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...
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...
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Compute the mean and then the variance from E(X squared) minus mean squared.
Updated On: Oct 1, 2026
- \(\frac{25}{(10)^6}\)
- \(\frac{35}{(10)^6}\)
- \(\frac{45}{(10)^6}\)
- \(\frac{55}{(10)^6}\)
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The Correct Option is D
Solution and Explanation
Step 1: Mean:
\(E(X) = 2(0.2) + 3(0.5) + 4(0.3) = 0.4 + 1.5 + 1.2 = 3.1\).
Step 2: Mean of squares:
\(E(X^2) = 4(0.2) + 9(0.5) + 16(0.3) = 0.8 + 4.5 + 4.8 = 10.1\).
Step 3: Standard deviation:
\[ \text{Var}(X) = E(X^2) - [E(X)]^2 = 10.1 - 9.61 = 0.49 \]
\[ \sigma = \sqrt{0.49} = 0.7 \]
Final Answer:
The standard deviation is 0.7, option (D). \[ \boxed{0.7} \]
\(E(X) = 2(0.2) + 3(0.5) + 4(0.3) = 0.4 + 1.5 + 1.2 = 3.1\).
Step 2: Mean of squares:
\(E(X^2) = 4(0.2) + 9(0.5) + 16(0.3) = 0.8 + 4.5 + 4.8 = 10.1\).
Step 3: Standard deviation:
\[ \text{Var}(X) = E(X^2) - [E(X)]^2 = 10.1 - 9.61 = 0.49 \]
\[ \sigma = \sqrt{0.49} = 0.7 \]
Final Answer:
The standard deviation is 0.7, option (D). \[ \boxed{0.7} \]
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