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Problem 95: A flu virus hits a company employing 180 people. Independent of the other employees, there is a probability of p = 0.35 that each person needs to take sick leave. What are the expectation and variance of the proportion of the workforce who need to take sick leave? In general, what value of the sick rate p produces the largest variance for this proportion?

Problem 95: A flu virus hits a company employing 180 people. Independent of the other employees, there is a probability of p = 0.35 that each person needs to take sick leave. What are the expectation and variance of the proportion of the workforce who need to take sick leave? In general, what value of the sick rate p produces the largest variance for this proportion?

Solution:


Given that $n=180, p=0.35$
Let the random variable X be the number of employees taking sick leave.
Therefore, the proportion of the workforce who need to take sick leave is
$Y = \frac{X}{180}$
so that
$E(Y) = \frac{E(X)}{180} = \frac{np}{180} = \frac{180\times0.35}{180} = 0.35$
$Var(Y) = \frac{Var(X)}{180^2} = \frac{np(1-p)}{180^2} = \frac{180\times0.35\times0.65}{180^2} = 0.0013$
In general,
$Var(Y) = \frac{Var(X)}{180^2} = \frac{np(1-p)}{180^2} = \frac{180\times p(1-p)}{180^2} = \frac{p(1-p)}{180} $
Condtion for maximum $Var(Y)$ is $\frac{d(Var(Y)}{dp}=0 $
$\frac{d(Var(Y)}{dp}=1-2p=0\Rightarrow p=0.5$

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