If the probability distribution is given by:
X 0 1 2 3 4 5 6 7 P(x) 0 k 2k 2k 3k k² 2k² 7k² + k
Then find: \( P(3 < x \leq 6) \)
If the probability distribution is given by:
| X | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|---|---|---|---|---|---|---|---|---|
| P(x) | 0 | k | 2k | 2k | 3k | k² | 2k² | 7k² + k |
Then find: \( P(3 < x \leq 6) \)
Show Hint
- 0.33
- 0.22
- 0.11
- 0.44
The Correct Option is A
Solution and Explanation
Step 1: Understanding the Question:
We are given a probability distribution for a discrete random variable X. First, we need to find the value of the constant k. Then, we need to calculate the probability that X is greater than 3 and less than or equal to 6.
Step 2: Key Property of Probability Distribution:
For any probability distribution, the sum of all probabilities must be equal to 1. \[ \sum P(x_i) = 1 \] Applying this to the given distribution: \[ P(0) + P(1) + P(2) + P(3) + P(4) + P(5) + P(6) + P(7) = 1 \] \[ 0 + k + 2k + 2k + 3k + k^2 + 2k^2 + (7k^2+k) = 1 \]
Step 3: Solving for k:
Combine the terms with k and k\(^2\): \[ (k + 2k + 2k + 3k + k) + (k^2 + 2k^2 + 7k^2) = 1 \] \[ 9k + 10k^2 = 1 \] Rearrange into a standard quadratic equation: \[ 10k^2 + 9k - 1 = 0 \] Factor the quadratic equation: \[ 10k^2 + 10k - k - 1 = 0 \] \[ 10k(k+1) - 1(k+1) = 0 \] \[ (10k-1)(k+1) = 0 \] This gives two possible values for k: \(k = 1/10\) or \(k = -1\).
Since probabilities cannot be negative (e.g., P(1) = k must be \(\geq 0\)), we must choose the positive value. \[ k = \frac{1}{10} = 0.1 \]
Step 4: Calculating the Required Probability:
We need to find P(3 \(<\) x \(\leq\) 6), which is the sum of probabilities for x=4, x=5, and x=6. \[ P(3<x \leq 6) = P(4) + P(5) + P(6) \] From the table: \[ P(3<x \leq 6) = 3k + k^2 + 2k^2 = 3k + 3k^2 \] Substitute the value of k = 0.1: \[ P(3<x \leq 6) = 3(0.1) + 3(0.1)^2 = 0.3 + 3(0.01) = 0.3 + 0.03 = 0.33 \]
Step 5: Final Answer:
The required probability is 0.33.
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