Q. 3.103 - Elementary Statistics

Question

Consider the following data set.

a. Draw a graph similar to Fig. $3.5$ on page $111$

b. Compare the percentage of the observations that actually lie within two standard deviations to either side of the mean with that given by Chebyshev's rule with $k=2$

c. Repeat part (b) with $k=3$

Step-by-Step Solution

Answer

(a) The required graph is given below.

(b) The Chebyshev's rule is, At least $75\%$ and the actual proportion is,  $90\%$

(c) The Chebyshev's rule is, At least $89\%$ and the actual proportion is, $100\%$

Part (a) Step 1: Given information

We need to plot the graph of the given data

Part (b) Step 2: Simplify

We know that, 

Chebyshev's rule is as follows,

$100\left(1-\frac{1}{k^2}\right)\%$ of the given data where $k$ is the standard deviation from mean.

The mean is calculated as the sum of all values divided by no. of values,

As a result mean is, 

$\overline{x}=\frac{82+85+65+91+81+78+94+84+86+84}{10}=\frac{830}{10}=83$

The standard deviation is the square root of variance and variance is the sum of squared deviations divided by no. of values$-1$.

$s=\sqrt{\frac{{\left(82-83\right)}^2+{\left(85-83\right)}^2+....+{\left(86-83\right)}^2+{\left(84-83\right)}^2}{10-1}}=7.8$

Therefore, 

$$\begin{gathered} \overline{x}-s=83-7.8=75.2 \\ \overline{x}+s=83+7.8=90.8 \\ \overline{x}-2s=83-\left(2\times 7.8\right)=67.4 \\ \overline{x}+2s=83+\left(2\times 7.8\right)=98.6 \\ \overline{x}-3s=83-\left(3\times 7.8\right)=59.6 \\ \overline{x}+3s=83+\left(3\times 7.8\right)=106.4 \end{gathered}$$

The graph is as follows, 

Part (b) Step 1: Given information

We are given that $k=2$

Part (b) Step 2: Simplify

We know that,

Chebyshev's rule is given as, 

   $100\left(1-\frac{1}{k^2}\right)\%=100\left(1-\frac{1}{2^2}\right)\%=100\left(1-\frac{1}{4}\right)\%=100\left(\frac{3}{4}\right)\%=75\%$

It is within the $2$ standard deviations from the mean.

We note in the graph that $9$ out of $10$ deviations lie between $\overline{x}-2s$ and $\overline{x}+2s$

Therefore, actual proportion is, $90\%$

Which is a lot higher than Chebyshev's rule.

Part (c) Step 1: Given information

We are given that  $k=3$

Part (c) Step 2: Simplify

By Chebyshev's rule,

$100\left(1-\frac{1}{k^2}\right)\%=100\left(1-\frac{1}{3^2}\right)\%=100\left(\frac{8}{9}\right)\%=89\%$

It is within the $3$ standard deviations from the mean.

We note in the graph that mainly deviations lie between  $\overline{x}-3s$ and $\overline{x}+3s$

Therefore, actual proportion is, $100\%$

Which is a lot higher than Chebyshev's rule.