Question

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

For \(f(x)=x^{2}/18,-3 < x < 3\), the probabilities are \(P(|X|<1) = 1/27\) and \(P(X^2 < 9) = 1\). For \(f(x)=(x+2) / 18,-2 < x < 4\), the probabilities are \(P(|X| < 1) = 1/12\) and \(P(X^2 < 9) = 1\).

Step-by-step solution

- Comprehend and Setup the Integral for Part (a)

For \(f(x)=\frac{x^{2}}{18}, -3 < x < 3\), when zero is elsewhere, the task is to find \(P(|X|<1)\) and \(P(X^2<9)\). First, let's set up the integral for \(P(|X|<1)\), which equals the probability that \(X\) lies between -1 and 1. The integral becomes \(\int_{-1}^{1} f(x) dx = \int_{-1}^{1} \frac{x^{2}}{18} dx\).

- Solve the Integral for Part (a)

Integrate \(\frac{x^{2}}{18}\) from -1 to 1. The integrated value is \(\frac{x^{3}}{54}\). Substituting the limits into the integral, we get \(\frac{1^{3}}{54}- \frac{(-1)^{3}}{54}= \frac{2}{54} =\frac{1}{27}\). Now for \(P(X^2<9)\), this is the same as \(P(-3< X < 3 )\), the entire interval on which our PDF is defined. Hence, \(P(X^2<9) = 1\).

- Setup the Integral for Part (b)

Now consider \(f(x)=\frac{x+2}{18}, -2 < x < 4\), when zero is elsewhere. Again, start with \(P(|X|<1)\), which equals the probability that \(X\) is in the interval \(-1 < X < 1\). This gives us the integral \(\int_{-1}^{1} \frac{x+2}{18} dx\).

- Solve the Integral for Part (b)

Integrate \(\frac{x+2}{18}\) from -1 to 1. The integrated function is \(\frac{x^2}{36} + \frac{x}{9}\). Substituting the limits into the integral results in \(\frac{1^2}{36} + \frac{1}{9}- ( \frac{(-1)^2}{36} - \frac{-1}{9}) =\frac{1}{12}\). Now for \(P(X^2<9)\), this is the same as \(P(-3< X < 3)\). Notice that in this range, the definition of PDF \(f(x)\) changes. So, finding \(P(X^2 < 9) = \int_{-2}^{3} \frac{x+2}{18} dx \) which results in \(1\).