Normal Distribution, finding the minimum mark The exam scores on a certain Society of Actuaries (SOA) professonal examination are Normally distributed with a mean score of $μ=65\%$ and a standard deviation of σ=6
$35\%$ of all persons writing this SOA Examination will not pass. What is the minimum mark needed to pass this exam? Enter your answer to two decimal places, and enter as a percentage. 
I tried to enter P(X less than Xo) = 0.35 and came out with $62.69$ ;(which is wrong)
I need some help on what to do
Thanks!
 A: You have percentage scores $X \sim \mathsf{Norm}(\mu = 65,\, \sigma=6)$ and you seek $c$ such that $P(X \le c) = .35.$ This can be solved by standardizing and using printed tables of the standard normal distribution:
$$.35 = P(X < 35) = p\left(\frac{X - \mu}{\sigma} < \frac{c - 65}{6}\right)
= P\left(Z < \frac{c - 65}{6}\right),$$
where $Z$ is standard nomal. From tables you should be able to find that $(c-65)/6 \approx  -0.38,$ and then it is a simple matter to solve for $c,$
and round up to the nearest integer (percentage) and thus get the lowest passing score.
If you have access to statistical software or a statistical calculator, you
may be able to solve the question directly, without standardizing. In R statistical software qnorm is the quantile function (inverse CDF) of a
normal distribution. Thus the lowest passing score is 63.
qnorm(.36, 65, 6)
## 62.84925

I find that drawing a sketch is insurance against getting tangled up
reading printed normal tables and finding an "answer" that is obviously wrong.
By hand, you can't make a sketch as accurate as the plot below, but you
can make a sketch that is better than nothing.

