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I have a rudimentary probability question that you will probably find amusing but hopefully you can also help me!

I am studying construction planning and this question is taken from an old exam paper.

## Homework Statement

"Risk analysis: At 5% level of significance, what is the probability of completing the project at or earlier than the employer's desired duration in the question above.

The question above requires that project is completed 7 days earlier than the normal duration, μ

Where:

Mean, μ = 50

Required duration, D = 43

Variance, σ² = 3.06

Standad deviation, σ = 1.749

## Homework Equations

I assume Z = D - μ / √σ²

## The Attempt at a Solution

I thought the answer would be z=43-50/√3.06

This gives 4.001, would this mean there is a less than 1% probability of the project being completed in 43 days?

On further reading I also discovered that a 5% level of significance is the same as a 95% confidence interval? If this is the case, doesn't this make the z value 1.645? In which case none of the variables for the equation are unknown.

As you can see I have got myself very confused :

Any and all help will be greatly appreciated. I'll have this forum open all day (and night)!