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Mean and variance of a random variable · Gujarat Board (GSEB) STD 12 Science (GSEB - English Medium). 17 questions.

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Question 12 Marks
The probability distribution function oif a random variable X is given by

Xi 0 1 2
Pi 3c3 4c - 10c2 5c - 1

Where c > 0

Find: P(X < 2).

Answer
$\text{P}(\text{X}<2)=\text{P}(0)+\text{P}(1)$
$=3\text{c}^3+4\text{c}-10\text{c}^2$
$=3\Big(\frac{1}{3}\Big)^3+4\Big(\frac{1}{3}\Big)-10\Big(\frac{1}{3}\Big)^2$
$=\frac{3}{27}+\frac{4}{3}-\frac{10}{9}$
$=\frac{1}{9}+\frac{4}{3}-\frac{10}{9}$
$=\frac{3}{9}$
$\therefore\ \text{P}(\text{x}<2)=\frac{1}{3}$
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Question 22 Marks
The probability distribution function oif a random variable X is given by

Xi 0 1 2
Pi 3c3 4c - 10c2 5c - 1

Where c > 0

Find: c.

Answer
We know that the sum of the probabilities in a probability distribution is always 1.
$\therefore$ P(X = 0) + P(X = 1) + P(X = 2) = 1
⇒ 3c3 + 4c - 10c2 + 5c - 1 = 1
⇒ 3c3 - 10c2 + 9c - 2 = 0
⇒ (c - 1)(3c2 - 7c + 2) = 0
⇒ (c - 1)(3c - 1)(c - 2) = 0
$\Rightarrow\text{c}=\frac{1}{3},1,2$
(Negleting 1 and 2 as individual probability should not be greater than one)
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Question 32 Marks
The probability distribution of random variable X is given below:

$\text{X}$
$0$
$1$
$2$
$3$
$\text{P}(\text{X})$
$\text{k}$
$\frac{\text{k}}{2}$
$\frac{\text{k}}{4}$
$\frac{\text{k}}{8}$

 Determine $\text{P}(\text{X}\leq2)$ and $\text{P}(\text{X}>2)$

Answer
$\text{P}(\text{X}\leq2)$
$=\text{P}(0)+\text{P}(1)+\text{P}(2)$
$=\text{k}+\frac{\text{k}}{2}+\frac{\text{k}}{4}$
$=\frac{8}{15}+\frac{8}{30}+\frac{8}{60}$
$=\frac{14}{15}$
$\text{P}(\text{X}>2)=\text{P}(3)=\frac{\text{k}}{8}=\frac{1}{15}$
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Question 42 Marks
The probability distribution of random variable X is given below:

$\text{X}$
$0$
$1$
$2$
$3$
$\text{P}(\text{X})$
$\text{k}$
$\frac{\text{k}}{2}$
$\frac{\text{k}}{4}$
$\frac{\text{k}}{8}$

 Determine the value of k.

Answer
We know that,
$\text{P}(0)+\text{P}(1)+\text{P}(2)+\text{P}(3)=1$
$\text{k}+\frac{\text{k}}{2}+\frac{\text{k}}{4}+\frac{\text{k}}{8}=1$
$15\text{k}=8$
$\text{k}=\frac{8}{15}$
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Question 52 Marks
The probability distribution function oif a random variable X is given by

Xi 0 1 2
Pi 3c3 4c - 10c2 5c - 1

Where c > 0

Find: $\text{P}(1<\text{X}\leq2)$

Answer
$\text{P}(1<\text{X}\leq2)$
$=\text{P}(\text{X}=2)$
$=5\text{c}-1$
$=\frac{5}{3}-1$
$=\frac{5-3}{3}$
$=\frac{2}{3}$
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Question 62 Marks
Which of the following distributions of a random variable X are the probability distributions?
X:
3
2
1
0
-1
P(X):
0.3
0.2
0.4
0.1
0.05
Answer
P(X = 3) + P(X = 2) + P(X = 1) + P(X = 0) + P(X = -1)
= 0.3 + 0.2 + 0.4 + 0.1 + 0.05
= 1.05 > 1
It is not the probability distribution of random variable X.
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Question 72 Marks
A discrete random variable X has the probability distribution given below:
X: 0.5 1 1.5 2
P(X): k k2 2k2 k
Find the value of k.
Answer
We know that,
P(0.5) + P(1) + P(1.5) + P(2) = 1
k + k2 + 2k2 + k = 1
3k2 + 2k - 1 = 0
3k2 + 3k - k - 1 = 0
(3k - 1)(k + 1) = 0
$\text{k}=\frac{1}{3}$ or $\text{k}=-1$
We know that $0\leq\text{P}(\text{X})\leq1$
$\therefore\ \text{k}=\frac{1}{3}$
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Question 82 Marks
If X denotesthe number on the uper face of a cubical die when it is thrown, find the mean of X.
Answer
A cubical die can show 1, 2, 3, 4, 5 or 6 on its face
$\text{x}_\text{i}$
 $\text{p}_\text{i}$
 $\text{p}_\text{i}\text{x}_\text{i}$
$1$
 $\frac{1}{6}$
 $\frac{1}{6}$
$2$
 $\frac{1}{6}$
 $\frac{2}{6}$
$3$
 $\frac{1}{6}$
 $\frac{3}{6}$
$4$
 $\frac{1}{6}$
 $\frac{4}{6}$
$5$
 $\frac{1}{6}$
 $\frac{5}{6}$
$6$
 $\frac{1}{6}$
 $\frac{6}{6}$
Mean $=\sum\text{p}_\text{i}\text{x}_\text{i}=\frac{1}{6}+\frac{2}{6}+\frac{3}{6}+\frac{4}{6}+\frac{5}{6}+\frac{6}{6}=\frac{21}{6}=3.5$
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Question 92 Marks
If the probability distribution of a random variable X is given below:
X = Xi 1 2 3 4
P(X = Xi) c 2c 4c 4c
Write the value of $\text{P}(\text{X}\leq2)$
Answer
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) = 1
⇒ c + 2c + 4c + 4c =1
⇒ 11c = 1
$\Rightarrow\text{c}=\frac{1}{11}$
Now,
$\text{P}(\text{X}\leq2)=\text{P}(\text{X}=1)+\text{P}(\text{X})=2=\frac{1}{10}+\frac{2}{10}=\frac{3}{10}$
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Question 102 Marks
Which of the following distributions of a random variable X are the probability distributions?
X: 0 1 2 3 4
P(X): 0.1 0.5 0.2 0.1 0.1
Answer
P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)
= 0.1 + 0.5 + 0.2 + 0.1 + 0.1
= 1
It is the probability distribution of random variable X.
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Question 112 Marks
A random variable has the following probability distribution:
X = Xi 1 2 3 4
P(X = Xi) k 2k 3k 4k
Write the value of $\text{P}(\text{X}\geq3).$
Answer
Here,

X = Xi 1 2 3 4
P(X = Xi) k 2k 3k 4k
Since, $\sum\text{P}(\text{X})=1$

⇒ P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 1

⇒ k + 2k + 3k + 4k =1

⇒ 10k = 1

$\Rightarrow\text{k}=\frac{1}{10}$

$\text{P}(\text{X}\geq3)$

= P(X = 3) + P(X = 4)

= 3k + 4k

= 7k

$=\frac{1}{10}$

$\text{P}(\text{X}\geq3)=\frac{7}{10}$

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Question 122 Marks
A random variable X has the following probability distribution:
Values of X: -2 -1 0 1 2 3
P(X) 0.1 k 0.2 2k 0.3 k
Find the value of k.
Answer
Here,
Values of X: -2 -1 0 1 2 3
P(X) 0.1 k 0.2 2k 0.3 k
We know that,
P(-2) + P(-1) + P(0) + P(1) + P(2) + P(3) = 1
⇒ 0.1 + k + 0.2 + 2k + 0.3 + k = 1
⇒ 4k + 0.6 = 1
⇒ 4k = 1 - 0.6
⇒ 4k = 0.4
$\Rightarrow\text{k}=\frac{0.4}{4}$
$\Rightarrow\text{k}=\frac{1}{10}$
$\Rightarrow\text{k}=0.1$
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Question 132 Marks
A random variable X has the following probability distribution:

Values of X: 0 1 2 3 4 5 6 7 8
P(X) a 3a 5a 7a 9a 11a 13a 15a 17a

Determine:

The Value of a.

Answer
Here,
Values of X: 0 1 2 3 4 5 6 7 8
P(X) a 3a 5a 7a 9a 11a 13a 15a 17a
Since $\sum\text{P}(\text{X})=1$
P(0) + P(1) + P(0) + P(2) + P(3) + P(4) + P(5) + P(6) + P(7) + P(8) = 1
⇒ a + 3a + 5a + 7a + 9a + 11a + 13a + 15a + 17a = 1
⇒ 81a = 1
$\Rightarrow\text{a}=\frac{1}{81}$
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Question 142 Marks
A discrete random variable X has the probability distribution given below:
X: 0.5 1 1.5 2
P(X): k k2 2k2 k
Determine the mean of the distribution.
Answer
Mean $=\sum\text{p}_\text{i}\text{x}_\text{i}=0.5\times\text{k}+1\times\text{k}^2+1.5\times2\text{k}^2+2\times\text{k}$
$=0.5\times\frac{1}{3}+1\times\Big(\frac{1}{3}\Big)^2+1.5\times2\Big(\frac{1}{3}\Big)^2+2\times\frac{1}{3}$
$=\frac{0.5}{3}+\frac{1}{9}+\frac{3}{9}+\frac{2}{3}$
$=\frac{1.5+1+3+6}{9}$
$=\frac{11.5}{9}$
$=\frac{115}{90}$
$=\frac{23}{18}$
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Question 152 Marks
Which of the following distributions of a random variable X are the probability distributions?
X: 0 1 2 3
P(X): 0.3 0.2 0.4 0.1
Answer
P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
= 0.3 + 0.2 + 0.4 + 0.1
= 1
It is the probability distribution of random variable X.
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Question 162 Marks
Which of the following distributions of a random variable X are the probability distributions?
X:
0
1
2
P(X):
0.6
0.4
0.2
Answer
P(X = 0) + P(X = 1) + P(X = 1) + P(X = 2)
= 0.6 + 0.4 + 0.2
= 1.2 > 1
It is not the probability distribution of random variable X.
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Question 172 Marks
Find the mean of the following probability distribution:
$\text{X}=\text{x}_\text{i}:$ $1$ $2$ $3$
$\text{P}(\text{X}=\text{x}_\text{i}):$ $\frac{1}{4}$ $\frac{1}{8}$ $\frac{5}{8}$
Answer

$\text{x}_\text{i}$

$\text{p}_\text{i}$

$\text{p}_\text{i}\text{x}_\text{i}$

$1$

$\frac{1}{4}$

$\frac{1}{4}$

$2$

$\frac{1}{8}$

$\frac{2}{8}$

$3$

$\frac{1}{8}$

$\frac{15}{8}$

Mean $=\sum\text{p}_\text{i}\text{x}_\text{i}=\frac{1}{4}+\frac{2}{8}+\frac{15}{8}=\frac{2+2+15}{8}=\frac{19}{8}$

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