# Business Statistics: Tchebysheff's Theorem

• Nov 17th 2009, 07:21 AM
lisa1984wilson
Business Statistics: Tchebysheff's Theorem
This is a three part question and I'm stuck on the third part, if someone could please help.

Consider the following set of sample data:
78 121 143 88 110 107 62 122 130 95 78 139 89 125

a. Compute the mean and the standard deviation for these sample data.

answer is mean=106.2142857 standard deviation=24.09854344

b. Calculate the coefficient of variation for these sample data and interpret its meaning.

answer is coefficient of variation=22.68860848
I don't know what it means by interpret it's meaning....

c. Using Tchebysheff's Theorem, determine the range of values that should include at least 89% of the data. Count the number of data values that fall into this range and comment on whether your interval range was conservative or not.(Headbang)
• Nov 20th 2009, 10:45 PM
mr fantastic
Quote:

Originally Posted by lisa1984wilson
This is a three part question and I'm stuck on the third part, if someone could please help.

Consider the following set of sample data:
78 121 143 88 110 107 62 122 130 95 78 139 89 125

a. Compute the mean and the standard deviation for these sample data.

answer is mean=106.2142857 standard deviation=24.09854344

b. Calculate the coefficient of variation for these sample data and interpret its meaning.

answer is coefficient of variation=22.68860848
I don't know what it means by interpret it's meaning....

c. Using Tchebysheff's Theorem, determine the range of values that should include at least 89% of the data. Count the number of data values that fall into this range and comment on whether your interval range was conservative or not.(Headbang)

You require the value of $\alpha$ such that $\Pr(|X - \mu| \leq \alpha) \geq 0.89 \Rightarrow \Pr(|X - \mu| \geq \alpha)\leq 0.11$ .

Tchebysheff's Theorem (one of the equivalent forms): $\Pr(|X - \mu| \geq \alpha) \leq \frac{\sigma^2}{\alpha^2}$.

So you require the value of $\alpha$ such that $\frac{\sigma^2}{\alpha^2} = 0.11$.