# Statistics, again

1. Jun 4, 2007

### Cyannaca

Hi,

I'm taking a distance learning course in statistics so there are a few concepts that are not clear to me. First I was wondering if it was possible to add up Z scores. I have a problem in which two students have two different z scores on two exams. Peter has 1,69 on the first one and -0,13 on the second, and Mary has -0,21 on the first one and 1,07 on the second one. Can I add up the Z scores to say that Peter is better?

My second interrogation is with this problem. Two detectors have a probability = to 0,7 of detecting an object. A clerck puts an object under both machines simultaneously. What is the probability that one machine or the other detects it?

So I did MxN to get all the possibilities, =100 and then 7x7 for the detections. I got p=49%, but when I do a tree of all the possibilities, I get 91%. So I'm wondering which way should I solve this problem?

2. Jun 4, 2007

### phoenixthoth

Since these are on two exams, the z-scores regarding the first exam are incompatible with z-scores on the second exam. This is because the mean and standard deviation are probably different among the two exams.

I think it's safe to assume the two events, machine A detects it and machine B detects it, are independent. What you want to find is p(A v B), the probability that one machine or the other detects it. There is a formula that breaks up p(A v B) into other probabilities, one of them involving A ^ B, the event that both machine detect it. This last probability can be further broken down using the independence of the events. I think 91% is right.

3. Jun 7, 2007

### Cyannaca

Is there a mathematical formula I can use that will give me 91%? I get this answer by drawing a tree of possibilities and MxN gives me 49%....

4. Jun 7, 2007

### phoenixthoth

Yes.

What you want to find is p(A v B), the probability that one machine or the other detects it. There is a formula that breaks up p(A v B) into other probabilities, one of them involving A ^ B, the event that both machine detect it. This last probability can be further broken down using the independence of the events. I think 91% is right.

5. Jun 8, 2007

### esalihm

answer to second question:

0,7*0,3*2=42 percent
because if one machine detects it and the other doesn't, this event's occuring has a probability of 0,7*(1-0,7)=0,7*0,3=0,21

then whatif the other detects it and this one doesn't, so 0,21*2=0,42

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