The probabilities that a batch of 4 computers will contain 0, 1, 2, 3, and 4 defective computers are 0.4096, 0.4096, 0.1536, 0.0256, and 0.0016, respectively. Round answer to the nearest hundredth.
A) $\mu = 0.80$
B) $\mu = 1.21$
C) $\mu = 0.70$
D) $\mu = 2.00$

Question

The probabilities that a batch of 4 computers will contain 0, 1, 2, 3, and 4 defective computers are 0.4096, 0.4096, 0.1536, 0.0256, and 0.0016, respectively. Round answer to the nearest hundredth. 
A) $\mu = 0.80$
B) $\mu = 1.21$
C) $\mu = 0.70$
D) $\mu = 2.00$

Quizplus · Accepted Answer

The expected value $\mu$ is calculated as $\mu = \sum (x \cdot P(x))$, which equals $0 \cdot 0.4096 + 1 \cdot 0.4096 + 2 \cdot 0.1536 + 3 \cdot 0.0256 + 4 \cdot 0.0016 = 0.80$.

The Probabilities That a Batch of 4 Computers Will Contain μ=0.80\mu = 0.80μ=0.80

The Probabilities That a Batch of 4 Computers Will Contain $\mu = 0.80$