Normal Distribution Question Given that X ~ N(20,16)

Question

Given that X ~ N(20,16), find the following probability:

P(X >= 17)

My Workings as follows:

P(X>=17) -> 1 - P(X=<16)

Z = X - μ/ σ

-> Z = 16-20/4 = -1

-> 1- Φ(-1)

-> 1 - (1 - Φ(1) )

-> 1 - (1- 0.8413)

= 0.8413

The answer is 0.7734

Answer 1

At at any, this ends up as a numerical computation. For any random variable, $X$, with density $f$ (here w.r.t. the Lebesgue measure), the probability of $X$ lying in some Borel set, $A$, is the integral $$ \mathbb{P}(X \in A) = \int_A f(x)~dx $$ The normal distribution with mean $\mu$ and variance $\sigma^2$ is given by

$$ f(x) = \frac{1}{\sqrt{2\pi\sigma^2}} \exp\left( -\frac{(x-\mu)^2}{2\sigma^2} \right) $$

So in order to find the probability you're interested in, you have to compute the integral of the density over the set $(-\infty,17]$. Problem is, though, that this integral has no closed form solution. So either way, you have to compute the probability numerically. This, however, is readily available in all statistical software (in R the command is $\texttt{pnorm()}$, for example).

In many of the introductory statistics courses I've taught, the point of exercises such as this one is however to test whether the student understand how to transform random variables that have a normal distribution. For $X \sim \mathcal{N}(\mu,\sigma^2)$ it holds that

$$ \forall a,b\in \mathbb{R}:\quad a+bX \sim \mathcal{N}(a+b\mu,b^2 \sigma^2) $$

Using this we may observe that

$$ \frac{X-\mu}{\sigma} \sim N(0,1) $$

And plenty of tables of probabilites are avaible for the standard normal. That is, computing your probability is equivalent to finding

$$ \mathbb{P} \left( Z \geq \frac{17-20}{\sqrt{16}} \right) $$ for $Z \sim \mathcal{N}(0,1)$ - this you should be able to find.

It is worth it to keep in mind that an exercise such as this is really only to make you understand how affine transformations of random variables work - because it is no more difficult to numerically calculate probabilities of any normal distribution that the standard normal.

Answer 2

You are computing the complimentary event in the wrong way. For a continous distribution: $$ P(X \geq 17) = 1 - P(X < 17) \neq 1 - P(X \leq 16) $$ $X < 17$ would equal $X \leq 16$ if $X$ could only take discrete values. But a Normal distribution is continuous.

If you are trying to approximate a discrete distribution with a Gauss have a look at continuity corrections (https://en.wikipedia.org/wiki/Continuity_correction)

Answer 3

You seem to have two difficulties. (1) Not realizing that the normal distribution is continuous and (2) not understanding the notation "$N(20,16),"$ in which it seems that the population mean is $\mu = 20$ and the population variance is $\sigma^2 = 16,$ so that $\sigma = 4.$

Then the required standardization can be done as follows: $$P(X < 17) = P\left(\frac{X - \mu}{\sigma} < \frac{17-20}{4}\right) = P(Z < -.75),$$ where $Z$ is standard normal. If you are using printed normal tables, you might not find an entry for the z-value $-.75,$ so you might need to use the symmetry of the normal distribution to get the answer. And then remember to subtract from 1.

I think it is a good idea to make sketches for such problems. By hand, you can't make as precise a sketch as the ones below, but you should be able to make a sketch that is better than nothing. The left-hand plot is on the original scale; the right-hand one is on the standardized z-scale. In each case, you want the area under the curve to the right of the vertical dashed line. The difference is that you can use a printed table to help you get the probability on the z-scale.

Note: Answer using R statistical software:

1-pnorm(17, 20, 4)
## 0.7733726
1 - pnorm(-.75, 0, 1)
## 0.7733726