# Thread: Quick Question on Probability and Normal Distribution tables

1. ## Quick Question on Probability and Normal Distribution tables

When a question is of the form 'what is the probability that a project will take "at least" (suppose) 23 days', how do you find the probability using normal distribution tables?

2. Hi

Simply by using $\displaystyle P(X\geq 23)=1-P(X\leq23)$ ?

3. Originally Posted by flyingsquirrel
Hi

Simply by using $\displaystyle P(X\geq 23)=1-P(X\leq23)$ ?
thanks. Also, how would you go about constructing a 95% confidence interval form (in my case) completion time of a project?

4. Originally Posted by flyingsquirrel
Hi

Simply by using $\displaystyle P(X\geq 23)=1-P(X\leq23)$ ?
You'd first need to convert X = 23 into a Z-score using the formula $\displaystyle Z = \frac{X - \mu}{\sigma}$.

Then, depending on whether the z-score is less than 0 or greater than zero, there might be some further fiddling around needed before a value can be read from the standard normal tables .....

If the Z-score is greater than zero, then you use 1 - Pr(Z < Z-score) since Pr(Z < Z-score) can be read straight from the table.

However, if the Z-score is less than 0, you actually use Pr(Z < -(Z-score)). The following example illustrates the chain of logic that's required:

Pr(Z > -1) = 1 - Pr(Z < -1) = 1 - Pr(Z > 1) = 1 - (1 - Pr(Z < 1)) = Pr(Z < 1).

All this is assuming of course that you're using a normal distribution .....

5. Originally Posted by DooBeeDoo
thanks. Also, how would you go about constructing a 95% confidence interval form (in my case) completion time of a project?
It is a totally recipe driven process. Even I can do it! Where are you stuck?