Probability Test

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Probability Test - 9

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Weekly Quiz Competition

Question 1
1 / -0.25

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 ……

A
√2/n

B
2

C
1/n

D
√2

Solution

Here, mean = [na+n(−a)]/2n=0
Hence variance = [na² +n(−a)² ]2n −(mean)²
⇒2 = a² −(0)²
⇒a² = 2
Question 2
1 / -0.25

Let X be a random variable whose possible values x₁ , x₂ , x₃ , …, x_n occur with probabilities p₁ , p₂ , p₃ ,…, p_n , respectively. The mean of X, denoted by

A

B

C

D
Question 3
1 / -0.25

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?

A
.50

B
.25

C
.10

D
.15

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
1 / -0.25

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

A
1/3

B
1/2

C
1/4

D
1/6

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
1 / -0.25

Let X be a random variable whose possible values x₁ , x₂ , x₃ , …, x_n occur with probabilities p₁ , p₂ , p₃ ,…, p_n , respectively. Also, μbe the mean of X. The variance of X, denoted by Var (X) is defined as

A

B

C

D
Question 6
1 / -0.25

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

A
0

B
35/12

C
91/36

D
1

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
1 / -0.25

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

A
3/2

B
1/6

C
6

D
3

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
1 / -0.25

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:

A
17.5

B
4.45

C
2.11

D
4.41
Question 9
1 / -0.25

Let X be a random variable whose possible values x₁ , x₂ , x₃ , …, x_n occur with probabilities p₁ , p₂ , p₃ ,…, p_n , respectively. Also, E(X) is the expectation of X, then E(X² ) - [E(X)]² is known as

A
Variance

B
Standard deviation

C
Mean x variance

D
Mean
Question 10
1 / -0.25

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).

A
0.6 and 0

B
0.6 and 0.44

C
0.6 and 0.24

D
0.4 and 0.24