Finding a constant $ z $ such that $ P(Z \leq z) = 0.95 $ when $ Z \sim \text{N}(0,1) $.

Question

This is for a homework assignment on normal distributions.

Question:

a) Find a constant $z$ such that $P(Z \leq z) = 0.95$

b) Find a constant $z$ such that $P(Z \geq z) = 0.95$

I'm having trouble understanding what a question wants in regards to normal distributions. I get that it wants me to find a constant $z$, but I don't know what that constant means or is.

For part a, from my notes, it states that $P(X \leq x_p) = p$ refers to the percentile. And since p = 0.95, it's asking me to find what the probability is below the 95th percentile? At the same time, my notes state this formula (with no explanation..) that $z = \frac{x - \mu}{\sigma}$, where $z$ is a $z-score$. But what is this and what does it have to even do with these problems?

I apologize if this question makes it seem like I'm trying to have someone do my homework, but I'm honestly confused about this entire section. I'd appreciate any help or even any links that could explain normal distributions in more understandable terms.

Answer 1

$P(Z\le z)=\int_{-\infty}^z \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}}\,dz=0.95$

Or if probability density function has not been covered, just consider the picture of a normal distribution. We know that $Pr(Z\le z)=0.5$ when $z=0$ since $0$ is in the middle. Now, move $z$ to the right, the probability will become larger. When should we stop?

Answer 2

At the same time, my notes state this formula (with no explanation..) that $z = \frac{x - \mu}{\sigma}$, where $z$ is a $z-score$. But what is this and what does it have to even do with these problems?

Suppose you have a random variable X, which is $ \mathcal N(\mu,\sigma^2 )$ distributed-with $\mu\neq 0$ and/or $\sigma^2\neq 1$. Now you want to calculate $P(X \leq x)$ and you have only a table of a standard normal distribution. Thus the table contains only values of a $\mathcal N(0,1 )$ distributed variable. The following equation shows a useful relation:

$P(X\leq x)=P(Z\leq z)=\Phi \left( \frac{x - \mu}{\sigma}\right)$

$\Phi \left( z\right)$ is the function of the standard normal distribution.

$Z\sim \mathcal N(0,1)$

Example:

Your variable X is $ \mathcal N(4, 9)$ distributed. And you want to calculate the probability, that X is smaller than 5. The equation becomes

$P(X\leq 5)=P(Z\leq \frac 13)=\Phi \left( \frac{5 - 4}{3 }\right)=\Phi (\frac 13)$

If you look at this table you will see, that $P(X\leq 5)\approx \Phi (0.33)=0.6230$

Answer 3

(a) If you are using at a normal table, first determine whether it gives areas $P(Z \le z)$ [or $P(0 < Z \le z)].$ In the first instance look in the body of the table to find the closest probability to .9500; in this case, probably a tie between .9496 and .0505. Then find the corresponding value of $z$ by looking at the margins; in this case, 1.64 or 1.65, so split the difference and say $z = 1.645.$ [In the second case, you'll look for the closest probability to .4500, eventually obtaining the same result.]

(b) By symmetry, it is $z = -1.645.$

If you are using statistical software, then you can get these using the 'inverse CDF' or 'quantile' function. You will get more places of accuracy than from normal tables. In R, it is:

 qnorm(.95)    # (a)
 ## 1.644854
 qnorm(.05)    # (b) Same as P(Z <= z) = .05
 ## -1.644854

In Minitab:

 MTB > invcdf .95;
 SUBC> norm.

 Inverse Cumulative Distribution Function 
 Normal with mean = 0 and standard deviation = 1

 P( X <= x )        x
        0.95  1.64485

Almost all statistical software packages have something like this, but each has its own syntax for input and formatting for output.

Addendum: I just noticed the nice example from @calculus. In software, you can sometimes work such problems without 'standardizing'; that is, without subtracting the mean and dividing the difference by the SD. The answer is a little different because software doesn't need to round to two places before table look-up. (Not that the small difference matters for most practical purposes.)

 pnorm(5, 4, sqrt(9))
 ## 0.6305587