Probability of Walnut Weight Difference >8g in Normal Distribution

[katastrophe]

Homework Statement

You measure the weight of each walnut in a big bag of walnuts and find that the probability distribution of walnut weight is a normal distribution with a mean weight <W> = 30 grams and a standard deviation σW = 5 grams. What is the probability that two randomly chosen walnuts will differ in weight by more than 8 grams? (Hint: find and use the uncertainty in the difference between the measured weights of the two walnuts.)

Homework Equations

Not entirely sure on this one which to use -- formulas would help greatly if there is one to use in this scenario!
I have Z scales with which to assign probabilities to sigma values that I have been using in other portions of the homework.

The Attempt at a Solution

I was thinking that the first step here would be to figure out how many standard deviations 8 grams would correspond to, which is 1.6. But the question from there is how to apply this: would the probability be same as a quantity being 1.6 standard deviations away from the mean, or would it be something else? I was thinking that it might be some form of integral inclusive of all sets of points greater than or equal to 1.6 standard deviations apart, but I wasn't sure how to set up this integral. Any help with formulas or the integral would be hugely appreciated!

 

[jbunniii]

The Attempt at a Solution

I was thinking that the first step here would be to figure out how many standard deviations 8 grams would correspond to, which is 1.6. But the question from there is how to apply this: would the probability be same as a quantity being 1.6 standard deviations away from the mean, or would it be something else? I was thinking that it might be some form of integral inclusive of all sets of points greater than or equal to 1.6 standard deviations apart, but I wasn't sure how to set up this integral. Any help with formulas or the integral would be hugely appreciated!

You are on the right track, but be careful when you talk about standard deviation. The key question here is "standard deviation of what?"

Let's say X and Y are the weights of the two walnuts.

Then X and Y are random variables, presumably independent. They are normally distributed, each with mean 30 and standard deviation 5. [Kind of cool that there is a nonzero, albeit tiny, chance that a walnut can have negative weight!]

Now you want to consider a new variable Z, which is the weight difference, so Z = X - Y.

What can you say about the type of distribution, the mean, and the standard deviation of Z?

 

[katastrophe]

So, if Z = X - Y, then I'd think the range of values be anywhere from around approximately -60 to 60 (just coming up with numbers at about 6 standard deviations from the original mean, which I know is not all encompassing but even those values have a very tiny nonzero probability in and of themselves). I'd assume would be a mean of zero. Since the standard deviation of the original set is 5 grams, would this also be the standard deviation of the new set? I'd think that they'd display similar variance patterns to the original set, since the new set compares numbers from the original set, but I could see this being a different deviation as well. I know that the standard deviation is going to be the most vital part of this problem, though -- I feel like it'd make sense for the new standard deviation to be the same as the old, but is that actually the case?

I'd think from there I would use the new curve describing Z and find the probability of obtaining values that are greater than 8 or less than -8, but of course, this depends entirely on the standard deviation. I'd think that the average variance in the Z curve would follow very closely the curve that it is derived from, but is this an incorrect notion?

 

[jbunniii]

So, if Z = X - Y, then I'd think the range of values be anywhere from around approximately -60 to 60 (just coming up with numbers at about 6 standard deviations from the original mean, which I know is not all encompassing but even those values have a very tiny nonzero probability in and of themselves). I'd assume would be a mean of zero. Since the standard deviation of the original set is 5 grams, would this also be the standard deviation of the new set? I'd think that they'd display similar variance patterns to the original set, since the new set compares numbers from the original set, but I could see this being a different deviation as well. I know that the standard deviation is going to be the most vital part of this problem, though -- I feel like it'd make sense for the new standard deviation to be the same as the old, but is that actually the case?

If X and Y are normally distributed, then Z = X - Y is also normally distributed.

A normally distributed random variable is completely characterized by its mean and standard deviation (or variance), so let's calculate those.

The mean of Z is E[Z] = E[X - Y] = E[X] - E[Y] = 30 - 30 = 0.

Assuming X and Y are statistically independent, we can also compute the variance of Z as follows:

var[Z] = var[X - Y] = var[X] + var[-Y] = var[X] + var[Y] = 5^2 + 5^2 = 50

and so

stdev[Z] = sqrt(var[Z]) = sqrt(50) [NOT five!]

So this is the standard deviation you want to work with.

The rest of your idea is good. You want to first determine, what multiple of the standard deviation is 8 grams, i.e., find \alpha such that \alpha \cdot \textrm{stdev }[Z] = 8. Then you want to find out what is the probability that the absolute value of a normally distributed zero mean random variable exceeds \alpha times its standard deviation.

You can express the answer as an integral of the probability density function, but there's no closed form for the answer, except in terms of the special functions erf and erfc. (Or you can look up the answer in an appropriate table.)

 

[katastrophe]

That makes perfect sense, thank you so much! I understand much better now, that was very helpful.