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QUESTION: 1

In a series of 2 n observations, half of them equal ‘a’ and remaining half equal – a. If the standard deviation of the observations is 2, then | a | equals ……

Solution:

Here, mean = [na+n(−a)]/2n=0

Hence variance = [na^{2}+n(−a)^{2}]2n − (mean)^{2}

⇒ 2 = a^{2} − (0)^{2}

⇒ a^{2 }= 2

QUESTION: 2

Let X be a random variable whose possible values x_{1}, x_{2}, x_{3}, …, x_{n} occur with probabilities p_{1}, p_{2}, p_{3},…, p_{n}, respectively. The mean of X, denoted by

Solution:

QUESTION: 3

The number of adults living in homes on a randomly selected city block is described by the following probability distribution.

What is the probability that 4 or more adults reside at a randomly selected home?

Solution:

The sum of all the probabilities is equal to 1.

Therefore, the probability that four or more adults reside in a home = 1 - (0.25 + 0.50 + 0.15)

= 0.10.

QUESTION: 4

Two dice are thrown simultaneously. If X denotes the number of sixes, then the expectation of X is:

Solution:

Here, X represents the number of sixes obtained when two dice are thrown simultaneously. Therefore, X can take the value of 0, 1, or 2.

∴ P(X=0)=P(not getting six on any of the dice) = 25/36

P(X=1)=P(sixo n first die and no six on second die) + P(no six on firstdie and six on second die)

=2(1/6 × 5/6)=10/36

P(X=2)=P(sixonboththedice)= 1/36

Therefore, the required probability distribution is as follows.

Then, expectation of X=E(X)=∑Xi

P(Xi)= 0(25/36)+1(10/36)+2(1/36)

= 1/3

QUESTION: 5

Let X be a random variable whose possible values x_{1}, x_{2}, x_{3}, …, x_{n} occur with probabilities p_{1}, p_{2}, p_{3},…, p_{n}, respectively. Also, μ be the mean of X. The variance of X, denoted by Var (X) is defined as

Solution:

QUESTION: 6

The variance of the number obtained on a throw of an unbiased dice is:

Solution:

The sample space of the experiment is S = {1, 2, 3, 4, 5, 6}.

Let X denote the number obtained on the throw.

Then X is a random variable which can take values X = 1, 2, 3, 4, 5, or 6. Also P(1) = P(2) = P(3) = P(4) = P(5) = ⅙

QUESTION: 7

The mean number of tails in three tosses of a fair coin is:

Solution:

Let X denotes the number of tails, then probability distribution is

X 0 1 2 3

P(X) 1/8 3/8 3/8 1/8

Mean (xipi) = 0(⅛) + 1(⅜) + 2(⅜) + 3(⅛)

= 3/2

QUESTION: 8

A class has 10 students whose ages are 15, 14, 16, 17, 19, 20, 16, 18, 20, and 20 years. One student is selected in such a manner that each has the same chance of being chosen and the age X of the selected student is recorded. The standard deviation of X is:

Solution:

QUESTION: 9

Let X be a random variable whose possible values x_{1}, x_{2}, x_{3}, …, x_{n} occur with probabilities p_{1}, p_{2}, p_{3},…, p_{n}, respectively. Also, E(X) is the expectation of X, then E(X^{2}) - [E(X)]^{2} is known as

Solution:

QUESTION: 10

In a meeting, 60% of the members favour and 40% oppose a certain proposal. A member is selected at random and we take X = 0 if he opposed, and X = 1 if he is in favour. Find E(X) and Var(X).

Solution:

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