2 Marks Questions

Question 12 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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Question 22 Marks

The probability distribution function oif a random variable $X$ is given by

$X_i$	$0$	$1$	$2$
$P_i$	$3c^3$	$4c - 10c^2$	$5c - 1$

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 32 Marks

The probability distribution function oif a random variable $X$ is given by

$X_i$	$0$	$1$	$2$
$P_i$	$3c^3$	$4c - 10c^2$	$5c - 1$

Where $c > 0$
Find: $c.$

Answer

We know that the sum of the probabilities in a probability distribution is always $1.$
$\therefore P(X = 0) + P(X = 1) + P(X = 2) = 1$
$\Rightarrow 3c^3 + 4c - 10c^2+ 5c - 1 = 1$
$\Rightarrow 3c^3 - 10c^2 + 9c - 2 = 0$
$\Rightarrow (c - 1)(3c^2 - 7c + 2) = 0$
$\Rightarrow (c - 1)(3c - 1)(c - 2) = 0$
$\Rightarrow\text{c}=\frac{1}{3},1,2$
$($Negleting $1$ and $2$ as individual probability should not be greater than one$)$

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Question 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 $\text{P}(\text{X}\leq2)$ 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 52 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 62 Marks

The probability distribution function oif a random variable $X$ is given by

$X_i$	$0$	$1$	$2$
$P_i$	$3c^3$	$4c - 10c^2$	$5c - 1$

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 72 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 82 Marks

A discrete random variable $X$ has the probability distribution given below:

$X:$	$0.5$	$1$	$1.5$	$2$
$P(X):$	$k$	$k^2$	$2k^2$	$k$

Find the value of $k.$

Answer

We know that,
$P(0.5) + P(1) + P(1.5) + P(2) = 1$
$k + k^2+ 2k^2 + k = 1$
$3k^2 + 2k - 1 = 0$
$3k^2 + 3k - k - 1 = 0$
$(3k - 1)(k + 1) = 0$
$\text{k}=\frac{1}{3}$ 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 92 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 102 Marks

If the probability distribution of a random variable $X$ is given below:

$X = X_i$	$1$	$2$	$3$	$4$
$P(X = X_i)$	$c$	$2c$	$4c$	$4c$

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,
$P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 1$
$\Rightarrow c + 2c + 4c + 4c =1$
$\Rightarrow 11c = 1$
$\Rightarrow\text{c}=\frac{1}{11}$
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 112 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 122 Marks

A random variable has the following probability distribution:

$X = X_i$	$1$	$2$	$3$	$4$
$P(X = X_i)$	$k$	$2k$	$3k$	$4k$

Write the value of $\text{P}(\text{X}\geq3).$

Answer

Here,

$X = X_i$	$1$	$2$	$3$	$4$
$P(X = X_i)$	$k$	$2k$	$3k$	$4k$

Since, $\sum\text{P}(\text{X})=1$
$\Rightarrow P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 1$
$\Rightarrow k + 2k + 3k + 4k =1$
$\Rightarrow 10k = 1$
$\Rightarrow\text{k}=\frac{1}{10}$
$\text{P}(\text{X}\geq3)= P(X = 3) + P(X = 4)$
$= 3k + 4k$
$= 7k =\frac{1}{10}$
$\text{P}(\text{X}\geq3)=\frac{7}{10}$

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Question 132 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 142 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 152 Marks

A discrete random variable $X$ has the probability distribution given below:

$X:$	$0.5$	$1$	$1.5$	$2$
$P(X):$	$k$	$k^2$	$2k^2$	$k$

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 162 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 172 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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