on, it was found that three observations were incorrect, which was recorded as 21, 21 and 18. Then the mean if the incorrect observations are omitted is
(a) 18
(b) 20
(c) 22
(d) 24
Ans: b
Solution Hint
Given mean of
100 observations is 20
Now  
∑ xi/100 = 20 (1 = i = 100)
⇒ ∑xi =
100×20                   
⇒ ∑xi = 2000
3 observations 21, 21 and 18 are recorded
in-correctly.
So ∑xi = 2000 – 21 – 21 – 18
⇒ ∑xi =
2000 – 60                

⇒ ∑xi = 1940
Now new mean is
∑ xi/100 = 1940/97 = 20
So, the new mean is 20

Q 29) The sum of 10 items is 12 and the sum of their squares is 18. The standard deviation is
(a) 1/5
(b) 2/5
(c) 3/5
(d) 4/5
Ans: c

Solution Hint

Given, ∑x = 12
and ∑x² = 18

Now, variance = ∑x²/n – (∑x/n)²

⇒
variance = 18/10 – (12/10)²             

⇒ variance = 9/5 – (6/5)²

⇒
variance = 9/5 – 36/25                    

⇒ variance = (9 × 5 – 36)/25

⇒
variance = (45 – 36)/25                   

⇒ variance = 9/25

⇒
Standard deviation = √(9/25)         

⇒ Standard deviation = 3/5

Q 30) If the mean of first n natural numbers is 5n/9, then n =
(a) 5
(b) 4
(c) 9
(d) 10
Ans: c

Solution Hint
Given mean of first n natural number is 5n/9

⇒
(n+1)/2 = 5n/9                         

⇒ n + 1 = (5n×2)/9

⇒ n + 1
= 10n/9                          

⇒ 9(n + 1) = 10n

⇒ 9n + 9
= 10n                           

⇒ 10n – 9n = 9              

⇒ n = 9

Q 31) The following is the data of wages per day: 5, 4, 7, 5, 8, 8, 8, 5, 7, 9, 5, 7, 9, 10, 8. The mode of the data is:
a. 7
b. 8
c. 5
d. 10 
Ans: b

Q 32) The average of 5 consecutive numbers is 16. The largest of these number is:
a. 18
b. 19
c. 21
d. 20 
Ans: a

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