# I Calculating expected number of measurements

1. Sep 20, 2017

### aron silvester

In my Physics lab, we divided into groups and our task was to throw darts on a target containing 13 bins. The ultimate goal is to hit bin 7. The bins look something like the image below. At the end, our class combined each groups average, standard deviation, and standard error (located below). There is a class table provided below where each group combined each of their data (how many darts landed on different bins) into one table. There are two questions (posted at the very bottom), problems 9 and 10, from the lab that puzzles me. If you need any more info, just ask me about it. I've been thinking about this for 3 hours and I can't make sense of it.

Here is the target. It was just a paper that we pinned to the wall. All 13 bins.

Here are data for each group and also the combined data from the class as a whole under the column "Class"

Here is the class table (how many darts landed on different bins for each group combined)

Problem 9:
I calculated the expected number for the bin #1. I plugged in 2.7 for the standard deviation and 7.22 for X (average) and got 1.09E-5. I thought this formula was supposed to calculate the "expected number of darts" for each bin? 1.09E-5 doesn't seem to make sense, or maybe I'm misunderstanding this formula?

Problem 10:
For this problem, do I just copy down whatever answers I get from problem 9? Though from the answer that I got for bin #1, it doesn't make sense to put 1.09E-5.

2. Sep 20, 2017

### Orodruin

Staff Emeritus
The formula you are using in 9 is the pdf of the normal distribution, not a number of events. It gives you the probability that a single dart will end up in a small region of size $dx$ as $P(x)dx$. Now, your distribution is discrete, not continuous, but if the discrete steps are small enough in comparison to the spread, you can use the size $\Delta x = 1$ of the bins in place of $dx$.

In addition, $X$ is the average position, not the average number of events, and $x$ is the position of your bin, not the number of events in it.

Now, you should be able to compute (at least approximatively) the probability of a single dart hitting a bin. If you know that, how many darts do you expect to hit a bin if you throw 1015 darts?

3. Sep 20, 2017

### Staff: Mentor

This formula gives the probability that a single dart goes into a given bin. How would you go from that number to the expected number of darts in each bin?

4. Sep 20, 2017

### aron silvester

What do you mean by average position and the position of your bin? Sorry, I'm confused.

5. Sep 20, 2017

### Orodruin

Staff Emeritus
Average position is the mean of the bin number. The position of your bin is the number of the bin you are considering.

6. Sep 20, 2017

### aron silvester

Here is the right formula that I found on my paper that calculates the expected number of darts at each bin location

Here is the calculation: Is this what you mean?

7. Sep 20, 2017

### Orodruin

Staff Emeritus
No. You have now said that the mean of the bin number is 78.08 and that the bin you are checking is bin number 19. Both are completely impossible as your highest bin number is 13. X refers to the mean bin number, not the mean number of darts in a bin when throwing 1015 darts, and x refers to the number of the bin you are interested in, not the number of darts in the bin you are interested in. The standard deviation should be the standard deviation in the bin number that a single dart lands in.