## Chapter 32 – Mean and variance of a random variable Exercise Ex. 32.1

Four balls are to be drawn without replacement from a

box containing 8 red and 4 white balls. If X denotes the number of red balls

drawn, find the probability distribution of X.

The probability distribution of a random variable X is given below:

(i) Determine the value of k

(ii) Determine P (X _{} 2) and P b(X > 2)

(iii) Find P (X _{} 2) + P(X > 2)

## Chapter 32 – Mean and variance of a random variable Exercise Ex. 32.2

Find the mean and standard deviation of each of the

following probability distributions:

x_{i} : 2 3 4

p_{i} : 2.2 0.5 0.3

Find the mean and standard deviation of each of the

following probability distributions:

A discrete random variable X has the probability

distribution given below:

X : 0.5 1 1.5 2

P(X) : k k^{2} 2k^{2} k

(i) Find the value of k.

(ii) Determine the mean of the distribution.

Three cards are drawn at random (without replacement) from a well shuffled pack of 52 cards. Find the probability distribution of number of red cards. Hence find the mean of the distribution.

An urn contains 5 are 2 black balls. Two balls are

randomly drawn, without replacement. Let X represent

the number of black balls drawn. What are the possible values of X? Is X a

random variable? If yes, find the mean and variance of X.

Two numbers are selected at random (without

replacement) from positive integers 2,3,4,5, 6 and 7. Let X denote the larger of the two number obtained. Find the

mean and variance of the probability distribution of X.

## Chapter 32 – Mean and variance of a random variable Exercise Ex. 32VSAQ

## Chapter 32 – Mean and variance of a random variable Exercise MCQ

If a random variable X has the following probability distribution:

X: | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |

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

then the value of a is

Correct option: (d)

A random variable X has the following probability distribution:

X: | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |

P(X): | 0.15 | 0.23 | 0.12 | 0.10 | 0.20 | 0.08 | 0.07 | 0.05 |

For the event E = {X:X is a prime number}, F = {X:X∪ F) is

a. 0.50

b. 0.77

c. 0.35

d. 0.87

Correct option: (b)

A random variable X takes the values 0, 1, 2, 3 and its mean is 1.3. If P(X=3)=2 P(X=1) and P (X=2)=0.3, then P(X=0) is

a. 0.1

b. 0.2

c. 0.3

d. 0.4

Correct option: (d)

A random variable has the following probability distribution:

X=x | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |

P(X=x | 0 | 2p | 2p | 3p | P | 2p | 7p | 2p |

The value of P is

a. 1/10

b. -1

c. -1/10

d. 1/5

Correct option: (a)

If X is a random -variable with probability distribution as given below:

X=x | 0 | 1 | 2 | 3 |

P(X=x | k | 3k | 3k | k |

The value of k and its variance are

a. 1/8, 22/27

b. 1/8, 23/27

c. 1/8, 24/27

d. 1/8, 3/4

Correct option: (d)

The probability distribution of a discrete random variable X is given below:

X: | 2 | 3 | 4 | 5 |

P(X): | 5/k | 7/k | 9/k | 11/k |

The value of E(x) is

a. 8

b. 16

c. 32

d. 48

Correct option: (c)

NOTE: Question is modified.

For the following probability distribution:

X: | -4 | -3 | -2 | -1 | 0 |

P(X): | 0.1 | 0.2 | 0.3 | 0.2 | 0.2 |

The value of E(X) is

a. 0

b. -1

c. -2

d. -1.8

Correct option: (d)

For the following probability distribution:

X: | 1 | 2 | 3 | 4 |

P(X): | 1/10 | 1/5 | 3/10 | 2/5 |

The value of E(X^{2}) is

a. 3

b. 5

c. 7

d. 10

Correct option: (d)

Let X be a discrete random variable. Then the variance of X is

a. E(X^{2})

b. E(X^{2}) + (E(X))^{2}

c. E(X^{2}) – (E(X))^{2}

d.

Correct option: (c)

Variance of discrete random variable is always E(X^{2}) – (E(X))^{2}