Do you know the answer, ive been stuck for hours
1 answer:
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Answer:
a)
And replacing we got:
b)
And then the expected value would be:
Step-by-step explanation:
We assume the following distribution given:
Y 0 1 2 3
P(Y) 0.60 0.25 0.10 0.05
Part a
We can find the expected value with this formula:
And replacing we got:
Part b
If we want to find the expected value of we need to find the expected value of Y^2 and we have:
And replacing we got:
And then the expected value would be:
Using the binomial distribution, it is found that there is a 0.8295 = 82.95% probability that at least 5 received a busy signal.
<h3>What is the binomial distribution formula?</h3>The formula is:
The parameters are:
- x is the number of successes.
- n is the number of trials.
- p is the probability of a success on a single trial.
In this problem:
- 0.54% of the calls receive a busy signal, hence p = 0.0054.
- A sample of 1300 callers is taken, hence n = 1300.
The probability that at least 5 received a busy signal is given by:
In which:
P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4).
Then:
Then:
P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0009 + 0.0062 + 0.0218 + 0.0513 + 0.0903 = 0.1705.
0.8295 = 82.95% probability that at least 5 received a busy signal.
More can be learned about the binomial distribution at brainly.com/question/24863377
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