exercise:0f095da318: Difference between revisions

Latest revision as of 22:47, 14 June 2024

Let [math]X[/math] be a continuous random variable with values normally distributed over [math](-\infty,+\infty)[/math] with mean [math]\mu = 0[/math] and variance [math]\sigma^2 = 1[/math].

Using Chebyshev's Inequality, find upper bounds for the following probabilities: [math]P(|X| \geq 1)[/math], [math]P(|X| \geq 2)[/math], and [math]P(|X| \geq 3)[/math].
The area under the normal curve between [math]-1[/math] and 1 is .6827, between [math]-2[/math] and 2 is .9545, and between [math]-3[/math] and 3 it is .9973 (see the table in Appendix A). Compare your bounds in (a) with these exact values. How good is Chebyshev's Inequality in this case?

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+Let <math>X</math> be a continuous random variable with values normally distributed over <math>(-\infty,+\infty)</math> with mean <math>\mu = 0</math> and variance <math>\sigma^2 = 1</math>.
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+<ul style="list-style-type:lower-alpha"><li> Using Chebyshev's Inequality, find upper bounds for the following
-\newcommand{\mat}[1]{{\bf#1}}
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-\newcommand{\mathds}{\mathbb}</math></div> Let <math>X</math> be a continuous random variable with values normally
-distributed over <math>(-\infty,+\infty)</math> with mean <math>\mu = 0</math> and variance <math>\sigma^2 = 1</math>.
-<ul><li> Using Chebyshev's Inequality, find upper bounds for the following
 probabilities: <math>P(|X| \geq 1)</math>, <math>P(|X| \geq 2)</math>, and <math>P(|X| \geq 3)</math>.
 </li>