EXERCISE 32.1
Question. 1(i)
Solution:
From the question distribution given,
x : 3 2 1 0 -1
P(x): 0.3 0.2 0.4 0.1 0.05
Then,
Sum of probabilities = P(x = 3) + P(x = 2) + P(x = 1) + P(x = 0) + P(x = -1)
= 0.3 + 0.2 + 0.4 + 0.1 + 0.05
= 1.05
So, 1.05 ≠ 1
Therefore, the given distribution is not a probability distribution.
Question. 1(ii)
Solution:
From the question distribution given,
x : 0 1 2
P(x): 0.6 0.4 0.2
Then,
Sum of probabilities = P(x = 0) + P(x = 1) + P(x = 2)
= 0.6 + 0.4 + 0.2
= 1.2
So, 1.2 ≠ 1
Therefore, the given distribution is not a probability distribution.
Question. 1(iii)
Solution:
From the question distribution given,
x : 0 1 2 3 4
P(x): 0.1 0.5 0.2 0.1 0.1
Then,
Sum of probabilities = P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4)
= 0.1 + 0.5 + 0.2 + 0.1 + 0.1
= 1
So, 1 = 1
Therefore, the given distribution is a probability distribution.
Question. 1(iv)
Solution:
From the question distribution given,
x : 0 1 2 3
P(x): 0.3 0.2 0.4 0.1
Then,
Sum of probabilities = P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3)
= 0.3 + 0.2 + 0.4 + 0.1
= 1
So, 1 = 1
Therefore, the given distribution is a probability distribution.
Question. 2
Solution:
From the question distribution given,
x : -2 -1 0 1 2 3
P(x) : 0.1 k 0.2 2k 0.3 k
Then, the Sum of the probabilities
[P(x = -2) + P(x = -1) + P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3)] = 10.1 + k + 0.2 + 2k + 0.3 + k = 1
4k + 0.6 = 1
4k = 1 – 0.6
4k = 0.4
K = 0.4/4
K = 4/40
K = 1/10
K = 0.1
Therefore, the value of K is 0.1.
Question. 3
Solution:
From the question distribution given,
x : 0 1 2 3 4 5 6 7 8
P(x) : a 3a 5a 7a 9a 11a 13a 15a 17a
(i)
Then, the Sum of the probabilities
P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4) + P(x = 5) + P(x = 6) + P(x = 7) + P(x = 8)= 1
a + 3a + 5a + 7a + 9a + 11a + 13a + 15a + 17a = 1
81a = 1
a = 1/81
Therefore, the value of a is 1/81.
(ii) P(x < 3) = P(0) + P(1) + P(2)
= a + 3a + 5a
= 9a
Now substitute the value of a, and we get
= 9(1/81)
So, P(x < 3) = 1/9
P(x ≥ 3) = 1 – P(x < 3)
= 1 – 1/9
= (9 – 1)/9
= 8/9
Then, P(0 < x < 5) = P(1) + P(2) + P(3) + P(4)
= 3a + 5a + 7a + 9a
= 24a
Substituting the value of a, we get,
= 24(1/81)
= 8/27
Therefore, P(0 < x < 5) = 8/27
Question. 4
Solution:
From the question distribution given,
x : 0 1 2
P(x) : 3c3 4c – 10c2 5c – 1
(i) First, we have to find the value of c,
Sum of probabilities = P(x = 0) + P(x = 1) + P(x = 2 = 1
3c3 + 4c – 10c2 + 5c – 1 = 1
3c3 – 10c2 + 9c – 1 – 1 = 0
3c3 – 10c2 + 9c – 2 = 0
Above terms can be written as,
3c3 – 3c2 – 7c2 + 7c + 2c – 2 = 0
Take out common in above terms, and we get,
3c2 (c – 1) – 7c(c – 1) + 2(c – 1) = 0
(c – 1) (3c2 – 7c + 2) = 0
(c – 1) (3c2 – 6c – c + 2) = 0
(c – 1) (3c (c – 2) – 1(c – 2)) = 0
(c – 1) (3c – 1) (c – 2) = 0
C – 1 = 0, 3c – 1 = 0, c – 2 = 0
C = 1, C = 1/3, C = 2
So, it is clear that c = 1/3 is possible because if c = 1, or c = 2, then P(2) will become negative.
(ii) Now, P(x < 2) = P(0) + P(1)
= 3c3 + 4c – 10c2
= 3(1/3)3 + 4(1/3) – 10(1/3)2
= 3/27 + 4/3 – 10/9
= 1/9 + 4/3 – 10/9
= 3/9
Therefore, the value of P(x < 2) is 3/9.
(iii) P(1< c ≤ 2) = P(2)
= 5c – 1
Substitute the value of c we get,
= 5(1/3) – 1
= 2/3
Therefore, the value P(1< c ≤ 2) is 2/3
Question. 5
Solution:
From the question, it is given that,
2P(x1) = 3P (x2) = P(x3) = 5P(x4)
So, let us assume P(x3) = a
Then, 2P(x1) = P(x3)
P(x1) = a/2
3P(x2) = P(x3)
P(x2) = a/3
5P(x4) = P(x3)
P(x4) = a/5
So, P(x1) + P(x2) + P(x3) + P(x4) = 1
a/2 + a/3 + a/1 + a/5 = 1
LCM of 2, 3, 1 and 5 is 30.
(15a + 10a + 30a + 6a)/30 = 1
61a = 30
a = 30/61
Therefore,
x : x1 x2 x3 x4
P(x) : 15/61 10/61 30/61 6/61
EXERCISE 32.2
Question. 1(i)
Solution:
We know that, Mean of any probability distribution = ∑xipi
| xi | pi | xipi | xi2pi |
| 2 | 0.2 | 0.4 | 0.8 |
| 3 | 0.5 | 1.5 | 4.5 |
| 4 | 0.3 | 1.2 | 4.8 |
| ∑xipi = 3.1 | ∑xi2pi = 10.1 |
Then, Variance = ∑pixi2 – (∑xipi)2
= 10.1 – (3.1)2
= 0.49
Therefore, Standard deviation = √Variance
= √0.49
= 0.7
Question. 1(ii)
Solution:
We know that, Mean of any probability distribution = ∑xipi
| xi | pi | xipi | xi2pi |
| 1 | 0.4 | 0.4 | 0.4 |
| 3 | 0.1 | 0.3 | 0.9 |
| 4 | 0.2 | 0.8 | 3.2 |
| 5 | 0.3 | 1.5 | 7.5 |
| ∑xipi = 3 | ∑xi2pi = 12 |
Then, Variance = ∑pixi2 – (∑xipi)2
= 12 – (3)2
= 3
Therefore, Standard deviation = √Variance
= √3
= 1.732
Question. 1(iii)
Solution:
We know that, Mean of any probability distribution = ∑xipi
| xi | pi | xipi | xi2pi |
| -5 | ¼ | -5/4 | 25/4 |
| -4 | 1/8 | -½ | 2 |
| 1 | ½ | ½ | ½ |
| 2 | 1/8 | ¼ | ½ |
| ∑xipi = -1 | ∑xi2pi = 37/4 |
Then, Variance = ∑pixi2 – (∑xipi)2
= 37/4 – (-1)2
= 33/4
Therefore, Standard deviation = √Variance
= √(33/4)
= 2.9
Question. 1(iv)
Solution:
We know that, Mean of any probability distribution = ∑xipi
| xi | pi | xipi | xi2pi |
| -1 | 0.3 | -0.3 | 0.3 |
| 0 | 0.1 | 0 | 0 |
| 1 | 0.1 | 0.1 | 0.1 |
| 2 | 0.3 | 0.6 | 1.2 |
| 3 | 0.2 | 0.6 | 1.8 |
| ∑xipi = 1 | ∑xi2pi = 3.4 |
Then, Variance = ∑pixi2 – (∑xipi)2
= 3.4 – (1)2
= 2.4
Therefore, Standard deviation = √Variance
= √(2.4)
= 1.5
Question. 1(v)
Solution:
We know that, Mean of any probability distribution = ∑xipi
| xi | pi | xipi | xi2pi |
| 1 | 0.4 | 0.4 | 0.4 |
| 2 | 0.3 | 0.6 | 1.2 |
| 3 | 0.2 | 0.6 | 1.8 |
| 4 | 0.1 | 0.4 | 1.6 |
| ∑xipi = 2 | ∑xi2pi = 5 |
Then, Variance = ∑pixi2 – (∑xipi)2
= 5 – (2)2
= 1
Therefore, Standard deviation = √Variance
= √1
= 1
Question. 1(vi)
Solution:
We know that, Mean of any probability distribution = ∑xipi
| xi | pi | xipi | xi2pi |
| 0 | 0.2 | 0 | 0 |
| 1 | 0.5 | 0.5 | 0.5 |
| 3 | 0.2 | 0.6 | 1.8 |
| 5 | 0.1 | 0.4 | 2.5 |
| ∑xipi = 1.6 | ∑xi2pi = 4.8 |
Then, Variance = ∑pixi2 – (∑xipi)2
= 4.8 – (1.6)2
= 4.8 – 2.56
= 2.24
Therefore, Standard deviation = √Variance
= √2.24
= 1.497
Question. 1(vii)
Solution:
We know that, Mean of any probability distribution = ∑xipi
| xi | pi | xipi | xi2pi |
| -2 | 0.1 | -0.2 | 0.4 |
| -1 | 0.2 | -0.2 | 0.2 |
| 0 | 0.4 | 0 | 0 |
| 1 | 0.2 | 0.2 | 0.2 |
| 2 | 0.1 | 0.2 | 0.4 |
| ∑xipi = 0 | ∑xi2pi = 1.2 |
Then, Variance = ∑pixi2 – (∑xipi)2
= 1.2 – (0)2
= 1.2 – 0
= 1.2
Therefore, Standard deviation = √Variance
= √1.2
= 1.095
Question. 1(viii)
Solution:
We know that, Mean of any probability distribution = ∑xipi
| xi | pi | xipi | xi2pi |
| -3 | 0.05 | -0.15 | 0.45 |
| -1 | 0.45 | -0.45 | 0.45 |
| 0 | 0.20 | 0 | 0 |
| 1 | 0.25 | 0.2 5 | 0.25 |
| 3 | 0.05 | 0.15 | 0.45 |
| ∑xipi = -0.2 | ∑xi2pi = 1.6 |
Then, Variance = ∑pixi2 – (∑xipi)2
= 1.6 – (-0.2)2
= 1.6 – 0.04
= 1.56
Therefore, Standard deviation = √Variance
= √1.56
= 1.249