A random variable X has the following probability distribution:
|
X |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
|
P(X) |
0 |
$k$ |
$2k$ |
$2k$ |
$3k$ |
$k^2$ |
$2k^2$ |
$7k^2+k$ |
The value of $P(4 <x<7)$ is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $\frac{3}{100}$
Given probability distribution:
$X: 0,1,2,3,4,5,6,7$
$P(X): 0, k, 2k, 2k, 3k, k^{2}, 2k^{2}, 7k^{2}+k$
Sum of probabilities is $1$.
$k+2k+2k+3k+k^{2}+2k^{2}+(7k^{2}+k)=1$
$9k+10k^{2}=1$
$10k^{2}+9k-1=0$
$k=\frac{-9+\sqrt{81+40}}{20}=\frac{-9+11}{20}=\frac{1}{10}$
Required probability:
$P(4<\text{ X }<7)=P(5)+P(6)$
$=k^{2}+2k^{2}=3k^{2}$
$=3\left(\frac{1}{10}\right)^{2}=\frac{3}{100}$
Final answer: $\frac{3}{100}$