A pizza chain monitors the total weight of pepperoni that...

A pizza chain monitors the total weight of pepperoni that goes on its deluxe pepperoni pizzas to make sure customers are satisfied and product isn't being wasted. Suppose that for pizzas in this population, these weights are strongly skewed to the left with a mean of \[250\,\text{g}\] and a standard deviation of \[8\,\text{g}\]. Management takes a random sample of \[64\] of these pizzas and calculates the mean weight of the pepperoni on the pizzas. Assume that the pizzas in the sample are independent. What is the probability that the mean weight of the pepperoni from the sample of \[64\] pizzas \[\bar x\] is within \[1\,\text{g}\] of the true mean? Choose 1 answer: Choose 1 answer: (Choice A) \[P(249 <\bar x < 251) \approx 0.09\] A \[P(249 <\bar x < 251) \approx 0.09\] (Choice B) \[P(249 <\bar x < 251) \approx 0.34\] B \[P(249 <\bar x < 251) \approx 0.34\] (Choice C) \[P(249 <\bar x < 251) \approx 0.68\] C \[P(249 <\bar x < 251) \approx 0.68\] (Choice D) \[P(249 <\bar x < 251) \approx 0.95\] D \[P(249 <\bar x < 251) \approx 0.95\] (Choice E) We cannot calculate this probability because the sampling distribution is not normal. E We cannot calculate this probability because the sampling distribution is not normal.

Transcript

00:01 All right, so we're told that the average weight of a ball of pizza dough at a pizza parlor is 451 grams.

00:10 So maybe it is 451.

00:12 And we are told that the owner weighed 355 pizza dough balls.

00:20 So it's 35.

00:23 And found the sample standard deviation was 28 .3.

00:35 And then we want to know the probability of this sample, of a random sample of 35 balls of pizza dough, as a mean weight of more than 459 grams.

00:47 So what we do is we're going to convert this to our extended normal distribution for the following reason.

00:54 N is greater than 30.

00:56 And if m is greater than 30 by the central limit theorem, we can take the sampling, the distribution about the sample mean to be approximately normal.

01:04 So here's our mean, 451, and the mean of the sampling distribution is the same as the mean of the population, so it's 451.

01:17 But the standard deviation of that sampling distribution is called the standard error and what is s e, and that's calculated as s over the square root of the sample size.

01:29 So that's what that is.

01:32 So what we're looking for is the probability of being greater than 459 grams.

01:37 So here's our normal distribution on which i've drawn the mean and this sample of this mean that we're looking for, 459.

01:45 And we want to know this probability, this blue area under the curve.

01:52 And the way we're going to do this to find the probability of x being greater than 459, we actually take one minus the probability of being less than 459.

02:07 So once we find that, we can get our probability.

02:11 And we're going to convert this to a standard normal distribution about the sample of mean, and this is our formula, z subx bar is equal to x bar minus mu subx bar, which is the mean that we're just talked about, divided by the standard error, which is s over root n.

02:36 So we're going to go ahead and substitute in our values and do this.

02:42 So the mean is 451.

02:44 Yeah, 451.

02:46 X bar that we're looking for is 459.

02:50 And then this is 28 .3 and about a square root of 35.

02:58 And just be careful with your order of operations here.

03:01 I've seen students, things incorrect to the voice.

03:06 Operation so just beware of that and we find the z score the formula gives us this 1 .67 239 but most z tables in a textbook only go to two decimal places so we're just going to round it to 1 .67 so if we convert these to z scores with our formulas will look like this or a notation will look like this you want z greater than 1 .67 and that's 1 minus the probability the z is less than 1 .67 and we do a table lookup, and we find the probability being less than 1 .67 is 0 .95244.

03:50 When minus that value is our desired probability, 0 .0475, if we go to four decimal places.

04:10 So there you go to that.

04:11 Let me write that out here for us.

04:12 And then we'll get to our next exercise.

04:14 So it ends up being 0 .0475.

04:21 Great.

04:22 So now we are another scenario where we have a random sample of size n.

04:28 I'll do this in a separate color.

04:32 Size of 28 is drawn.

04:38 And we're told that is drawn from a normally distributed population with a mean of 70 .98.

04:47 And we're told the standard deviation of the distribution is 11 .48.

05:01 And we want to know the expected value for the sampling distribution of the sample mean, and that is our mean.

05:08 I mean, that's the mean of the population distribution.

05:10 So it's 70.

05:12 It's that thing, same thing, 70 .98.

05:15 So the expected value about the sample mean is 70 .98.

05:23 So that's nice.

05:29 Now we want to know the standard error for the sampling distribution of the sample mean.

05:34 And another little bit of notation about this, you might see it as sigma subx bar.

05:41 That's the same thing as the standard error.

05:43 And the way we calculate that is taking the standard deviation, divided by the square of the sample size.

05:48 So it's 11 .48 divided by the square root of 28, which ends up being 2 .1 .1 .2 .2 .2...

Question

Please give Ace some feedback