Probability of Walnut Weight Difference >8g in Normal Distribution

In summary, the problem involves finding the probability that two randomly chosen walnuts will differ in weight by more than 8 grams, given that the weight of each walnut is normally distributed with a mean of 30 grams and a standard deviation of 5 grams. To solve this, we first create a new variable Z which represents the weight difference between the two walnuts. We can then determine the mean and standard deviation of Z, which are 0 and sqrt(50), respectively. Using this standard deviation, we can calculate the probability of Z being greater than 8 or less than -8, which will give us the desired probability for the weight difference between the two walnuts.
  • #1
katastrophe
5
0

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!
 
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  • #2


katastrophe said:

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?
 
  • #3


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?
 
  • #4


katastrophe said:
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 [itex]\alpha[/itex] such that [itex]\alpha \cdot \textrm{stdev }[Z] = 8[/itex]. Then you want to find out what is the probability that the absolute value of a normally distributed zero mean random variable exceeds [itex]\alpha[/itex] 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.)
 
  • #5


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

1. What is the normal distribution?

The normal distribution is a probability distribution that is commonly used to model continuous random variables. It is characterized by a bell-shaped curve, with the mean, median, and mode all being equal.

2. How is the probability of walnut weight difference calculated?

The probability of walnut weight difference >8g in a normal distribution can be calculated using the standard normal distribution table or by using a statistical software. The formula for calculating this probability is P(X > 8) = 1 - P(Z < 8), where Z is the standard normal variable.

3. What does a probability of walnut weight difference >8g in normal distribution indicate?

A probability of walnut weight difference >8g in normal distribution means that there is a chance that the difference in weight between two walnuts chosen at random from a population will be greater than 8g. This could be due to natural variation in the weight of walnuts or other factors.

4. What does the mean and standard deviation represent in a normal distribution?

The mean of a normal distribution represents the center or average value of the data. The standard deviation represents the measure of spread or variability of the data around the mean. The higher the standard deviation, the more spread out the data is.

5. How does the sample size affect the probability of walnut weight difference >8g in normal distribution?

The sample size can affect the probability of walnut weight difference >8g in normal distribution in two ways. First, a larger sample size will provide more accurate estimates of the mean and standard deviation of the population, which will in turn affect the probability calculation. Second, a larger sample size can decrease the probability of a large weight difference, as it reduces the variability in the data.

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