Refer to Example 7.2 . The amount of fill dispensed by a...

Key Concepts

Sampling Distribution of the Sample Mean

This concept refers to the probability distribution of the sample mean computed from repeated samples. When individual observations are normally distributed, the sample mean is also normally distributed, with the same mean but a reduced variance, given by ?²/n. This forms the basis for making inference about the population mean based on sample data.

Standard Error

The standard error is the standard deviation of the sampling distribution of the sample mean, calculated as ?/?n. It quantifies the amount of variability one can expect in the sample mean from sample to sample, and it decreases as the sample size increases, making the sample mean a more accurate estimate of the true population mean.

Z-Score and Normal Probabilities

Converting the condition on the sample mean to a z-score involves standardizing the difference between the sample mean and the population mean using the standard error. This transformation allows one to use the standard normal distribution to determine probabilities for how far the sample mean is from the true mean.

Effect of Sample Size on Variability

This concept highlights that as the sample size increases, the variability of the sample mean decreases because the standard error decreases. Consequently, the probability that the sample mean falls within a fixed interval around the true mean increases with the sample size, leading to more precise estimates.

Central Limit Theorem

The Central Limit Theorem states that, regardless of the underlying distribution of the data, the sampling distribution of the mean will tend to be normal if the sample size is sufficiently large. In this problem, since the original distribution is already normal, the theorem reinforces that the sample mean is normally distributed, providing a robust framework for probability calculations.

Transcript

00:01 The amount of fill dispensed by a bottling machine is normally distributed, so we could model this with that normal shaped curve.

00:11 The standard deviation of the population is going to be one ounce.

00:16 And if we were to select a sample size of nine bottles from this machine, we would find that the probability that the sample mean is, within 0 .3 ounces of the true mean would be equal to .6318.

00:43 So for part a, we want to take and change the sample size of 9 to a sample size of 16.

00:52 And we want to answer that same question.

00:55 What's the probability that our sample mean is within 0 .3 of our population.

01:04 Mean.

01:06 So we would have to rewrite this absolute value inequality into a compound inequality.

01:16 So we're going to say equals the probability of negative point three, which is less than or equal to x bar minus mu, which is less than or equal to positive point three.

01:29 Now i want you to recall what the z score formula looks like when you're working with sampling distributions.

01:38 And the z score would be x bar minus mu all over sigma divide by the square root of n.

01:46 So if you look at the middle part of our compound inequality, it's really close to the z score.

01:57 We've got the numerator of the z score.

01:59 So if i could just put a sigma under or sigma over square root of n underneath, i could then turn.

02:07 This whole thing into a z score.

02:10 But i can't just put it underneath.

02:11 So what i'm technically doing is i'm dividing all the parts of the compound inequality by that variable expression.

02:21 So now i have a new one.

02:23 And the new one is going to read the probability of negative point three over sigma divided by the square root of n is less than or equal to.

02:34 We're going to replace that center part with a z score, which is less than or equal to positive 0 .3 over sigma divide by square root of n.

02:44 And i have the values that i could replace sigma and n with.

02:49 So now i will have the probability of negative 0 .3 over 1 divided by the square root of 16 is less than or equal to z, which is less than or equal to 0 .3 over 1 over the square root of 16.

03:06 I could now simplify those radicals and say the probability of negative 0 .3 over 1 fourth is less than or equal to z, which is less than or equal to 0 .3 over 1 fourth.

03:22 Now i don't like fractions within fractions, the compound fractions.

03:26 So what i'm going to do is i'm going to multiply top and bottom by a 4 over 4.

03:32 So technically i'm multiplying by a 1, so i'm not changing the value any.

03:37 Changing the appearance.

03:39 And in doing so, i will now have the probability of negative 1 .2 is less than or equal to z, which is less than or equal to positive 1 .2.

03:51 So let's go back to our bell shape picture.

03:54 So if i think of this line here as being a positive 1 .2 and this line here as being a negative 1 .2, and they're symmetric with respect to the 0 being smacked.

04:07 Dab in the middle, i'm technically looking for the area in between here.

04:15 I'm going to have to rely on table four in the back of the book, which is your normal distribution table.

04:22 And the normal distribution table is set up to go and talk about the area in the right tail.

04:30 So technically, the area in our right tail is when my x or sorry, when my z is greater, than 1 .2.

04:41 And the area in the left tail is identical in size and shape due to the fact that this curve is symmetric.

04:50 So what i can do for this last statement that the probability that z is between a negative 1 .2 and a positive 1 .2, i could say that that is the same as 1 minus 2 times the probability of being in the right tail because the right tail and the left tail are symmetric, so i can say two times the right tail would be the right and the left put together.

05:20 And the reason i said one minus is because the area under a complete normal curve is always one.

05:26 So if i do one take away the right tail and the left tail, i'll be left with that area or probability in the center.

05:34 So what i'm really looking for is 1 minus 2 times the probability that our z score is greater than a 1 .2.

05:46 So at that point, you're going to actually look at that table.

05:50 And when you look at the table, on the left side, you're going to find a column titled z score, and you're always going to find a z score to its 10th place.

06:03 So you're going to go down this column until you'll find a z score.

06:06 Find 1 .2.

06:07 And across the top represents the hundreds place.

06:11 And if you think about the fact that we really have a zero in the hundreds place, you're going to find the point zero across the top.

06:19 And in doing so, you're going to find a value of 0 .1151, which is the probability that z is greater than 1 .2.

06:31 So we'll now have 1 minus 2 of those.

06:38 If we go back to our picture, we're really saying that in here is 0 .1151 and in the other tail is 0 .1151.

06:50 So if we take away 0 .1151 twice from 1, we will be left with the blue area, which is, again, the probability that our sample mean is within 0 .1151 twice from 1, we will be left with the blue area, which is, again, the probability that our sample mean is within 0.

07:06 Of our populations mean.

07:10 And in doing so, we will get 1 minus 0 .2302 or an overall probability of 0 .7698.

07:24 So in here is a 0 .7698.

07:30 When we move on to part b, we want to do the same thing multiple times and just change the sample size.

07:40 So in part b, we want to start with n being 25.

07:48 So i'm going to say part b subsection 1.

07:51 We want n to be 25, and we want to answer the same question.

07:55 What's the probability that the sample mean is within 0 .3 ounces of the population mean.

08:07 Again we're going to rewrite the absolute value inequality as a compound inequality.

08:21 Again, we're going to divide all the parts of that compound inequality by a sigma over square root n, which is going to turn this middle section into a z score.

08:44 So we could say the probability of negative 0 .3 over sigma divide by square root of n, is less than or equal to z, which is less than or equal to positive 0 .3 over sigma divided by the square root of n.

09:00 And for the remaining parts here, we're going to continue to use this, and we'll just change out our n values.

09:09 So let's continue with that.

09:12 So we're going to get the probability of negative 0 .3 over 1 divided by the square root of 25, is less than or equal to z, which is less than or equal to 0 .3 over 1 divided by square root of 25.

09:34 And the square root of 25 will simplify into 1 over 5...

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