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Maths M.C.Q (1 Marks)

Mean and variance of a random variable · Gujarat Board (GSEB) STD 12 Science (GSEB - English Medium). 9 questions.

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M.C.Q (1 Marks)

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Question 11 Mark
The probability distribution of a discrete random variable X is given below:

$\text{X}:$

$2$

$3$

$4$

$5$

$\text{P}(\text{X}):$

$\frac{5}{\text{k}}$

$\frac{7}{\text{k}}$

$\frac{9}{\text{k}}$

$\frac{11}{\text{k}}$

The value of k is:

  1. 8
  2. 16
  3. 32
  4. 48
Answer
  1. 32

Solution:

$\sum\limits_2^5\text{P}(\text{x})=1$

$\frac{5}{\text{k}}+\frac{7}{\text{k}}+\frac{9}{\text{k}}+\frac{11}{\text{k}}=1$

$\text{k}=32$

NOTE: Question is modified.

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Question 21 Mark
Let X be a discrete random variable. Then the variance of X is:

  1. E(X2)

  2. E(X2) + (E(X))2

  3. E(X2) - (E(X))2

  4. $\sqrt{\text{E}(\text{X}^2)-(\text{E}(\text{X}))^2}$

Answer
  1. E(X2) - (E(X))2

Solution:

Since, the variance of a discrete random variable X is given by:

Var(X) = E(X2) - (E(X))2

Hence, the correct alternative is option (c).

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Question 31 Mark
If X is a random variable with probability distribution as given below:

X = xi 0 1 2 3
P(X = Xi) k 3k 3k k

The value of k and its variance are:

  1. $\frac{1}{8},\frac{22}{27}$

  2. $\frac{1}{8},\frac{23}{27}$

  3. $\frac{1}{8},\frac{24}{27}$

  4. $\frac{1}{8},\frac{3}{4}$

Answer
  1. $\frac{1}{8},\frac{3}{4}$

Solution:

$\sum\limits_0^3\text{P}(\text{x})=1$

$\text{k}+3\text{k}+3\text{k}+\text{k}=1$

$\text{k}=\frac{1}{8}$

$\text{x}$

$\text{P}(\text{x})$

$\text{x}\text{P}(\text{x})$

$\text{x}^2\text{P}(\text{x})$

$0$

$\frac{1}{8}$

$0$

$0$

$1$

$\frac{3}{8}$

$\frac{3}{8}$

$\frac{3}{8}$

$2$

$\frac{3}{8}$

$\frac{6}{8}$

$\frac{12}{8}$

$3$

$\frac{1}{8}$

$\frac{3}{8}$

$\frac{9}{8}$

$\text{Total}$

 

$\text{E(x)}=\frac{12}{8}=1.5$

$\text{E}(\text{x}^2)=3$

$\text{V(x)}=\text{E}(\text{x}^2)-[\text{E}(\text{x})^2]$

$\text{V(x)}=3-(1.5)^2$

$\text{V(x)}=0.75=\frac{3}{4}$

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Question 41 Mark
The probability distribution of a discrete random variable X is given below:

$\text{X}:$
$1$
$2$
$3$
$4$
$\text{P}(\text{X}):$
$\frac{1}{10}$
$\frac{1}{5}$
$\frac{3}{10}$
$\frac{2}{5}$

The value of E(X2) is:

  1. 3
  2. 5
  3. 7
  4. 10
Answer
  1. 10

Solution:

$\text{X}$

$1$

$2$

$3$

$4$

 

$\text{P}(\text{X})$

$\frac{1}{10}$

$\frac{1}{5}$

$\frac{3}{10}$

$\frac{2}{5}$

 

$\text{X}^2\text{P(X)}$

$\frac{1}{10}$

$\frac{4}{5}$

$\frac{27}{10}$

$\frac{32}{5}$

$\text{E}(\text{X}^2)=10$
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Question 51 Mark
A random variable X takes the values 0, 1, 2, 3 and its mean is 1.3. If P(X = 3) = 2P(X = 1) and P(X = 2) = 0.3, then P(X = 0) is:
  1. 0.1
  2. 0.2
  3. 0.3
  4. 0.4
Answer
  1. 0.4

Solution:

Let:

P(X = 0) = m

P(X = 1) = k

Now,

P(X = 3) = 2k

xi pi pixi
0 m 0
1 k k
2 0.3 0.6
3 2k 6k

Mean $=\sum\text{p}_\text{i}\text{x}_\text{i}$

⇒ 0 + k + 0.6 + 6k = 1.3

⇒ 7k = 1.6 - 0.6

$\Rightarrow\text{k}=\frac{0.7}{7}$

⇒ 0.1

We know that the sum of probabilities in a probability distribution is always 1.

$\therefore$ P(X = 0) + P(X = 1) + P(X = 3) = 1

⇒ m + 0.1 + 0.3 + 0.2 = 1

⇒ m + 0.6 = 1

⇒ m = 0.4

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Question 61 Mark
A random variable has the following probability distribution:

X = xi 0 1 2 3 4 5 6 7
P(X = Xi) 0 2p 2p 3p p2 2p2 7p2 2p

  1. $\frac{1}{10}$

  2. $-1$

  3. $-\frac{1}{10}$

  4. $\frac{1}{5}$

Answer
  1. $\frac{1}{10}$

Solution:

We know that the sum of probabilities in a probability distribution is always 1.

$\therefore$ P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) +P(X = 6) + P(X = 7) +P(X = 8) = 1

⇒ 0 + 2p + 2p + 3p + p2 + 2p2 + 7p2 + 2p = 1

⇒ 10p2 + 9p - 1 = 0

⇒ (10p - 1)(p + 1) = 0

$\Rightarrow\text{p}=\frac{1}{10}\text{ or }-1$ (Negleting -1 as the value of the probability cannot be negative)

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Question 71 Mark
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

Find the events E = {X : X is a prime number}, F{X : X < 4}, the probability $\text{P}(\text{E}\cup\text{F})$ is:

  1. 0.50
  2. 0.77
  3. 0.35
  4. 0.87
Answer
  1. 0.77

Solution:

P(E) = P(2) + P(3) + P(5) + P(7)

P(E) = 0.23 + 0.12 + 0.20 + 0.07

P(E) = 0.62

And

P(F) = P(1) + P(2) + P(3)

P(F) = 0.15 + 0.23 + 0.12

P(F) = 0.5

Also,

$\text{P}(\text{E}\cap\text{F})=\text{P}(2)+\text{P}(3)$

$\text{P}(\text{E}\cap\text{F})=0.23+0.12$

$\text{P}(\text{E}\cap\text{F})=0.35$

$\text{P}(\text{E}\cup\text{F})=\text{P}(\text{E})+\text{P(F)}-\text{P}(\text{E}\cap\text{F})$

$\text{P}(\text{E}\cup\text{F})=0.62+0.5-0.35$

$\text{P}(\text{E}\cup\text{F})=0.77$

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Question 81 Mark
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:

  1. 0
  2. -1
  3. -2
  4. -1.8
Answer
  1. -1.8

Solution:

The probability distribution of X is given below:

X: -4 -3 -2 -1 0
P(X): 0.1 0.2 0.3 0.2 0.2

E(X) = (-4) × 0.1 + (-3) × 0.2 + (-2) × 0.3 + (-1) × 0.2 + 0 × 0.2

= -0.4 - 0.6 - 0.6 - 0.2

= -1.8

Hence, the correct alternative is option (d).

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Question 91 Mark
If the random variable X has the following 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:
  1. $\frac{7}{81}$
  2. $\frac{5}{81}$
  3. $\frac{2}{81}$
  4. $\frac{1}{81}$
Answer
  1. $\frac{1}{81}$

Solution:

We know that the sum of probsabilities in a probability distribution is always 1.

$\therefore$ P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) = 1

⇒ a + 3a+ 5a+ 7a+ 9a + 11a + 13a + 15a + 17a = 1

⇒ 81a = 1

$\Rightarrow\text{a}=\frac{1}{81}$

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