Probability in my exam

[newbiestats]

Here is the scenario for my practice exam:

"You have been hired by a firm which produces customised engineering machinery to provide some information for their staff planning. They have informed you that the time taken to assemble a customised engineering unit is normally distributed with a mean of 20 hours and a standard deviation of 4 hours.
What is the probably that a particular unit will take more than 26 hours to complete? "

And here is the solution: (26-20)/4= 1.5, Probablitiy = 0.5 - 04332 = 0.0668 (now my question is where did 0.5 come from??)

[Dason]

Some normal tables provide a table of values such you get P(0 <= Z <= z) (ie the area between 0 and whatever value you're interested in). Looks like the solution was using such a table and was using the fact that P(Z >= 1.5) = P(0 <= Z <= infinity) - P(0 <= Z <= 1.5) = 0.5 - 0.4332

[newbiestats]

Some normal tables provide a table of values such you get P(0 <= Z <= z) (ie the area between 0 and whatever value you're interested in). Looks like the solution was using such a table and was using the fact that P(Z >= 1.5) = P(0 <= Z <= infinity) - P(0 <= Z <= 1.5) = 0.5 - 0.4332

Here is the normal distribution table that the exam been using but I still don't get where 0.5 come from
http://img560.imageshack.us/img560/4508/normaldistributiontable.jpg

[Dason]

Some normal tables provide a table of values such you get P(0 <= Z <= z) (ie the area between 0 and whatever value you're interested in). Looks like the solution was using such a table and was using the fact that P(Z >= 1.5) = P(0 <= Z <= infinity) - P(0 <= Z <= 1.5) = 0.5 - 0.4332

The bolded part is what gives you 0.5. The area to the right of 0 is 0.5.

[newbiestats]

The bolded part is what gives you 0.5. The area to the right of 0 is 0.5.

I still don't underestand the bold part "P(0 <= Z <= infinity)"

[Dason]

The area to the right of 0 is 0.5. The normal distribution is symmetric. The total area is 1. So the area to the right of 0 needs to be 1/2 and the area to the left of 0 needs to be 1/2. I'm not sure how much more direct I can say it.

[newbiestats]

Thanks a lot for your help Dason, also for the same scenario another question is asked "Above what time will the slowest 10.03% of units take to complete? "
And the solution is this: "Lookup 0.3997 -> z = 1.28
x = 20 + 4 x 1.28 = 25.12 hours" but my question is where did "Lookup 0.3997 " come from?

[Dason]

0.50 - .1003 = .3997

Hopefully you can figure out the rest.

[newbiestats]

Thanks once again, one more thing on this same scenario:"If 50% of the jobs require 6 staff, 30% of jobs require 8 staff and the remainder require 10 staff, what is the expected value for the number of staff required?" and the solution is: 0.5 x 6 + 0.3 x 8 + 0.2 x 10 = 7.4 staff but my question is where did 0.2 come from.

[Dason]

I think if you think about this one for a little while longer you should be able to get it...

[newbiestats]

I think if you think about this one for a little while longer you should be able to get it...

I read the question again and it said "remainder require 10 staff" but it doesnt say anything about 0.2

[Dason]

... Ok so it doesn't explicitly give you the number 0.2. But it gives you everything you need. Think about it.

[newbiestats]

... Ok so it doesn't explicitly give you the number 0.2. But it gives you everything you need. Think about it.

But it hasn't given me 0.2 in the scenario so how would I know the 0.2??

[Dason]

Thanks once again, one more thing on this same scenario:"If 50% of the jobs require 6 staff, 30% of jobs require 8 staff and the remainder require 10 staff, what is the expected value for the number of staff required?" and the solution is: 0.5 x 6 + 0.3 x 8 + 0.2 x 10 = 7.4 staff but my question is where did 0.2 come from.
What do you possibly think the remainder is?

[newbiestats]

What do you possibly think the remainder is?

remainder is what is left over?

[Dason]

Which would be...

[newbiestats]

Thanks a lot, I appreciate all your help. You have been very helpful to me. Thanks :)

[ryan8200]

Here is the scenario for my practice exam:

"You have been hired by a firm which produces customised engineering machinery to provide some information for their staff planning. They have informed you that the time taken to assemble a customised engineering unit is normally distributed with a mean of 20 hours and a standard deviation of 4 hours.
What is the probably that a particular unit will take more than 26 hours to complete? "

And here is the solution: (26-20)/4= 1.5, Probablitiy = 0.5 - 04332 = 0.0668 (now my question is where did 0.5 come from??)


Here is my solution for your reference :)