Not really optics, but....

[Aarlan]

Hello all,

I figured I'd post here rather than the regular section since most don't venture here.

I have a statistics question.

Is it possible to estimate approximately where a number falls staistically given the following info...

Sample Size: 87
Mean: .4763333
Median .4633333
High .96
Low .19

Number in question .736666

Any idea without knowing the actual 87 inputs where the number in question falls (approximately?)

I don't have the SD

Thanks for any input

AA

[Darryl Meister]

My guess is, probably not. For starters, your distribution is probably skewed to the right. Secondly, if you're interested in something like a probability value or Z score (the percentage of the population that should fall below 0.736666), you'd really need to know the standard deviation, and the distribution should be normal, not skewed.

That said, you could probably "guesstimate" using the high and low values and the fact that with a normal distribution nearly 100% of the values fall within +/-3 standard deviations. Consequently, we can split the average difference between the high and low values from the mean (giving us 0.385), and divide this by 3, so that 1 standard deviation is equal to 0.385 / 3 = 0.12833.

This would give you a z score of (0.736666 - 0.4763333) / 0.12833 = 2.0286. In a normal distribution, this would be a probability of around 0.9788%. That is to say, nearly 98% of the population should be less than 0.736666. Of course, your distribution is skewed and may or may not be a Gaussian (bell curve-type) distribution in the first place, so this is only a guesstimate. A simple linear (non-Gaussian) guesstimate would be (0.736666 - 0.4763333) / (0.96 - 0.4763333) * 50 + 50% = 76.9%. This would probably be a better estimate for random data.

[Darryl Meister]

Consequently, we can split the average difference between the high and low values from the mean (giving us 0.385), and divide this by 3, so that 1 standard deviation is equal to 0.385 / 3 = 0.12833.

On second thought, since the distribution is skewed, it might make more sense to take the one-sided difference from the mean, instead of the average of both sides from the mean. This would give you 0.96 - 0.4763333 = 0.4836667.

This would then give a standard deviation of 0.4836667 / 3 = 0.1612, a z score of (0.736666 - 0.4763333) / 0.1612 = 1.615, and a probability value of closer to 94.6%.

[Aarlan]

Thank you. This helps a lot.

I will find out the actual inputs in a few weeks to find out exactly where I fell.


AA

[Darryl Meister]

My guess is that you probably did better than at least 85% of the sample population.

Best regards,
Darryl

[Aarlan]

This would then give a standard deviation of 0.4836667 / 3 = 0.1612, .

The professor just emailed me. The actual SD was 0.177. You were right on. Thank you for your help.

AA