Question

On average, one out of 10 persons is busy. If six persons are selected at random, then the probability that at least 5 of them will be busy is...

• A. \(\frac{25}{(10)^6}\)
• B. \(\frac{35}{(10)^6}\)
• C. \(\frac{45}{(10)^6}\)
• D. \(\frac{55}{(10)^6}\)

Hint:

Compute the mean and then the variance from E(X squared) minus mean squared.

Answer 1

The Correct Option is D

Step 1: Use deviations directly:
Mean = 3.1. Deviations: $-1.1, -0.1, 0.9$.

Step 2: Variance:
$\text{Var} = 0.2(1.21) + 0.5(0.01) + 0.3(0.81) = 0.242 + 0.005 + 0.243 = 0.49$. So $\sigma = 0.7$ (D).

Final Answer:
0.7. \[ \boxed{0.7} \]