Q77 / 180 · COMEDK 2024 (Morning Shift) · Mathematics
A random variable X with probability distribution is given below
$X = x_i$ 2 3 4 5 $P(X = x_i)$ $\dfrac{5}{k}$ $\dfrac{7}{k}$ $\dfrac{9}{k}$ $\dfrac{11}{k}$
| $X = x_i$ | 2 | 3 | 4 | 5 |
|---|---|---|---|---|
| $P(X = x_i)$ | $\dfrac{5}{k}$ | $\dfrac{7}{k}$ | $\dfrac{9}{k}$ | $\dfrac{11}{k}$ |
- A$\dfrac{61}{8}$
- Correct answerB$\dfrac{61}{16}$
- C7
- D$\dfrac{61}{4}$
Solution
For a probability distribution, the sum of all probabilities must be equal to 1.
$\sum P(X=x_i) = \dfrac{5}{k} + \dfrac{7}{k} + \dfrac{9}{k} + \dfrac{11}{k} = 1$
$\dfrac{5+7+9+11}{k} = 1$
$\dfrac{32}{k} = 1 \Rightarrow k = 32$
The mean of the distribution is given by $E(X) = \sum x_i P(X=x_i)$.
$E(X) = 2 \times \dfrac{5}{32} + 3 \times \dfrac{7}{32} + 4 \times \dfrac{9}{32} + 5 \times \dfrac{11}{32}$
$E(X) = \dfrac{10}{32} + \dfrac{21}{32} + \dfrac{36}{32} + \dfrac{55}{32}$
$E(X) = \dfrac{10 + 21 + 36 + 55}{32} = \dfrac{122}{32}$
Simplifying the fraction by dividing numerator and denominator by 2:
$E(X) = \dfrac{61}{16}$
Answer: $\dfrac{61}{16}$
$\sum P(X=x_i) = \dfrac{5}{k} + \dfrac{7}{k} + \dfrac{9}{k} + \dfrac{11}{k} = 1$
$\dfrac{5+7+9+11}{k} = 1$
$\dfrac{32}{k} = 1 \Rightarrow k = 32$
The mean of the distribution is given by $E(X) = \sum x_i P(X=x_i)$.
$E(X) = 2 \times \dfrac{5}{32} + 3 \times \dfrac{7}{32} + 4 \times \dfrac{9}{32} + 5 \times \dfrac{11}{32}$
$E(X) = \dfrac{10}{32} + \dfrac{21}{32} + \dfrac{36}{32} + \dfrac{55}{32}$
$E(X) = \dfrac{10 + 21 + 36 + 55}{32} = \dfrac{122}{32}$
Simplifying the fraction by dividing numerator and denominator by 2:
$E(X) = \dfrac{61}{16}$
Answer: $\dfrac{61}{16}$
Hence, the correct answer is option: 'B'.