MCQ Questions for Class 11 Maths Chapter 15 Statistics with Answers

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

Answer: (b) 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

Answer: (d) 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

Answer: (a) 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

Answer: (b) S.D./Mean
The coefficient of variation = S.D./Mean

Answer: (b) 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

Answer: (b) 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

Answer: (c) 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

Answer: (b) 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

Answer: (b) 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

Answer: (d) 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

Answer: (b) 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

Answer: (a) both origin and the scale
The coefficient of correlation between two variables is independent of both origin and the scale.

Answer: (b) 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

Answer: (c) Mode = 3 Median – 2 Mean
We can calculate the mode as
Mode = 3 Median – 2 Mean

Answer: (c) coincident
If the correlation coefficient between two variables is 1, then the two least square lines of regression are coincident

Answer: (b) 6, 8
Given mean and variance of 7 observations are 8 and 16.
Five observations are 2, 4, 10, 12, 14.
Let the other two observations are x and y.
So 7 observations are : 2, 4, 10, 12, 14 ,x ,y
Now
Mean = (2 + 4 + 10 + 12 + 14 + x + y)/7
⇒ 8 = (2 + 4 + 10 + 12 + 14 + x + y)/7
⇒ 8 × 7 = 2 + 4 + 10 + 12 + 14 + x + y
⇒ 56 = 42 + x + y
⇒ x + y = 56 – 42
⇒ x + y = 14 ………….. 1
Again Given varience = 16
⇒ (1/7) × ∑ (xi – mean)² = 16 (7 <= i <= 1)
⇒ ∑ (xi – mean)² = 16 × 7
⇒ ∑ (xi – mean)² = 112
⇒ {(2 – 8)² + (4 – 8)² + (10 – 8)² + (12 – 8)² + (14 – 8)² + (x – 8)² + (y – 8)²} = 112
⇒ {(-6)² +(-4)² + (2)² + (4)² + (6)² + x² + 64 – 16x + y² + 64 – 16y } = 112
⇒ {36 + 16 + 4 + 16 + 36 + x² + y² + 64 + 64 – 16(x + y) } = 112
⇒ {108 + x² + y² + 128 – (16 × 14)} = 112 (since x + y = 14)
⇒ {108 + x² + y² + 128 – 224} = 112
⇒ x² + y² + 236 – 224 = 112
⇒ x² + y² + 12 = 112
⇒ x² + y² = 12 – 12
⇒ x² + y² = 100…………….2
Squaring equation 1, we get
(x + y)² = 196
⇒ x² + y² + 2xy = 196
⇒ 100 + 2xy = 196
⇒ 2xy = 196 – 100
⇒2xy = 96
⇒ xy = 96/2
⇒ xy = 48 …………. 3
Now (x – y)² = x² + y² – 2xy
= 100 – 2 × 48
= 100 – 96
= 4
⇒ x – y = √2
⇒ x – y = 2, -2
case 1: when x – y = 2 and x + y = 14
After solving it, we get x = 8, y = 6
case 2: when x – y = -2 and x + y = 14
After solving it, we get x = 6, y = 8
So, the two numbers are 6 and 8

Answer: (a) Range = Max Value – Min Value
Range of a data is calculated as
Range = Max Value – Min Value

Answer: (b) {∑|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)

Answer: (b) 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