<p> Chebyshev’s Inequality, and an Application</p><p>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</p><p> or, equivalently, .</p><p>Proof: (Continuous case; the proof for the discrete case is parallel.) By definition,</p><p>The interval of integration may be partitioned as follows:</p><p>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</p><p>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</p><p>Therefore, we have </p><p>We may then factor the constants out of each integral to obtain</p><p>Now, the first integral in the above equation is </p><p> while the second integral is </p><p>Substituting these two equalities into the previous inequality, we obtain</p><p> or If we divide through the equality by the common factor we obtain the result</p><p>Equivalently, by the complement rule, we may write this result as</p><p>These last two statements are equivalent forms of Chebyshev’s Inequality. ■</p><p>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%?</p><p>From the latter form of Chebyshev’s Inequality, we have or</p><p>Thus, we write </p><p> or</p><p> or </p>