Question 1191546: The weight of potato chips in a medium-sized bag is said to be 10 ounces. The amount that the packaging machine puts in a bag of potato chips can be modeled by a normal curve with a mean 10.2 ounces and standard deviation 0.2 ounces.
Response area (a) What fraction of all bags sold are underweight?
Response area (c) What is the probability that the average weight of the 3 bags is below the stated amount?
Response area (d) What is the chance the average weight of a 24-bag case of potato chips is below 10 ounces?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the weight of potato chips is assumed to be 10 ounces.
the machine spits out an average weight of 10.2 with a standard deviation of .2
i think that you have to look at the output of the machine as your population that you want to compare against, since that same machine is used for all bags of potato chips.
under that assumption, the population mean is 10.2 and the population standard deviation is .2.
if you look at each bag of chips individually, then the z-score formula can be used.
that formula is z = (x - m) / s
z = the z-score
x = the raw score
m = the mean
s = the standard deviation.
the formula becomes z = (10 - 10.2) / .2 = -1.
the area to the left of a z-score of -1 is equal to .1587 rounded to 4 decimal places.
that tells you that approximately 15.87% of the bags of chips coming out of the machine will be less than 10 ounces.
that should answer response area (a).
if you are measuring the average weight of 3 bags, then you would probably use the t-score formula, based on the rules of the following reference.
https://math.stackexchange.com/questions/1817980/how-to-know-when-to-use-t-value-or-z-value
the t-score formula is the same as the z-score formula.
the difference is in the area to the left or right of the t-score versus the area to the left or right of the z-score.
with the t-score formula, there is an additional requirement to determine the degrees of freedom.
this affects the area to the left or right of the t-score.
the degrees of freedom are equal to the sample size minus 1 in these types of problems.
with a sample size of 3, the degrees of freedom would be equal to 2.
the t-score formula is t = (x - m) / s
since you are dealing with the mean of a sample, rather than an individual sample, you need to use the standard error, rather than the standard deviation.
the formula for standard error is s = sd / sqrt(n)
s is the standard error
sd is the standard deviation
n is the sample size.
with a sample size of 3 and a standard deviation of .2, the standard error would be equal to .2 / sqrt(3) = .1155 rounded to 4 decimal places.
the t-score formula of t = (x - m) / s becomes t = (10 - 10.2) / .1155 = -1.73 rounded to 2 decimal places.
the area to the left of that t-score, with 2 degrees of freedom, is equal to .1129, rounded to 4 decimal places.
that tells you that approximately 11.29% of samples of 3 bags each will have an average weight that is less than 10 ounces.
that should be your answer to response area (c).
if you are looking for the average weight of 24 bags being under 10 ounces, then the standard error becomes .2 / sqrt(24) = .0408, rounded to 4 decimal places.
the t-score formula of t = (x - m) / s becomes t = (10 - 10.2) / .0804 = -4.90, rounded to 2 decimal places.
the area to the left of a t-score of -4.90 with 23 degrees of freedom (24 sample size minus 1 = 23 degrees of freedom) is equal to .00002989.
if you round that to 4 decimal places, it becomes 0.
that should be your answer to response area (d).
i believe this is the way this problem should be analyzed.
let me know if you have any questions or concerns.
theo
|
|
|
