You are given a normal distribution of a population. You want to know things about a sample distribution, so you should start with constructing a normal distribution for your sample of 465 M&Ms.
The mean of this sample remains the same. The standard deviation is the population standard deviation over the square root of your sample size, so your sample distribution is N(.8565, .00240).
This problem now becomes a simple cdf problem. Using your sample distribution, you want to know P(X < .8535), which is .106.
Individual M&M plain candies have weights that are normally distributed with
a mean weight of 0.8565 g and a standard deviation of 0.0518 g.
If you randomly select 465 M&M plain candies, what is the probability
that their mean weight will be less than 0.8535 g?
what is the formula I need to use?
6 answers
I wrote P(X<.8535) but how did that become .106?
and where did you find sample size? is it 465 times something? I'm still confused.
and where did you find sample size? is it 465 times something? I'm still confused.
So the population distribution-- the distribution of ALL M&Ms-- is given in the problem: N(0.8565, 0.0518). However, we want to know what the mean weight will be if, instead of pulling from ALL M&Ms, we "randomly select 465 M&M plain candies." This is our sample. Since any sample of a normal distribution yields a normal distribution, we can use the formulas I mentioned in the first post.
P(X < .8535) is what we're looking for. At this point, we have a smple distribution with a mean of .8565 and a stndard deviation of .00240, so we'd throw this into a calculator's normalcdf function (or equivalent function for some other calculator). Be sure to use the sample distribution! (Technical note, in this probability statement this is technically an x-bar, since this is a sample MEAN.)
P(X < .8535) is what we're looking for. At this point, we have a smple distribution with a mean of .8565 and a stndard deviation of .00240, so we'd throw this into a calculator's normalcdf function (or equivalent function for some other calculator). Be sure to use the sample distribution! (Technical note, in this probability statement this is technically an x-bar, since this is a sample MEAN.)
oh okay thank you. I have another question about a claim problem. I don't know how to calculate the claim.
"Suppose you are going to test the claim that more than half of adults
prefer a window seat when they fly.
The results of a poll indicated that 492 of 806 adults preferred a
window seat when they were flying.
Calculate the test statistic for this hypothesis test."
Do I use p(a or B) or something else?
"Suppose you are going to test the claim that more than half of adults
prefer a window seat when they fly.
The results of a poll indicated that 492 of 806 adults preferred a
window seat when they were flying.
Calculate the test statistic for this hypothesis test."
Do I use p(a or B) or something else?
I divided 492 by 806. I want to know if that's the correct way to do it
The problem wants the test statistic for a one-proportion z-test, which we know because we're only given data for 1 test and that "492 / 806" implies a proportion.
The sample statistic for this test is given by sqrt(pq / n).
p = 492/806 (probability of success)
q = 1 - p (complement of p)
n = 806 (sample size-- we polled 806 adults)
The sample statistic for this test is given by sqrt(pq / n).
p = 492/806 (probability of success)
q = 1 - p (complement of p)
n = 806 (sample size-- we polled 806 adults)
1 answer