 # RD Sharma Solutions Class 12 Mean And Variance Of A Random Variable Exercise 32.1

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EXERCISE 32.1

Question. 1(i)

Solution:

From the question distribution is 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 is 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 is 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 is 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 is given,

x : -2 -1 0 1 2 3

P(x) : 0.1 k 0.2 2k 0.3 k

Then, Sum of probabilities

[P(x = -2) + P(x = -1) + P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3)] = 1

0.1 + 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, value of K is 0.1.

Question. 3

Solution:

From the question distribution is given,

x : 0 1 2 3 4 5 6 7 8

P(x) : a 3a 5a 7a 9a 11a 13a 15a 17a

(i)

Then, Sum of 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, 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, 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 is 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 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, 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