Number 10.

Mathematics

1 answer:

Usimov [2.4K]3 years ago

6 0

Answer:

712:17800

Step-by-step explanation:

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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, 0.0

Zinaida [17]

Answer:

$E(X) = \sum_{i=1}^n X_i P(X_i)$

And if we replace we got:

$E(X) = 0*0.4096 +1*0.4096+ 2*0.1536+ 3*0.0256 +4*0.0016 = 0.8$

So we expect about 0.8 defective computes in a batch of 4 selected.

Step-by-step explanation:

Previous concepts

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

Solution to the problem

For this case we have the following distribution given:

X 0 1 2 3 4

P(X) 0.4096 0.4096 0.1536 0.0256 0.0016

And we satisfy that $P(X_i) \geq 0$ and $\sum P(X_i) =1$ so we have a probability distribution. And we can find the expected value with the following formula: