Question

Take the following sample of student grades from a class:

70, 88, 91, 75, 79, 77, 70, 74, 81, 90

If these are assumed to represent the general pattern of the grades for the full class, use Chebychev’s inequality to give a lower bound on the probability that a given grade will be within 10 points of the average?

Answer 1

The computations are made using the following table:

X	(X - Mean(X))^2
70	90.25
88	72.25
91	132.25
75	20.25
79	0.25
77	6.25
70	90.25
74	30.25
81	2.25
90	110.25
795	554.5

The sample mean here is computed as:

$\bar X = \frac{\sum X_i}{n} = \frac{795}{10} = 79.5$

The sample standard deviation here is computed as:

$s = \sqrt{\frac{\sum (X_i - \bar X)^2}{n-1}} = \sqrt{\frac{554.5}{10-1}} = 7.849$

Now according to chebyshev's inequality theorem, at least 1 - 1/k² of the observations lies within k standard deviation of the mean.

k = 10/s = 10/7.849 = 1.274

Therefore 1 - 1/1.274² = 0.3839

Therefore at least 0.3839 proportion of the total values lies within 10 points of the average here. Therefore 0.3839 is the required probability here.