Need help finding the 80th percentile of random variable X with a normal distribution with $μ_X = 80, σ_X = 10,$ using standard normal distribution and also the 90th percentile when $μ_X = 12, σ_X = 1.$

I get that you start off with $x = μ+zσ,$ but then I have no clue where to go with since I don't know what $X$ and $Z$ are.

  • $\begingroup$ My 2 cents... start with $ P(X \le a ) = .8$ which as you noted is equal to $ P(\mu+Z\sigma \le a ) = .8$ and now get the $Z$ by itself on the LHS and use the Z table to figure out what the right side should be. $\endgroup$ – HJ_beginner Apr 6 at 19:06

You can standardize the random variable: $Z=\frac{X-\mu_x}{\sigma_x}$, with $Z\sim \mathcal N(0,1)$

Then the equation is $P(X\leq x)=\Phi\left(\frac{x-\mu_x}{\sigma_x}\right)=\Phi\left(\frac{x-80}{10}\right)=0.8$

Inverse function


Now you take a table of a standard normal distribution (cummulative) and look for the value of $z$ where $\Phi(z)\approx 0.8$. This is at $z=0.84$ I have circled the number.

enter image description hereThus the equation becomes


I think you can proceed. Just solve the equation for $x$.

  • $\begingroup$ Thanks, I think my mistake was trying to find a way to do it without a software or table, lack of research on my side. Thanks $\endgroup$ – DJBay Apr 7 at 22:33
  • $\begingroup$ @DJBay Yes, without a table it is mostly impossible to obtain a sufficient accurate result. One exeption is $P(X>\mu)=P(X<\mu)=0.5$, where $X\sim \mathcal N(\mu,\sigma^2)$ $\endgroup$ – callculus Apr 8 at 12:19

With technology. Using software instead of printed tables, you can get exact answers. In R, the function qnorm is the inverse CDF (also called the quantile function) of a specified normal distribution. So answer (a) is about 88.416 and (b) is about 13.282.

qnorm(.8, 80, 10);  qnorm(.90, 12, 1)
[1] 88.41621
[1] 13.28155

Some statistical calculators and many kinds of statistical software will do such computations.

With printed standard normal CDF tables. Some precision is lost using printed tables because you will usually not be able to find exactly the desired value in the body of a normal table.

For (b), if $X \sim \mathsf{Norm}(\mu = 12,\, \sigma=1),$ then $$P(X \le c) = P\left(\frac{X-\mu}{\sigma} \le \frac{c - 12}{1}\right) = P(Z < c - 12) = 0.9,$$

where $Z$ is standard normal.

Find the (four-place) value in the body of the table that cuts (as nearly as possible to) 10% of the area from the upper tail of $\mathsf{Norm}(0,1).$

That will be the value in the table corresponding to $z = 1.28$ in the margins of the table. Then $1.28 = c - 12,$ so $c = 13.28,$ to two places. (By interpolation, you might be able to get an accurate digit in the third decimal place.)


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