Let $X\sim N(3,4)$. Find $\mathbb{P}(X<7)$, $\mathbb{P}(X \ge 9)$

Question

Let $X \sim N(3,4)$. Find $\mathbb{P}(X\lt7)$, $\mathbb{P}(X \ge 9)$, and $\mathbb{P}(|x-3|\lt 2) $

Okay lets figure out the PDF.

$\mu=3$, $\sigma=4$.

$$f(X)= \frac{e^\left(\frac{-(x-\mu)^2}{2 \sigma^2}\right)}{\sigma\sqrt{2\pi}} $$

Plugging it all in I get:

$$f(X)= \frac{e^\left(\frac{-(x-3)^2}{2\cdot4^2}\right)}{4\sqrt{2\pi}} $$

$$f(X)= \frac{e^\left(\frac{-(x-3)^2}{32}\right)}{4\sqrt{2\pi}} $$

Now that we have our PDF, we can calculate $\mathbb{P}(X<7)$ and the rest, which is what I'm having difficulties with.

Am I correct in saying that $\mathbb{P}(X<7) = f(x=7)$, or would it be $1-f(x=7)$?

Answer 1

If $f(x)$ is the probability density function, then $P(X<X_0)$ is an integral of $f(x)$:

$$P(X<X_0) = \int_{-\infty}^{X_0} f(x)\,dx$$

But for the probability density function $f$ for the normal distribution this integral is difficult to express in terms of commonly used functions. It is usually tabulated or calculated numerically.

For the other intervals you can use complementary probability: $$P(X<X_0) + P(X\geq X_0) = 1 \Leftrightarrow\\\text{or}\\P(X\geq X_0) = 1-P(X<X_0)$$

and the fact that $P(X<X_1 \& \text{not}(X<X_0)) = P(X<X_1) - P(X<X_0)$

You can derive this fact from the integral definition with some rules for adding integrals:

$$\int_a^bf(x)\,dx + \int_b^cf(x)\,dx = \int_a^cf(x)\,dx$$

Answer 2

You are partially correct. You need to integrate the PDF so you get the CDF, then it's: $$P(X<7)=F(7)$$and$$P(X≥7)=1-F(7)$$

Note: $F$ here is the CDF. Normally a lowercase $f$ indicates the PDF.

As you have a normal distribution though, you can get around integrating that expression, and instead use the table for the normalized version as Steven Ramos said.

Answer 3

Just normalized X to the N(0,1) and used a Z table! $$P(X<7)=P(X-3/\sqrt4<7-3/\sqrt4)=P(Z< \frac{7-3}{\sqrt4})$$ Just use a Z table now for this number.Same for the others, let me know if this helped.