# P(X</=x) given mean, 10% of values less than Y, and assuming a normal distribution?

sorry for my ignorance!

is it possible to calculate a cumulative probability for an event to occur knowing a mean, that a certain percent of values will fall below a threshold, and assuming a normal distribution?

for example:

say we had a mean of 35000, and knew that 10% of values were below 1000. can we calculate the probability that x is less than or equal to 1350?

• Not answering your question, but just a comment: If your variable X is always positive (thus 0% of values below 0) then your assumption of a normal distributed variable would be very wrong. – Martijn Weterings Mar 5 '18 at 11:04

Say the RV $X$ is $\sim N(35000, \sigma^2)$. Then $\frac{X - 35000}{\sigma} \sim N(0, 1)$.
Note that $$X < 1000 \Leftrightarrow \frac{x - 35000}{\sigma} < {-34000 \over \sigma},$$
so 0.1 is the probability of a standard normal variable to be less than ${-34000 \over \sigma}$.
$$\sigma \simeq {34000 \over 1.2816} .$$
Your variable is therefore approximately $X \sim N(35000, ({34000 \over 1.2816})^2)$. Any numerical software will tell you the probability of its being over 1350.