Question:
The probability that A wakes up before the alarm rings is 0.4. Then, the mean and variance of the number of times A wakes up before the alarm rings in the next 7 days, respectively are:
The probability that A wakes up before the alarm rings is 0.4. Then, the mean and variance of the number of times A wakes up before the alarm rings in the next 7 days, respectively are:
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For repeated independent events with fixed probability, use binomial distribution: \(\text{Mean} = n p\), \(\text{Variance} = n p (1-p)\).
Updated On: Jul 18, 2026
- 0.4, 0.6
- 2.8, 0.6
- 2.8, 1.68
- 7, 0.6
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The Correct Option is C
Solution and Explanation
Step 1: Identify the distribution.
Number of times A wakes up before alarm in 7 days follows a binomial distribution \(X \sim B(n=7, p=0.4)\).
Step 2: Mean formula for binomial.
\(\text{Mean} = n p = 7 \cdot 0.4 = 2.8\)
Step 3: Variance formula for binomial.
\(\text{Variance} = n p (1-p) = 7 \cdot 0.4 \cdot 0.6 = 1.68\)
Step 4: Apply numbers.
\[ \text{Mean} = 2.8, \quad \text{Variance} = 1.68 \]
Step 5: Final conclusion.
Hence, the mean and variance are \[ \boxed{2.8, 1.68} \]
Number of times A wakes up before alarm in 7 days follows a binomial distribution \(X \sim B(n=7, p=0.4)\).
Step 2: Mean formula for binomial.
\(\text{Mean} = n p = 7 \cdot 0.4 = 2.8\)
Step 3: Variance formula for binomial.
\(\text{Variance} = n p (1-p) = 7 \cdot 0.4 \cdot 0.6 = 1.68\)
Step 4: Apply numbers.
\[ \text{Mean} = 2.8, \quad \text{Variance} = 1.68 \]
Step 5: Final conclusion.
Hence, the mean and variance are \[ \boxed{2.8, 1.68} \]
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