Question

For any set of n measurements, the fraction included in the interval y − ks to y + ks is at least 1 − 1 k2 . This result is known as Tchebysheff's theorem. A personnel manager for a certain industry has records of the number of employees absent per day. The average number absent is 7.5, and the standard deviation is 2.5. Because there are many days with zero, one, or two absent and only a few with more than ten absent, the frequency distribution is highly skewed. The manager wants to publish an interval in which at least 75% of these values lie. Use Tchebysheff's theorem to find such an interval.

Answer 1

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  

$100(1-\frac{1}{k ^2 })\%$

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

$100(1-\frac{1}{k ^2 })\%= 75\%$

or the equivalent equation

$(1-\frac{1}{k ^2 })= 0.75$

Finding the value of k

$(1-\frac{1}{k ^2 })=0.75\Rightarrow \frac{1}{k^2}=0.25\Rightarrow k^2=\frac{1}{0.25}\Rightarrow k^2=4\Rightarrow k=2$

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]