Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Random Variables and Probability Distributions Solutions Exercise 10(a) will help students to clear their doubts quickly.

## Intermediate 2nd Year Maths 2A Random Variables and Probability Distributions Solutions Exercise 10(a)

I.

Question 1.

A p.d.f of a discrete random variable is zero except at the points x = 0, 1, 2. At these points it has the value P(0) = 3c^{3}, P(1) = 4c – 10c^{2}, P(2) = 5c – 1 for some c > 0. Find the value of c.

Solution:

P(X = 0) + P(X = 1) + P(X = 2) = 1

3c^{3} + 4c – 10c^{2} + 5c – 1 = 1

3c^{3} – 10c^{2} + 9c – 2 = 0

Put c = 1, then 3 – 10 + 9 – 2 = 12 – 12 = 0

c = 1 satisfies the above equation

c = 1 ⇒ P(X = 0) = 3 which is not possible

Dividing with c – 1, we get

3c^{2} – 7c + 2 = 0

(c – 2) (3c – 1) = 0

∴ c = 2 or c = \(\frac{1}{3}\)

c = 2

⇒ P(X = 0) = 3 . 2^{3} = 24 which is not possible

∴ c = \(\frac{1}{3}\)

Question 2.

Find the constant C, so that F(x) = \(C\left(\frac{2}{3}\right)^x\), x = 1, 2, 3,……… is the p.d.f of a discrete random variable X.

Solution:

Question 3.

is the probability distribution of a random variable X. Find the value of K and the variance of X.

Solution:

Sum of the probabilities = 1

⇒ 0.1 + k + 0.2 + 2k + 0.3 + k = 1

⇒ 4k + 0.6 = 1

⇒ 4k = 1 – 0.6 = 0.4

⇒ k = \(\frac{0.4}{4}\) = 0.1

Mean = (-2) (0.1) + (-1) (k) + 0 (0.2) + 1 (2k) + 2(0.3) + 3k

= – 0.2 – k + 0 + 2k + 0.6 + 3k

= 4k + 0.4

= 4(0.1) + 0.4

= 0.4 + 0.4

= 0.8

μ = 0.8

Variance (σ^{2}) = \(\sum_{i=1}^n x_i^2 P\left(x=x_i\right)-\mu^2\)

∴ Variance = 4(0.1) + 1(k) + 0(0.2) + 1 (2k) + 4 (0.3) + 9k – μ^{2}

= 0.4 + k + 0 + 2k + 4(0.3) + 9k – μ^{2}

= 12k + 0.4 + 1.2 – (0.8)^{2}

= 12(0.1) + 1.6 – 0.64

= 1.2 + 1.6 – 0.64

= 2.8 – 0.64

= 2.16

∴ σ^{2} = 2.16

Question 4.

is the probability distribution of a random variable X. Find the variance of X.

Solution:

Question 5.

A random variable X has the following probability distribution.

Find (i) k (ii) the mean and (iii) P(0 < X < 5)

Solution:

Sum of the probabilities = 1

⇒ 0 + k + 2k + 2k + 3k + k^{2} + 2k^{2} + 7k^{2} + k = 1

⇒ 10k^{2} + 9k = 1

⇒ 10k^{2} + 9k – 1 = 0

⇒ 10k^{2} + 10k – k – 1 = 0

⇒ 10k(k + 1) – 1(k + 1) = 0

⇒ (10k – 1) (k + 1) = 0

⇒ k = \(\frac{1}{10}\), -1 Since k > 0

∴ k = \(\frac{1}{10}\)

(i) k = \(\frac{1}{10}\)

(ii) Mean = \(\sum_{i=1}^n x_i P\left(x=x_i\right)\)

Mean (μ) = 0(0) + 1(k) + 2(2k) + 3(2k) + 4(3k) + 5(k2) + 6(2k^{2}) + 7 (7k^{2} + k)

= 0 + k + 4k + 6k + 12k + 5k^{2} + 12k^{2} + 49k^{2} + 7k

= 66k^{2} + 30k

= 66(\(\frac{1}{100}\)) + 30(\(\frac{1}{10}\))

= 0.66 + 3

= 3.66

(iii) P(0 < x < 5)

P(0 < x < 5) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

= k + 2k + 2k + 3k

= 8k

= 8(\(\frac{1}{10}\))

= \(\frac{4}{5}\)

II.

Question 1.

The range of a random variable X is {0, 1, 2}. Given that P(X = 0) = 3c^{3}, P(X = 1) = 4c – 10c^{2}, P(X = 2) = 5c – 1

(i) Find the value of c

(ii) P(X < 1), P(1 ≤ X < 2) and P(0 < X ≤ 3)

Solution:

P(X = 0) + P(X = 1) + P(X = 2) = 1

3c^{3} + 4c – 10c^{2} + 5c – 1 = 1

3c^{3} – 10c^{2} + 9c – 2 = 0

c = 1 satisfy this equation

c = 1 ⇒ P(X = 0) = 3 which is not possible

Dividing with c – 1, we get

3c^{2} – 7c + 2 = 0

(c – 2) (3c – 1) = 0

c = 2 or c = \(\frac{1}{3}\)

c = 2 ⇒ P(X = 0) = 3 . 2^{3} = 24 which is not possible

∴ c = \(\frac{1}{3}\)

(i) P(X < 1) = P(X = 0)

= 3 . c^{3}

= 3 . \(\left(\frac{1}{3}\right)^3\)

= 3 . \(\frac{1}{2}\)

= \(\frac{1}{9}\)

(ii) P(1 < X ≤ 2) = P(X = 2)

= 5c – 1

= \(\frac{5}{3}\) – 1

= \(\frac{2}{3}\)

(iii) P(0 < X ≤ 3) = P(X = 1) + P(X = 2)

= 4c – 10c^{2} + 5c – 1

= 9c – 10c^{2} – 1

= 9 . \(\frac{1}{3}\) – 10 . \(\frac{1}{9}\) – 1

= 3 – \(\frac{10}{9}\) – 1

= \(\frac{8}{9}\)

Question 2.

The range of a random variable X is {1, 2, 3, …..} and P(X = k) = \(\frac{C^K}{K !}\), (k = 1, 2, 3, ……), Find the value of C and P(0 < X < 3)

Solution:

Sum of the probabilities = 1