MCQ Questions for Class 11 Maths Chapter 15 Statistics with Answers

Statistics Class 11 Maths MCQs Questions with Answers

Check the below NCERT MCQ Questions for Class 11 Maths Chapter 15 Statistics with Answers Pdf free download. MCQ Questions for Class 11 Maths with Answers were prepared based on the latest exam pattern. We have Provided Statistics Class 11 Maths MCQs Questions with Answers to help students understand the concept very well.

Class 11 Maths Chapter 15 Quiz

Class 11 Maths Chapter 15 MCQ Online Test

You can refer to NCERT Solutions for Class 11 Maths Chapter 15 Statistics to revise the concepts in the syllabus effectively and improve your chances of securing high marks in your board exams.

Statistics Class 11 Maths MCQ online test

Q1.	If the varience of the data is 121 then the standard deviation of the data is
	A.121
	B.11
	C.12
	D.21

Ans: 11
Given, varience of the data = 121
Now, the standard deviation of the data = √(121)
= 11

Q2.	The mean deviation from the mean for the following data: 4, 7, 8, 9, 10, 12, 13 and 17 is
	A.2
	B.3
	C.4
	D.5

Ans: 3
Mean = (4 + 7 + 8 + 9 + 10 + 12 + 13 + 17)/10 = 80/10 = 8
|x_i – mean|= |4 – 10| + |7 – 10| + |8 – 10| + |9 – 10| + |10 – 10| + |12 – 10| + |13 – 10| + |17 – 10|
= 6 + 3 + 2 + 1 + 0 + 2 + 3 + 7 = 24
Now, mean deviation form mean = 24/8 = 3

Q3.	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.4
	B.5
	C.6
	D.7

Ans: 7
The mean of 1, 3, 4, 5, 7, 4 is m
⇒ (1 + 3 + 4 + 5 + 7 + 4)/6 = m
⇒ m = 24/6
⇒ m = 4
The numbers 3, 2, 2, 4, 3, 3, p have mean m – 1
⇒ (3 + 2 + 2 + 4 + 3 + 3 + p)/7 = m – 1
⇒ (17 + p)/7 = 4 – 1
⇒ (17 + p)/7 = 3
⇒ 17 + p = 7 × 3
⇒ 17 + p = 21
⇒ p = 21 – 17
⇒ p = 4
The numbers 3, 2, 2, 4, 3, 3, p have median q.
⇒ The numbers 2, 2, 3, 3, 3, 4, 4 have median q
⇒ (7 + 1)/2 th term = q
⇒ 4th term = q
⇒ q = 3
Now p + q = 4 + 3 = 7

Q4.	If the difference of mode and median of a data is 24, then the difference of median and mean is
	A.12
	B.24
	C.8
	D.36

Ans: 12
Given the difference of mode and median of a data is 24
⇒ Mode – Median = 24
⇒ Mode = Median + 24
Now, Mode = 3 × Median – 2 × Mean
⇒ Median + 24 = 3 × Median – 2 × Mean
⇒ 24 = 3 × Median – 2 × Mean – Median
⇒ 24 = 2 × Median – 2 × Mean
⇒ Median – Mean = 24/2
⇒ Median – Mean = 12

Q5.	The coefficient of variation is computed by
	A.S.D/.Mean × 100
	B.S.D./Mean
	C.Mean./S.D × 100
	D.Mean/S.D.

Ans: S.D./Mean
The coefficient of variation = S.D./Mean

Q6.	The geometric mean of series having mean = 25 and harmonic mean = 16 is
	A.16
	B.20
	C.25
	D.30

Ans: 20
The relationship between Arithmetic mean (AM), Geometric mean (GM) And Harmonic mean (HM) is
GM² = AM × HM
Given AM = 25
HM = 16
So GM² = 25 × 16
⇒GM = √(25 × 16)
= 5 × 4
= 20
So, Geometric mean = 20

Q7.	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: 1446
Given, lives (in hours) of 5 bulbs were noted as follows: 1357, 1090, 1666, 1494, 1623
Now, mean = (1357 + 1090 + 1666 + 1494 + 1623)/5
= 7230/5
= 1446

Q8.	Mean of the first n terms of the A.P. a + (a + d) + (a + 2d) + ……… is
	A.a + nd/2
	B.a + (n – 1)d
	C.a + (n − 1)d/2
	D.a + nd

Ans: a + (n − 1)d/2
Mean of the first n terms of the A.P. {a + (a + d) + (a + 2d) + ……… a + (n-1)d}/n
= (n/2){2a + (n – 1)d}/n
= (1/2){2a + (n – 1)d}
= a + (n – 1)d/2

Q9.	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: 20
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 incorrectly.
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

Q10.	If covariance between two variables is 0, then the correlation coefficient between them is
	A.nothing can be said
	B.0
	C.positive
	D.negative

Ans: 0
The relationship between the correlation coefficient and covariance for two variables as shown below:
r_{(x, y)} = COV (x, y)/{s_x × s_y}
r_{(x, y)} = correlation of the variables x and y
COV (x, y) = covariance of the variables x and y
s_x = sample standard deviation of the random variable x
s_y = sample standard deviation of the random variable y
Now given COV (x, y) = 0
Then r_{(x, y)} = 0

Q11.	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.4
	B.5
	C.6
	D.7

Ans: 7
The mean of 1, 3, 4, 5, 7, 4 is m
⇒ (1 + 3 + 4 + 5 + 7 + 4)/6 = m
⇒ m = 24/6
⇒ m = 4
The numbers 3, 2, 2, 4, 3, 3, p have mean m – 1
⇒ (3 + 2 + 2 + 4 + 3 + 3 + p)/7 = m – 1
⇒ (17 + p)/7 = 4 – 1
⇒ (17 + p)/7 = 3
⇒ 17 + p = 7 × 3
⇒ 17 + p = 21
⇒ p = 21 – 17
⇒ p = 4
The numbers 3, 2, 2, 4, 3, 3, p have median q.
⇒ The numbers 2, 2, 3, 3, 3, 4, 4 have median q
⇒ (7 + 1)/2th term = q
⇒ 4th term = q
⇒ q = 3
Now p + q = 4 + 3 = 7

Q12.	In a series, the coefficient of variation is 50 and standard deviation is 20 then the arithmetic mean is
	A.20
	B.40
	C.50
	D.60

Ans: 40
Given, in a series, the coefficient of variation is 50 and standard deviation is 20
⇒ (standard deviation/AM) × 100 = 50
⇒ 20/AM = 50/100
⇒ 20/AM = 1/2
⇒ AM = 2 × 20
⇒ AM = 40
So, the arithmetic mean is 40

Q13.	The coefficient of correlation between two variables is independent of
	A.both origin and the scale
	B.scale but not origin
	C.origin but not scale
	D.neither scale nor origin

Ans: both origin and the scale
The coefficient of correlation between two variables is independent of both origin and the scale.

Q14.	The geometric mean of series having mean = 25 and harmonic mean = 16 is
	A.16
	B.20
	C.25
	D.30

Ans: 20
The relationship between Arithmetic mean (AM), Geometric mean (GM) And Harmonic mean (HM) is
GM² = AM × HM
Given AM = 25
HM = 16
So GM² = 25 × 16
⇒ GM = √(25 × 16)
= 5 × 4
= 20
So, Geometric mean = 20

Q15.	One of the methods of determining mode is
	A.Mode = 2 Median – 3 Mean
	B.Mode = 2 Median + 3 Mean
	C.Mode = 3 Median – 2 Mean
	D.Mode = 3 Median + 2 Mean

Ans: Mode = 3 Median – 2 Mean
We can calculate the mode as
Mode = 3 Median – 2 Mean

Q16.	If the correlation coefficient between two variables is 1, then the two least square lines of regression are
	A.parallel
	B.none of these
	C.coincident
	D.at right angles

Ans: coincident
If the correlation coefficient between two variables is 1, then the two least square lines of regression are coincident

Q17.	The mean and variance of 7 observations are 8 and 16, respectively. If five of the observations are 2, 4, 10, 12, 14. then the remaining two observations are
	A.4, 6
	B.6, 8
	C.8, 10
	D.10, 12

Ans: 6, 8

Q18.	Range of a data is calculated as
	A.Range = Max Value – Min Value
	B.Range = Max Value + Min Value
	C.Range = (Max Value – Min Value)/2
	D.Range = (Max Value + Min Value)/2

Ans: Range = Max Value – Min Value
Range of a data is calculated as
Range = Max Value – Min Value

Q19.	Mean deviation for n observations x₁, x₂, ……….., x_n from their mean x is given by
	A.∑(x_i – x) where (1 ≤ i ≤ n)
	B.{∑\|x_i – x\|}/n where (1 ≤ i ≤ n)
	C.∑(x_i – x)² where (1 ≤ i ≤ n)
	D.{∑(x_i – x)²}/n where (1 ≤ i ≤ n)

Ans: {∑|x_i – x|}/n where (1 ≤ i ≤ n)
Mean deviation for n observations x₁, x₂, ……….., x_n from their mean x is calculated as
{∑|x_i – x|}/n where (1 ≤ i ≤ n)

Q20.	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: 20
Mean = ∑ f_i × xi /∑ f_i
⇒ 20.6 = (10 × 3 + 15 × 10 + p × 25 + 25 × 7 + 35 × 5)/(3 + 10 + 25 + 7 + 5)
⇒ 20.6 = (30 + 150 + 25p + 175 + 175)/50
⇒ 20.6 = (530 + 25p)/50
⇒ 530 + 25p = 20.6 × 50
⇒ 530 + 25p = 1030
⇒ 25p = 1030 – 530
⇒ 25p = 500
⇒ p = 500/25
⇒ p = 20
So, the value of p is 20