# Thread: Probability in my exam

1. ## Probability in my exam

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??)

2. ## Re: Probability in my exam

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

3. ## Re: Probability in my exam

Originally Posted by 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
Here is the normal distribution table that the exam been using but I still don't get where 0.5 come from

4. ## Re: Probability in my exam

Originally Posted by 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.

5. ## Re: Probability in my exam

Originally Posted by Dason
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)"

6. ## Re: Probability in my exam

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.

7. ## The Following User Says Thank You to Dason For This Useful Post:

newbiestats (11-13-2011)

8. ## Re: Probability in my exam

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?

9. ## Re: Probability in my exam

0.50 - .1003 = .3997

Hopefully you can figure out the rest.

10. ## The Following User Says Thank You to Dason For This Useful Post:

newbiestats (11-13-2011)

11. ## Re: Probability in my exam

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.

12. ## Re: Probability in my exam

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

13. ## Re: Probability in my exam

Originally Posted by Dason
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

14. ## Re: Probability in my exam

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

15. ## Re: Probability in my exam

Originally Posted by Dason
... 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??

16. ## Re: Probability in my exam

Originally Posted by 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.
What do you possibly think the remainder is?

17. ## Re: Probability in my exam

Originally Posted by Dason
What do you possibly think the remainder is?
remainder is what is left over?