Question ID: #728
If a random variable $x$ has the probability distribution
| $X$ | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|---|---|---|---|---|---|---|---|---|
| $P(x)$ | 0 | 2k | k | 3k | $2k^{2}$ | 2k | $k^{2}+k$ | $7k^{2}$ |
then $P(3\lt x \le6)$ is equal to
- (1) 0.34
- (2) 0.22
- (3) 0.64
- (4) 0.33
Solution:
For a probability distribution, the sum of all probabilities must be 1.
$$ \sum P(x_i) = 1 $$
$$ 0 + 2k + k + 3k + 2k^2 + 2k + k^2 + k + 7k^2 = 1 $$
$$ 10k^2 + 9k = 1 $$
$$ 10k^2 + 9k – 1 = 0 $$
Solving the quadratic equation for $k$:
$$ 10k^2 + 10k – k – 1 = 0 $$
$$ 10k(k + 1) – 1(k + 1) = 0 $$
$$ (10k – 1)(k + 1) = 0 $$
So, $k = \frac{1}{10}$ or $k = -1$.
Since probability cannot be negative, $k \ne -1$.
Thus, $k = \frac{1}{10} = 0.1$.
We need to find $P(3 < x \le 6)$. This includes $x = 4, 5, 6$. $$ P(3 < x \le 6) = P(4) + P(5) + P(6) $$ $$ P(3 < x \le 6) = 2k^2 + 2k + (k^2 + k) $$ $$ = 3k^2 + 3k $$
Substitute $k = 0.1$:
$$ = 3(0.1)^2 + 3(0.1) $$
$$ = 3(0.01) + 0.3 $$
$$ = 0.03 + 0.30 = 0.33 $$
Ans. (0.33)
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