Answer to Question #120920 in Statistics and Probability for gwein Juspain

Question

A4. Using the Chebyshev's inequality, nd the value of k that will guarantee that the
probability is 0:95 when the deviation of X from its mean is not more than k.

Answer 1

Chebyshev's inequality

$$P(|X-\mu|\geq k\sigma)\leq{1\over k^2}$$

Thus, the probability that a random variable takes a value more than k standard deviations away from its mean is at most $1/k^2.$

$$k=4: {1\over k^2}={1\over 16}=0.0625(6.25\ \%)$$

$$100\ \%-6.25 \ \%=93.75\ \%<95\ \%$$

$$k=5: {1\over k^2}={1\over 25}=0.04(4\ \%)$$

$$100\ \%-4 \ \%=96\ \%>95\ \%$$

$k=5.$