For a Binomial distribution with mean 6 and standard deviation \( \sqrt{2} \), what is the value of P(X = 0) ?
- A. \left(\frac{1}{3}\right)^9 ✓
- B. \left(\frac{2}{3}\right)^9
- C. \frac{1}{3}\left(\frac{2}{3}\right)^8
- D. \frac{2}{3}\left(\frac{1}{3}\right)^8
Correct Answer: A. \left(\frac{1}{3}\right)^9
Explanation
Mean np = 6 and variance npq = 2. Dividing them gives q = 2/6 = 1/3, meaning p = 2/3. Substituting p into np=6 gives n(2/3) = 6, so n = 9. P(X=0) = ^9C_0 p^0 q^9 = (1/3)^9.
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