Bigger bag Suppose the class in Exercise 12 buys bigger bags of candy, with 200 M M’s each. Ag

step 1 Once again, welcome to a new problem. This time we're dealing with normal distributions. Normal

Bigger bag Suppose the class in Exercise 12 buys bigger bags...

Key Concepts

68–95–99.7 Empirical Rule

This rule describes the spread of data in a normal distribution, specifying that approximately 68% of observations lie within one standard deviation, 95% within two, and 99.7% within three. It provides a practical way to understand how much variability to expect from sample to sample when the underlying model is normal.

Impact of Increased Sample Size on Variability

As the number of observations increases, the variability (or standard error) of the sample proportion decreases, resulting in a distribution that is more tightly clustered around the mean. This reflects increased precision and a narrower confidence interval around the expected proportion, while the overall normal approximation remains valid with these adjusted parameters.

Central Limit Theorem

This theorem states that as the number of independent trials increases, the distribution of the sample proportion (or sum) of outcomes will tend towards a normal distribution regardless of the underlying distribution. In the context of counting successes or specific outcomes, a sufficiently large number of trials justifies using the normal approximation for the binomial distribution.

Normal Approximation to the Binomial Distribution

When dealing with large sample sizes, a binomial random variable representing counts of a specific outcome can be approximated by a normal distribution. This is particularly useful when calculating probabilities, as it simplifies the analysis of proportions of outcomes by using the parameters (mean and standard deviation) derived from the binomial setting.

Transcript

00:01 Once again, welcome to a new problem.

00:04 This time we're dealing with normal distributions.

00:07 Normal distributions are typically bell -shaped.

00:11 So this is a normal distribution.

00:16 This is normal distribution.

00:18 And the two ways you can see a normal distribution.

00:22 You can see it in terms of quartiles where the 50th percentile is the, is the second quartile or q2 or you could also see it in terms of this would be the first quartile and this is the third quartile such that each one of these parts represents 25 % of the data distribution so that's a normal distribution you could also have the same distribution where you don't have portals but you have certain specific percentages.

01:05 Your mean is still in the middle.

01:08 Your mean is still in the middle and here you could have the mean of distribution, the sampling distribution of sample means, or you could have the sampling distribution of sample proportion.

01:28 So sampling distribution of sample proportions, sampling distribution of sample proportions, sampling distribution of sample proportions simply means that you have a bunch of samples and in each case you pick a certain proportion within the sample.

01:52 Remember defining a sample proportion is you take the number of subjects divided by the sample size.

01:59 So this is sample size and this is the number of subjects, sample size and the number of subjects.

02:11 And then this is the, well actually this is the number of subjects.

02:16 The denominator is the sample size.

02:19 This is a sample size.

02:22 So this denominator is the sample size and this is the number of subjects which is x and and then p -hat is the sample proportion.

02:38 And for the most part, the sample proportion is different from the mean of the sampling distribution of sample proportions.

02:47 So this means that you're taking a bunch of samples like in this case up until the nth sample.

02:56 And then you're dividing by the number of samples that you're dealing with.

03:02 For example in this case we do have three samples so if i wanted to get the mean of this sampling distribution i would have to sum up all the uh sample proportions the p huts since there are three of them and then divide by three uh for the most part the central limit theorem central limit theorem simply says that the main of the sampling distribution of sample proportion is the same as the population mean.

03:36 So mu p -hat is the mean of the sampling distribution of sample proportions, meaning you're taking a bunch of samples, you're taking a bunch of samples, p -hat 1, p -hat 2, p -hut 3 all the way up until your nth sample, and then you're getting distribution of that.

04:08 Ideally you want to have a normal distribution.

04:10 This is the population proportion that you're estimating.

04:18 That's the population proportion.

04:21 So if we have a sampling distribution of sample proportion, our goal is to estimate the population proportion.

04:27 Since it's a distribution, again by the central limit theorem, you're going to have a standard deviation of the sampling distribution of sample proportions and that's going to be p 1 minus p all over n or sometimes if you're estimating based off of the sample you have p ht 1 minus p ht all over n that's the standard deviation of the sampling distribution of sample proportions and we're using the sampling distribution to estimate the population um standard error of the distribution.

05:14 It's a distribution of proportions of the proportion.

05:19 So in this particular problem that we're dealing with, in this particular problem we're dealing with, we do have bags, we have bags of candy.

05:33 So these are bags of candy.

05:36 And they happen to be a couple of them that happen to be a couple of them couple of bags of candy and maybe end bags of candy and inside of them we have 200 m and m's we have 200 m and ebs so that's candy it's a bag of candy so we go on and on and on up until the end bag and each bag, each bag is going to have a certain number of green candy.

06:25 So p hat would be x, which represents the number of green candy for each bag and then divide by n, which is the sample size of m and and this is equivalent to 200 of them.

06:49 We have 200 of them.

06:51 So three problems coming up in part a.

06:56 The question is why is it appropriate to use normal distributions to describe to describe the proportion of green m and m's in the back.

07:40 And then the second part is we want to use the empirical rule, which is the 68, 95, 99 .7 % rule, to explain the to explain the variation of this proportion, the variation of this proportion from one bag to the other.

08:28 And then patsi was saying what would happen, what would happen to the model, if you increase the, if you increase what would happen to the model, if you increase, or rather, we want to say if you change, and change simply means increase or decrease, you change the number of bags.

09:28 If you change the, or rather, i think we should say what would happen to the model, if you increase the amount of candy per bag, yeah, that's what we're going to say, if you increase the amount of candy per bag, we're saying what's going to happen to the model.

10:07 So we jump right into the first part of the problem where we are explaining the reasons for using the normal distribution.

10:15 So the first thing we're saying is the standard to determine the effectiveness of the normal distribution is based out of the normal distribution is based out of the expected value per bag and its complement and its complement the complement of the expected value and its complement so if p equals to 0 .10 and this is assuming assuming the proportion of green candy, green m &m's candy, is 10%.

12:03 That's the assumption.

12:04 If p is 10%, so, and that's the probability of success, probability of finding green candy, so we call that success.

12:18 So the 1 minus p would be 0 .90, which is probability of failure and by failure we mean not finding eminence for a normal model or distribution the expected value the expected value np has to be greater than or equal to 10 and n1 minus p the complement also has to be greater than or equal to 10 that's why we mentioned the standard so using these values our n is 200 times our p is 0 .1 and that gives us 20 which is greater than equal to 10 or n again is 200 our 1 minus p is 90 and this gives us so 90 this gives us 180 this gives us 180 this gives us 180 this gives us 180 which is greater than equal to 10.

13:47 So we're going to conclude that the normal model is appropriate for describing the distribution, for describing the distribution of the proportion of green m &m...

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