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Maths MCQ

Mean and variance of a random variable · Maharashtra Board STD 12 (MSBSHSE - Semi English Medium). 9 questions.

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9 questions · auto-graded multiple-choice test.

MCQ 11 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 $E(X^2)$ is:
  • A
    $3$
  • B
    $5$
  • C
    $7$
  • ✓
    $10$
Answer
Correct option: D.
$10$
$\text{X}$ $1$ $2$ $3$ $4$  
$\text{P}(\text{X})$ $\frac{1}{10}$ $\frac{1}{5}$ $\frac{3}{10}$ $\frac{2}{5}$  
$\text{X}^2\text{P(X)}$ $\frac{1}{10}$ $\frac{4}{5}$ $\frac{27}{10}$ $\frac{32}{5}$ $\text{E}(\text{X}^2)=10$
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MCQ 21 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:
  • A
    $8$
  • B
    $16$
  • ✓
    $32$
  • D
    $48$
Answer
Correct option: C.
$32$
$\sum\limits_2^5\text{P}(\text{x})=1$
$\frac{5}{\text{k}}+\frac{7}{\text{k}}+\frac{9}{\text{k}}+\frac{11}{\text{k}}=1$
$\text{k}=32$
NOTE: Question is modified.
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MCQ 31 Mark
Let $X$ be a discrete random variable. Then the variance of $X$ is:
  • A
    $E(X^2)$
     
  • B
    $E(X^2) + (E(X))^2$
     
  • ✓
    $E(X^2) - (E(X))^2$
     
  • D
    $\sqrt{\text{E}(\text{X}^2)-(\text{E}(\text{X}))^2}$
Answer
Correct option: C.
$E(X^2) - (E(X))^2$
 
Since, the variance of a discrete random variable $X$ is given by:
$Var(X) = E(X^2) - (E(X))^2$
Hence, the correct alternative is option $(c).$
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MCQ 41 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:
  • A
    $0$
  • B
    $-1$
  • C
    $-2$
  • ✓
    $-1.8$
Answer
Correct option: D.
$-1.8$
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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MCQ 51 Mark
If $X$ is a random variable with probability distribution as given below:
$X = x_i$ $0$ $1$ $2$ $3$
$P(X = X_i)$ $k$ $3k$ $3k$ $k$
The value of $k$ and its variance are:
  • A
    $\frac{1}{8},\frac{22}{27}$
  • B
    $\frac{1}{8},\frac{23}{27}$
  • C
    $\frac{1}{8},\frac{24}{27}$
  • ✓
    $\frac{1}{8},\frac{3}{4}$
Answer
Correct option: D.
$\frac{1}{8},\frac{3}{4}$
$\frac{1}{8},\frac{3}{4}$
$\sum\limits_0^3\text{P}(\text{x})=1$
$\text{k}+3\text{k}+3\text{k}+\text{k}=1$
$\text{k}=\frac{1}{8}$
 
$\text{x}$ $\text{P}(\text{x})$ $\text{x}\text{P}(\text{x})$ $\text{x}^2\text{P}(\text{x})$
$0$ $\frac{1}{8}$ $0$ $0$
$1$ $\frac{3}{8}$ $\frac{3}{8}$ $\frac{3}{8}$
$2$ $\frac{3}{8}$ $\frac{6}{8}$ $\frac{12}{8}$
$3$ $\frac{1}{8}$ $\frac{3}{8}$ $\frac{9}{8}$
$\text{Total}$   $\text{E(x)}=\frac{12}{8}=1.5$ $\text{E}(\text{x}^2)=3$
$\text{V(x)}=\text{E}(\text{x}^2)-[\text{E}(\text{x})^2]$
$\text{V(x)}=3-(1.5)^2$
$\text{V(x)}=0.75=\frac{3}{4}$
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MCQ 61 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:
  • A
    $\frac{7}{81}$
  • B
    $\frac{5}{81}$
  • C
    $\frac{2}{81}$
  • ✓
    $\frac{1}{81}$
Answer
Correct option: D.
$\frac{1}{81}$
We know that the sum of probsabilities in a probability distribution is always $1.$
$\therefore P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$
$+ P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) = 1$
$\Rightarrow a + 3a+ 5a+ 7a+ 9a + 11a + 13a + 15a + 17a = 1$
$\Rightarrow 81a = 1$
$\Rightarrow\text{a}=\frac{1}{81}$
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MCQ 71 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:
  • A
    $0.1$
  • B
    $0.2$
  • C
    $0.3$
  • ✓
    $0.4$
Answer
Correct option: D.
$0.4$
Let:
$P(X = 0) = m$
$P(X = 1) = k$
Now,
$P(X = 3) = 2k$
$x_i$ $p_i$ $p_ix_i$
$0$ $m$ $0$
$1$ $k$ $k$
$2$ $0.3$ $0.6$
$3$ $2k$ $6k$
Mean $=\sum\text{p}_\text{i}\text{x}_\text{i}$
$\Rightarrow 0 + k + 0.6 + 6k = 1.3$
$\Rightarrow 7k = 1.6 - 0.6$
$\Rightarrow\text{k}=\frac{0.7}{7}$
$\Rightarrow 0.1$
We know that the sum of probabilities in a probability distribution is always $1.$
$\therefore P(X = 0) + P(X = 1) + P(X = 3) = 1$
$\Rightarrow m + 0.1 + 0.3 + 0.2 = 1$
$\Rightarrow m + 0.6 = 1$
$\Rightarrow m = 0.4$
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MCQ 81 Mark
A random variable has the following probability distribution:
$X = x_i$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$
$P(X = X_i)$ $0$ $2p$ $2p$ $3p$ $p^2$ $2p^2$ $7p^2$​​​​​​​ $2p$
  • ✓
    $\frac{1}{10}$
  • B
    $-1$
  • C
    $-\frac{1}{10}$
  • D
    $\frac{1}{5}$
Answer
Correct option: A.
$\frac{1}{10}$
We know that the sum of probabilities in a probability distribution is always $1$.
$\therefore P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) +P(X = 6) + P(X = 7) +P(X = 8) = 1$
$\Rightarrow 0 + 2p + 2p + 3p + p^2 + 2p^2 + 7p^2 + 2p = 1$
$\Rightarrow 10p^2+ 9p - 1 = 0$
$\Rightarrow (10p - 1)(p + 1) = 0$
$\Rightarrow\text{p}=\frac{1}{10}\text{ or }-1 ($Negleting $-1$ as the value of the probability cannot be negative$)$
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MCQ 91 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:
  • A
    $0.50$
  • ✓
    $0.77$
  • C
    $0.35$
  • D
    $0.87$
Answer
Correct option: B.
$0.77$
$P(E) = P(2) + P(3) + P(5) + P(7)$
$P(E) = 0.23 + 0.12 + 0.20 + 0.07$
$P(E) = 0.62$
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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