Probability Problem

[tomtk]

Hi from Melbourne.

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)!

 

[Mark44]

Hi from Melbourne.

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

You need parentheses here.

z = (x - μ )/ σ

The Attempt at a Solution

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

You need parentheses here, too, otherwise this will be interpreted as 43 - (50/√3.06
).

The answer to the question is not a value of z - it's the probability of completing the project at or earlier than the employer's desired duration (43 days).

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

Wouldn't this give you a z value of -4.001? Where are you getting a probability of 1%?

On further reading I also discovered that a 5% level of significance is the same as a 95% confidence interval?

Yes.

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.

The mean length of jobs is 50 days, with a standard deviation of 3.06 days. What fraction of jobs are done in 43 days or fewer? To answer this question look at the standard normal distribution to see how much area under the curve corresponds to your (corrected) z value.