Answer:
The interval [2.5, 12.5]
Step-by-step explanation:
Tchebysheff’s Theorem states that for any distribution of probabilities if k ≥ 1, at least
[tex]100(1-\frac{1}{k ^2 })\%[/tex]
of the data will lie within k standard deviations of the mean.
We need then to find a k greater or equal than 1 such that
[tex]100(1-\frac{1}{k ^2 })\%= 75\%[/tex]
or the equivalent equation
[tex](1-\frac{1}{k ^2 })= 0.75[/tex]
Finding the value of k
[tex](1-\frac{1}{k ^2 })=0.75\Rightarrow \frac{1}{k^2}=0.25\Rightarrow k^2=\frac{1}{0.25}\Rightarrow k^2=4\Rightarrow k=2[/tex]
So, applying Tchebysheff’s Theorem, the manager can assure that the number of employees that can be absent in a given day, lies in the interval
[7.5-2*2.5, 7.5+2*2.5] = [2.5, 12.5]
