91影视

For a real-valued random variable, a normal distribution is a sort of continuous probability distribution.

We begin by obtaining the general probability formula\(P(|X - \mu | \ge k\sigma )\). Since both\(k\)and\(\sigma \)as a result, we may simplify this expression to

\(P(|X - \mu | \ge k\sigma ) = P(X - \mu \le - k\sigma ) + P(|X - \mu | \ge k\sigma )\)

We also know that if both sides are divided by the same positive amount, equality is unaffected.

\(P(|X - \mu | \ge k\sigma ) = P\left( {\frac{{X - \mu }}{\sigma } \le - k} \right) + P\left( {\frac{{X - \mu }}{\sigma } \ge k} \right)\)

Standardizing gives:

\(P(|X - \mu | \ge k\sigma ) = P(Z \le - k) + P(Z \ge k)\)

\(Z\)is a normal distribution rv cdf with a standard deviation of\(\phi (z).\)As a result, we may write

\(P(|X - \mu | \ge k\sigma ) = \phi ( - k) + (1 - \phi (k))\)

\( = (1 - \phi (k)) + (1 - \phi (k))\)

\(P(|X - \mu | \ge k\sigma ) = 2 \cdot (1 - \phi (k))\)

Let Z be a continuous rv with cdf as a proposition\(\phi (z)\). Then for any\(a\),

\(\begin{array}{l}P(Z \le a) = \phi (a)\\P(a \le Z) = 1 - \phi (a)\end{array}\)

We may now calculate the probability using the expression of equation (l) and Appendix A-3\(P(|X - \mu | \ge k\sigma )\)for\(k = 1,2,3\)

\(P(|X - \mu | \ge \sigma ) = 2(1 - \phi (1)) = 2(1 - 0.8413) = 0.3174\)

\(P(|X - \mu | \ge 2\sigma ) = 2(1 - \phi (2)) = 2(1 - 0.9772) = 0.0456\)

\(P(|X - \mu | \ge 3\sigma ) = 2(1 - \phi (3)) = 2(1 - 0.9987) = 0.0026\)

According to Chebyshev's inequality, for any\(k\)fulfilling,\(k \ge 1\)

\(P(|X - \mu | \ge k\sigma ) \le \frac{1}{{{k^2}}}\)

The upper bounds for\(k = 1,2,3\)are

\(\begin{array}{l}P(|X - \mu | \ge \sigma ) \le 1\\P(|X - \mu | \ge 2\sigma ) \le 0.25\\P(|X - \mu | \ge 3\sigma ) \le 0.1111\end{array}\)

The probability are well within the upper ranges determined using the Chebyshev's inequality, as can be shown.