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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 pdf (a) the probabilities \(P(|X|<1)\) and \(P\left(X^{2}<9\right)\) are \(\frac{1}{27}\) and 1. For pdf (b) the probabilities \(P(|X|<1)\) and \(P\left(X^{2}<9\right)\) are \(\frac{5}{18}\) and 1.

Step by step solution

01

Problem (a) Step 1: Find \(P(|X|

Integrate the pdf \(f(x)\) from -1 to 1 because the problem is about |X|. We get the following: \(\int_{-1}^{1}f(x)dx = \int_{-1}^{1}\frac{x^2}{18}dx = \frac{x^3}{54} \Biggr|_{-1}^{1} = \frac{1}{54}-(-\frac{1}{54}) = \frac{2}{54} = \frac{1}{27}\).
02

Problem (a) Step 2: Find \(P\left(X^{2}

Since the square root of 9 is 3 (and -3), and the pdf only goes from -3 to 3, this is the total probability and thus equals 1 since the integral of a pdf over its entire range is 1.
03

Problem (b) Step 1: Find \(P(|X|

This time our pdf is \(f(x)=(x+2) / 18,-2<x<4\). For \(P(|X|<1)\), we now need to integrate from -1 to 1 again, but with the new pdf. The result is: \(\int_{-1}^{1}f(x)dx = \int_{-1}^{1}\frac{x+2}{18}dx = \frac{(x^{2}+4x)}{36} \Biggr|_{-1}^{1} = \left(\frac{1+4}{36} - \frac{1+4*-1}{36} \right) = \left(\frac{5}{36} + \frac{5}{36} \right) = \frac{10}{36} = \frac{5}{18}\).
04

Problem (b) Step 2: Find \(P\left(X^{2}

X^2 is less than 9 between -3 and 3. The lower limit of our pdf is greater than -3 and the upper limit is less than 3. Hence, this is again the total probability for our pdf and the result is 1.

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