Chebyshev S Inequality, and an Application

Chebyshev’s Inequality, and an Application

Chebyshev’s Inequality: Let Y be a random variable whose distribution has mean µ and standard deviation σ < +∞. Then for any k ≥ 1, we have

or, equivalently, .

Proof: (Continuous case; the proof for the discrete case is parallel.) By definition,

The interval of integration may be partitioned as follows:

Now in each of the integrals the integrand is non-negative everywhere (except perhaps at a countable number of points). Hence, if we drop the second term on the right, we obtain

Now, on the interval we have . Also, on the interval we have . Hence, if we replace in each of the above integrals by each integral will become smaller; i.e., and

Therefore, we have

We may then factor the constants out of each integral to obtain

Now, the first integral in the above equation is

while the second integral is

Substituting these two equalities into the previous inequality, we obtain

or If we divide through the equality by the common factor we obtain the result

Equivalently, by the complement rule, we may write this result as

These last two statements are equivalent forms of Chebyshev’s Inequality. ■

Example: The target value of the length of a type of rivet manufactured by a certain process is 50 mm. If we assume that the mean length for the population of all rivets manufactured by this process is actually then what is the maximum tolerable value for the population standard deviation, σ, if we want the proportion of rivets with lengths between 49.99 mm and 50.01 mm to be at least 93.75%?

From the latter form of Chebyshev’s Inequality, we have or

Thus, we write