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Chapter 1: Problem 6

For each of the following pdfs of \(X\), find \(P(|X|<1)\) and \(P\left(X^{2}<9\right)\). (a) \(f(x)=x^{2} / 18,-3

Short Answer

Expert verified
For the pdf \(f(x)=x^{2} / 18,-3<x<3\), \(P(|X|<1)\) is \(\frac{2}{9}\) and \(P(X^{2}<9)\) is 1. For the pdf \(f(x)=(x+2) / 18,-2<x<4\), \(P(|X|<1)\) is \(\frac{1}{3}\) and \(P(X^{2}<9)\) is \(\frac{5}{6}\).

Step by step solution

01

Finding \(P(|X|

This can be done by integrating the pdf from -1 to 1: \(P(|X|<1) = \int_{-1}^{1} f(x) dx = \int_{-1}^{1} \frac{x^2}{18} dx\). Evaluating this integral gives \(P(|X|<1) = \frac{2}{9}\)
02

Finding \(P(X^{2}

This is done by integrating the pdf from -3 to 3, which is \(P(X^{2}<9) = \int_{-3}^{3} f(x) dx = \int_{-3}^{3} \frac{x^2}{18} dx\). Evaluating this integral gives \(P(X^{2}<9) = 1\) as the entire range [-3,3] is covered.
03

Finding \(P(|X|

This is done by integrating the pdf from -1 to 1: \(P(|X|<1) = \int_{-1}^{1} f(x) dx = \int_{-1}^{1} \frac{x+2}{18} dx\). Evaluating this integral gives \(P(|X|<1) = \frac{1}{3}\).
04

Finding \(P(X^{2}

This is done by integrating the pdf from -2 to 3, as it is the overlap of range [-2,4] that x can take and the range [-3,3] from the condition \(X^{2}<9\): \(P(X^{2}<9) = \int_{-2}^{3} f(x) dx = \int_{-2}^{3} \frac{x+2}{18} dx\). Evaluating this integral gives \(P(X^{2}<9) = \frac{5}{6}\).

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