Finding the probability of loss from standard deviation in normal distribution I am unsure how to approach the following question.
The returns from a project are normally distributed with a mean of \$220,000 and a standard deviation of \$160,000. If the project loses more than \$80,000 the company will be in financial difficulties. What is the probability of the project losing more than \$80,000? Give your answer to the nearest whole number.
The answer given is 3%, but I am at a complete loss on how to reach this... If anyone could give me some steps, or even let me know how to approach this, it would be much appreciated.
 A: The usual technique is to convert a normal variable $X$, which has mean $m$ and standard deviation $\sigma$, to a normal variable with mean $0$ and standard deviation $1$, usually denoted by $Z$. You do this by defining $Z=\frac{X-m}{\sigma}$. Here $m=220000$ and $\sigma=160000$. Then your inequalities change. Let's say your inequality was $X \leq x$, then you now have $Z \leq \frac{x-220000}{160000}$. This is now in a form that can be looked up in a standard normal distribution table. (If this isn't for a class, then this can be done with software).
Do you see what $x$ should be in your case?
A: $$-80,000 - 220,000=-300,000$$
I.e. an $\$80,000$ loss is $\$300,000$ below the average.
$$\dfrac{-\$300,000}{\$160,000} = -1.875 $$
I.e. this is $1.875$ standard deviations below the mean.
The probability that a standardized normally distributed random variable is less than $-1.875$ is $\Phi(-1.875)\approx 0.030396$ if I can believe the software I'm using.  In lots of books there is a table in the back where you would look up $-1.875$ and find $0.03$.
A: If $X\sim N(220000, 160000^2)$ describes the profits of the project, then the probability of a loss greater than $80000$ is $\mathbb{P}(X\leq -80000)$. I.e. the probability of a profit less than $-80000$. Now this can be calculated by transforming $X$ into a standard normal random variable $Z$, and looking this probability up in a table as @Ian suggested. Or you could calculate it directly using the standard normal cdf. Or, if any method is acceptable, even use WolframAlpha.
