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.
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?