The probability distribution of a random variable X is as given below X - 2 - 1 0 1 2 3 P X 0 . 1 k 0 . 2 2 k 0 . 3 k Find the value of k and the variance of X .

The probability distribution of a random variable X is as given below X - 2 - 1 0 1 2 3 P X 0 . 1 k 0 . 2 2 k 0 . 3 k We know that, Sum of the probabilities is equal to 1 ⇒ P ( - 2 ) + P ( - 1 ) + P ( 0 ) + P ( 1 ) + P ( 2 ) + P 3 = 1 ⇒ 0 . 1 + k + 0 . 2 + 2 k + 0 . 3 + k = 1 ⇒ k = 0 . 4 4 = 0 . 1 Therefore, the probability distribution of a random variable X X - 2 - 1 0 1 2 3 P X 0 . 1 0 . 1 0 . 2 0 . 2 0 . 3 0 . 1 Now, Mean ∑ X P X = ( - 2 ) ( 0 . 1 ) + ( - 1 ) k + 0 ( 0 . 2 ) + 1 ( 2 k ) + 2 ( 0 . 3 ) + 3 k = - 0 . 2 - k + 0 + 2 k + 0 . 6 + 3 k ∴ Mean = 0 . 8 Variance σ 2 = ∑ X 2 P X - ∑ X P X 2 Therefore, variance = 4 ( 0 . 1 ) + 1 ( k ) + 0 ( 0 . 2 ) + 1 ( 2 k ) + 4 ( 0 . 3 ) + 9 k - 0 . 8 2 = 0 . 4 + k + 0 + 2 k + 1 . 2 - 9 k - ( 0 . 8 ) 2 = 12 k + 0 . 4 + 1 . 2 - 0 . 64 = 12 ( 0 . 1 ) + 1 . 6 - 0 . 64 = 1 . 2 + 1 . 6 - 0 . 64 = 2 . 8 - 0 . 64 = 2 . 16 Therefore, variance = 2 . 16 .

The probability distribution of a random variable X is as given below X = x - 3 - 2 - 1 0 1 2 3 P X = x 1 9 1 9 1 9 1 3 1 9 1 9 1 9 Find the variance of X .

The probability distribution of a random variable X is as given below X = x - 3 - 2 - 1 0 1 2 3 P X = x 1 9 1 9 1 9 1 3 1 9 1 9 1 9 Mean ( μ ) = ∑ x = - 3 3 X = x P X = x = - 3 1 9 - 2 1 9 - 1 1 9 + 0 1 3 + 1 1 9 + 2 1 9 + 3 1 9 = - 3 9 - 2 9 - 1 9 + 2 9 + 3 9 = 0 Variance σ 2 = ∑ x = - 3 3 X = x 2 P X = x - ∑ x = - 3 3 X = x P X = x 2 = ( - 3 ) 2 1 9 + ( - 2 ) 2 1 9 + ( - 1 ) 2 1 9 + 0 2 1 3 + ( 1 ) 2 1 9 + ( 2 ) 2 1 9 + 3 2 1 9 - 0 2 = 9 9 + 4 9 + 1 9 + 0 + 1 9 + 4 9 + 9 9 - 0 2 = 28 9 - 0 = 28 9 Therefore, variance σ 2 = 28 9 .

The probability distribution of a random variable X is as given below X = x 0 1 2 3 4 5 6 7 P X = x 0 k 2 k 2 k 3 k k 2 2 k 2 7 k 2 + k Find k .

The probability distribution of a random variable X is as given below X = x 0 1 2 3 4 5 6 7 P X = x 0 k 2 k 2 k 3 k k 2 2 k 2 7 k 2 + k Sum of the probabilities = 1 ⇒ ∑ x = 0 7 P X = x = 1 ⇒ P ( X = 0 ) + P ( X = 1 ) + P ( X = 2 ) + P ( X = 3 ) + P ( X = 4 ) + P ( X = 5 ) + P ( X = 6 ) + P ( X = 7 ) ⇒ 0 + k + 2 k + 2 k + 3 k + k 2 + 2 k 2 + 7 k 2 + k = 1 ⇒ 10 k 2 + 9 k = 1 ⇒ 10 k 2 + 9 k - 1 = 0 ⇒ 10 k 2 + 10 k - k - 1 = 0 ⇒ 10 k ( k + 1 ) - 1 ( k + 1 ) = 0 ⇒ ( 10 k - 1 ) ( k + 1 ) = 0 ⇒ k = 1 10 or - 1 Here, k should be greater than zero, since Probabilities cannot be negative. ∴ k = 1 10

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

The range of a random variable, X is ( 1 , 2 , 3 , … . . ) . P ( X = k ) = c k k ! ( k = 1 , 2 , 3 , … . . ) We know that, sum of the probabilities = 1 P ( X = 1 ) + P ( X = 2 ) + P ( X = 3 ) + … … = 1 c 1 ! + c 2 2 ! + c 3 3 ! + … … = 1 e x = 1 + x 1 ! + x 2 2 ! + x 3 3 ! + … … 1 + c 1 ! + c 2 2 ! + c 3 3 ! + … … = 2 ⇒ e c = 2 ⇒ c = log e 2 . Now, P ( 0 < X < 3 ) = P X = 1 + P X = 2 = log e 2 1 1 ! + log e 2 2 2 ! = log e 2 1 1 ! + log e 2 2 2 ! = log e 2 1 + log e 2 2 Therefore, P 0 < X < 3 = log e 2 1 + log e 2 2 .

Telangana Board Solutions for Exercise 1: Exercise 10(a)

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Telangana Board Mathematics Solutions for Exercise - Telangana Board Solutions for Exercise 1: Exercise 10(a)

Attempt the free practice questions from Exercise 1: Exercise 10(a) with hints and solutions to strengthen your understanding. Intermediate Second Year Mathematics Paper 2A solutions are prepared by Experienced Embibe Experts.

Questions from Telangana Board Solutions for Exercise 1: Exercise 10(a) with Hints & Solutions

EASY

12th Telangana Board

IMPORTANT

Intermediate Second Year Mathematics Paper 2A>Chapter 10 - Random Variables and Probability Distributions>Exercise 10(a)>Q I

The probability distribution of a random variable $X$ is as given below

$X$	$- 2$	$- 1$	$0$	$1$	$2$	$3$
$P (X)$	$0.1$	$k$	$0.2$	$3$	$0.3$	$0.1$

Find the value of $0.1$ and the variance of $X$ .

EASY

12th Telangana Board

IMPORTANT

Intermediate Second Year Mathematics Paper 2A>Chapter 10 - Random Variables and Probability Distributions>Exercise 10(a)>Q I

The probability distribution of a random variable $X$ is as given below

$X = x$	$- 3$	$- 2$	$- 1$	$0$	$1$	$2$	$3$
$P (X = x)$	$\frac{1}{9}$	$\frac{1}{9}$	$\frac{1}{9}$	$\frac{1}{3}$	$\frac{1}{9}$	$\frac{1}{9}$	$\frac{1}{9}$

Find the variance of $X$ .

EASY

12th Telangana Board

IMPORTANT

Intermediate Second Year Mathematics Paper 2A>Chapter 10 - Random Variables and Probability Distributions>Exercise 10(a)>Q I

The probability distribution of a random variable $X$ is as given below

$X = x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P (X = x)$	$0$	$k$	$2 k$	$2 k$	$3 k$	$k^{2}$	$2 k^{2}$	$7 k^{2} + k$

Find $k$ .

EASY

12th Telangana Board

IMPORTANT

Intermediate Second Year Mathematics Paper 2A>Chapter 10 - Random Variables and Probability Distributions>Exercise 10(a)>Q I

The probability distribution of a random variable $X$ is as given below

$X = x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P (X = x)$	$0$	$k$	$2 k$	$2 k$	$3 k$	$k^{2}$	$2 k^{2}$	$7 k^{2} + k$

Find the mean.

EASY

12th Telangana Board

IMPORTANT

Intermediate Second Year Mathematics Paper 2A>Chapter 10 - Random Variables and Probability Distributions>Exercise 10(a)>Q I

The probability distribution of a random variable $X$ is as given below

$X = x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P (X = x)$	$0$	$k$	$2 k$	$2 k$	$3 k$	$k^{2}$	$2 k^{2}$	$7 k^{2} + k$

Find $P (0 < X < 5)$ .

EASY

12th Telangana Board

IMPORTANT

Intermediate Second Year Mathematics Paper 2A>Chapter 10 - Random Variables and Probability Distributions>Exercise 10(a)>Q II

The range of random variable $X$ is ${0, 1, 2}$ . Given that $P (X = 0) = 3 c^{3}, P (X = 1) = 4 c - 10 c^{2}, P (X = 2) = 5 c - 1$ . Find the value of $c$ .

EASY

12th Telangana Board

IMPORTANT

Intermediate Second Year Mathematics Paper 2A>Chapter 10 - Random Variables and Probability Distributions>Exercise 10(a)>Q II

The range of random variable $X$ is ${0, 1, 2}$ . Given that $P (X = 0) = 3 c^{3}, P (X = 1) = 4 c - 10 c^{2}, P (X = 2) = 5 c - 1$ . Find $P (X < 1), P (1 < X \leq 2) and P (0 < X \leq 3)$ .[Answer in the order of $P (X < 1), P (1 < X \leq 2), P (0 < X \leq 3)$ ]

EASY

12th Telangana Board

IMPORTANT

Intermediate Second Year Mathematics Paper 2A>Chapter 10 - Random Variables and Probability Distributions>Exercise 10(a)>Q II

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

Telangana Board Solutions for Exercise 1: Exercise 10(a)

Telangana Board Mathematics Solutions for Exercise - Telangana Board Solutions for Exercise 1: Exercise 10(a)

Questions from Telangana Board Solutions for Exercise 1: Exercise 10(a) with Hints & Solutions

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