Sampling distribution of the sample mean

The central limit theorem and the sampling distribution of the sample mean

Sampling distribution of the sample mean

Discussion and questions for this video
Where on the onlinestatbook site is this little software toy?

Thanks, John
If we know the mean and the standard deviation of the population, then why are we taking samples, if we already have the data?

Thanks in advance.
Learning statistics can be a little strange. It almost seems like you're trying to lift yourself up by your own bootstraps. Basically, you learn about populations working under the assumption that you know the mean/stdev, which is silly, as you say, but later you begin to drop these assumptions and learn to make inferences about populations based on your samples.

Once you have some version of the Central Limit Theorem, you can start answering some interesting questions, but it takes a lot of study just to get there!
I have a practice question that I just can't figure out. It is: "Eighteen subjects are randomly selected and given proficiency tests. The mean for this group is 492.3 and the standard deviation is 37.6. Construct the 98% confidence interval for the population standard deviation."

I don't know how to figure out the confidence interval for a standard deviation. Can you please help. Thanks. Katie
We already know that:
A range from -1 to 1 contains 68.3% of outcomes.
A range from -2 to 2 contains 95.4% of outcomes.
A range from -3 to 3 contains 99.7% of outcomes.

So the question is, how many Std.Dev's do we have to move away from the mean in both directions on the graph to contain 98% of outcomes. Not 95.4%, Not 99.7%, exactly 98%. Right away you know the answer will be between 2 and 3's, as 98% is between 95.4% and 99.7%

Is there any difference if I take 1 "sample" with 100 "instances", or I take 100 "samples" with 1 "instance"?
(By sample I mean the S_1 and S_2 and so on. With instances I mean the numbers, [1,1,3,6] and [3,4,3,1] and so on.)
Do your sample sizes have to be the same size? E.G, at 1:05(ish) there are a bunch of samples with a sample size of four. Would it mess up any calculations if you took a sample of four and then, say, a sample of ten?
Yes, the sample sizes should be the same. The sample size is not considered to be a variable, it's considered to be a constant. The sampling distribution of the sample mean can be thought of as "For a sample of size n, the sample mean will behave according to this distribution." Any random draw from that sampling distribution would be interpreted as the mean of a sample of n observations from the original population.
So if every distribution approaches normal when do I employ say a Poisson or uniform or a Bernoulli distribution? I suppose it's a concept I haven't breached yet but how do I know when or which distribution to employ so I appropriately analyze the data? End goal = solve real world problems!
Not every distribution goes to the Normal. the distribution of the sample mean does, but that's as the sample size increases. If you have smaller sample sizes, assuming normality either on the data or the sample mean may be wholly inappropriate.

In terms of identifying the distribution, sometimes it's a matter of considering the nature of the data (e.g. we might think "Poisson" if the data collected are a rate, number of events per some unit/interval), sometimes it's a matter of doing some exploratory data analysis (histograms, boxplots, some numerical summaries, and the like).

For actually analyzing data: I would suggest hiring someone with more extensive training in Statistics to actually do such. Taking one course in Stats, which is basically what KhanAcademy goes through, isn't really enough to prepare someone to be a data analyst. I see the primary goal of taking one or two stats courses as giving you enough information to allow you to understand the results of statistical analyses. You can better tell the statistician what you want in his/her own terms, and you can better understand what s/he gives back to you.
why can we say that the sampling distribution of mean follows a normal distribution for a large enough sample size even though the population is may not be normally distributed?
Properly, the sampling distribution APPROXIMATES a normal distribution for a sufficiently large sample (sometimes cited as n > 30). A coin flip is not normally distributed, it is either heads or tails. But 30 coin flips will give you a binomial distribution that looks reasonably normal (at least in the middle).
Is it possible to determine the sample variance without the population variance? I have an assignment that requires me to show the sampling distribution of the mean with only a population proportion and sample size.
If a question talks about a "population proportion" then you are dealing with a binomial distribution, except that you divide by the sample size to get sample proportion rather than the sample count. If the population proportion is p, then the mean value of sample proportions will be also be p (as usual, the mean of the sampling distribution is just the same as for the whole population), and the variance will be p(1 - p)/n, where n is the size of the sample. You can read about this distribution here (note they use the letter pi for population proportion. It does NOT mean 3.14159...):
What is the difference between "sample distribution" and "sampling distribution"?
The sample distribution is what you get directly from taking a sample. You plot the value of each item in the sample to get the distribution of values across the single sample. When Sal took a sample in the previous video at 2:04 and got S1 = {1, 1, 3, 6}, and graphed the values that were sampled, that was a sample distribution. The 2nd graph in the video above is a sample distribution because it shows the values that were sampled from the population in the top graph.

The sampling distribution is what you get when you compare the results from several samples. You plot the mean of each sample (rather than the value of each thing sampled). In the previous video, Sal did that starting at 4:29, when he plotted the mean of each sample. The 3rd and 4th graphs above are sampling distributions because each shows a distribution of means from the many samples of a particular size. also has an explanation.
Me and my friend Callum have been experimenting with sampling distribution progran on online stat book used by Sal ( However we found a result we cannot explain nor rationalise: When we ask for a sample size of 2 for the median disribution of any population it aproximates the population distribution and not a 'bell curve'. I am very disturbed by this because surely the median of 2 numbers is the same as the mean of 2 numbers and according to the central limit theorem should approximate a normal distribution. Is this assumption correct? Is the programme wrong? Or is there something we fail to understand?
The distribution of the sample median is not normal even if you take a larger sample size, such as n=5,10,or 25. The distribution of the sample median seems to be more related to the distribution of the population.
But I don't know why.
Some sources state that Kurtosis for ND is 0 and other books state that it is 3.
I am confused about this and I kindly ask for your advise since I don`t want to go too deep in the formulas for kurtosis.
Using kurtosis as it's properly defined, the Normal distribution has kurtosis of 3. Sometimes though, people want the Standard Normal distribution to represent the most basic values possible (mean zero, standard deviation 1, skewness 0, etc) so that it's a baseline (in some manner). So they'll define "excess kurtosis" which is just "kurtosis - 3" so that the Normal distribution has 0 kurtosis. Apparently, doing this also makes some other calculations more convenient.

By the way, you don't need to delve into formulas. Wikipedia has a lot of the information on distributions labeled. For instance, check out:

Along the right-hand side, there is a table with all sorts of information, one of which is "Ex. Kurtosis." When you click on it, there's a short description of what excess kurtosis means.
At around the 8 minute mark the meaning behind a negative or positive kurtosis is explained. What exactly is the relevance behind knowing whether the kurtosis is negative or positive? As it is explained in the video It seems knowing the kurtosis only gives us a more specific idea of the shape of the the curve, I'd like to know if it says anything about the data set? If so what? Thanks!
In the display Sal was working with, there was a field that tracked kurtosis - how close it was to 3 (normal distributions have a kurtosis of 3) as the number of samples increases. I don't know the answer to your questions, but I think he talked about it so we would get some clue about what it is, because the closer the distribution gets to normal, the closer the kurtosis gets to 3.
I believe the tool can generate the same sample more that once, right?

Any special behavior if we plot ALL possible combinations of the population(for a particular sample size N) only once?

Will it produce a "more normal" distribution if we plot the same number of samples but with the possibility of generating the same sample more than once (and therefore leaving out some other sample combination which wont be generated?)
"Any special behavior if we plot ALL possible combinations of the population(for a particular sample size N) only once?"
I'm going to make a guess that it depends on the distribution of the population. Kind of interesting because they're not random samples. If you made up some finite number of samples that approximated a normal sampling distribution, would the central limit theorem apply to them? In my opinion, no.
"Will it produce a "more normal" distribution if ..." How do you satisfy your conditions?
What is the difference between X-bar and mu? Like when do you know which to use what?
X-bar is the mean of a sample (as Sal says at 4:29 in You use X-bar for the mean calculated from data that was only gathered from part of the population (such as a survey of 1000 adults out of the entire US population).

Mu is the mean of the entire actual population. You only use mu to describe the mean if you are talking about data gathered from every element in the population, such as the 2010 census or every porcupine in the zoo.
How would one answer a question such as "what is the sampling distribution of the sample mean? Explain." after being given a problem where the only info given is the mean of a (normal) distribution and its standard deviation? There is also a number that is being randomly computed and averaged. Is the sample mean the mean of the normal distribution?
The sampling distribution of a normal distribution is itself normally distributed. The mean of the sampling distribution is the mean of the original distribution (by symmetry there is no other possible result), and the standard deviation of the sampling distribution shrinks by the square root of the sample size.

This derives from the properties of the variance. When you add two random variables, the variance of the sum adds. Thus when you add n identical random variables, the variance of the sum is n times the original variance and the standard deviation (square root of the variance) is sqrt(n) times the original standard deviation. Divide this by n, to AVERAGE n identical random variables, and you get the above result.
@ 9:15 two distributions are shown and compared (N=5 and N=25) and Sal explains in terms of skew and Kurzweillosis (or something) that the N=25 distribution is more normal. But wait... it does not LOOK more normal to me. Specifically, it looks a lot lumpier... as if it were composed of less data. Each bin is fatter and there are less bins. Am I making sense? Can someone explain?
The lumpier look you're seeing is exactly because of the fewer number of bins. If we wanted, we could go in and specify how we wanted the bins formed, but typically there's just a computer algorithm that chooses the bins in some fashion. If we chose a few more bins there, it would looks much more smooth.

The bottom histogram looks more normal because of the general behavior of the distribution. The one for n=5 is like a normal distribution that was smashed down a bit. It's too short in the middle and has too "fat" of tails. If you think back to, say, the Empirical Rule, the top one would probably have less than 68% of the data within 1 standard deviation of the mean.

p.s. the word is "kurtosis," it's a way to describe the "peakedness" of the graph. A graph with high kurtosis will have much sharper peak (picture 1 below), a graph with low kurtosis will have much more of a rolling hill look to it.

Picture 1:

Picture 2:
Frankly, I'm not sure if my question actually belongs here... but this seems to be in the ballpark. It's been 30 yrs since my statistics class and I'm more than a little rusty! lol If this is the wrong forum for this question, I would appreciate if someone would point me in the right direction.

So, what I'm trying to do is find how to figure out a possible -distribution of scores- given the following info... 120 scores, ranging from 879 - 900 w/ a mean of 886.

Is that enough info to produce a sample distribution of possible scores and if it is.. is there a lesson/video that can explain the procedure to calculate this.
What do you mean by the following:
"figure out a possible -distribution of scores"
"produce a sample distribution of possible scores"
I have a question that I cant figure out please help:
Identify the class width, class midpoints, & class boundaries for the given frequency distribution
Daily low temp (F) Frequency Daily low temp (F) Frequency
32-35 1 48-51 7
36-39 3 52-55 7
40-43 5 56-59 1
44-47 11
The class widths are the width of each interval which in this case is 4 (e.g. {32, 33, 34, 35} has 4 items),

the mid points are the mid point of each class, (top + bottom)/2, 33.5 in the case of the first one.

The boundaries between the ranges except you want to include the data that gets rounded up or down, so you add 0.5 to the top boundary or subtract it from the lower. So they would be 31.5, 35.5, 39.5, ..., 59.5
I'm a little confused about what you're doing at 04:40. Lets say the PDF represents the 32 species of animals on a small island. So that application selects 5 types of animals lets say zebras, goats, penguins, gorillas and porcupines and plots their mean on the graph below. How the hell can you get the mean of a set of 5 species of animals? I don't get it.
@cnidoblast, selecting 5 types of animals invalidates the CLT. One of the assumptions of the most common CLT (there are actually many versions, this one is the most common) is that the observations, what Mr. Khan calls samples, are independent and identically distributed instances of a random variable. A random variable is a function that converts an observation from a random process in to a number. Your animals are not numbers, so it's meaningless to sum them much less find the mean. If you're talking about averaging their weights then it still fails the CLT assumptions because the weights that you're averaging do not come from an identical distribution. That is, the distribution of weights of zebras is very different from the distribution of weights of goats. Hope this helps! :)
Could you define a measure of skewness as (mean-median)/standard deviation? An advantage of this would be that it is easier to calculate, and it can only take values between -1 and 1
I'm having some issues with this question.

3. For the general population, mean IQ is 100 with a standard deviation of 15. A sample of 100 people is selected at random from the population, with a sample mean of 102. This sample mean comes from a distribution of sample means with the following properties:

a. a mean of 100 and a standard error of 1.5
b. a mean of 102 and a standard error of 1.5
c. a mean of 100 and a standard error of 15
d. a mean of 102 and a standard error of 15

I think that the answer is either a or b, because you would divide the SD 15 by the square root of the original mean 10, which gives 1.5. But I have no idea what to do about the mean 100/102? Can anyone explain why it is one or the other?
THe general population is known to have a mean IQ of 100. That means that the distribution of sample means also has a mean of 100.
Excuse me can any one explain for me what is the difference between sampling distribution and population distribution and can explain by example for each of them
At 513 pm: For some reason, I understand this when it comes to means but in Sampling distribution of the sample proportion- Using population (4,5,9), sample size n = 2- I am struggling to construct a table that represents the sampling distribution of the sample proportion of odd numbers. Can you please explain?
The mean of a set of data is 25 with a standard deviation of 2. What is the interval about the mean of the data within one standard deviation?
I have a question that I dont quite understand and it goes like this: "Assume the weights of eggs produced on an egg farm have a normal distribution with mean 64 grams and standard deviation 7 grams. and it also says "describe the distribution of weights of 12 (randomly chosen) mixed grade eggs?
9:08, how do you get five samples from the non-normally distributed probability function? How do you get a set of data from the probability function?
Computers can quite easily simulate uniform distributions (for example the rand() function in matlab that gives a number between 0 and 1 accordingly to an uniform distribution). With that number you can simulate all sorts of other distributions.
For example if you want to simulate a fair dice you do :
x = rand(1)
if (x<1/6) then y = 1
elseif (x<2/6) then y = 2
elseif (x<3/6) then y = 3
elseif (x<4/6) then y = 4
elseif (x<5/6) then y = 5
else y=6

This is how you can simulate easily discrete distributions.
My professor said the answer to the problem is "NOT" 0. I take meticulous notes, record lectures, online research, etc. Why can't I figure this out. Do I need to somehow calculate a sample proportion? Not sure what else to do. If the sample proportion is not given, how do I find it. The problem is the Z scores are above 3 and our Standard Normal Distribution Table stops at 3. Again, he said the answer is not 0. Below are some problems directly pasted here:

1) Given a normal distribution with a µ = 100 and σ = 10, if you select a random sample of n = 25, what is the probability that the sample mean is between 90 and 97.5?

2) Given a normal distribution with a µ = 50 and σ = 8, if you select a random sample of n = 100, what is the probability that the sample mean is between 47 and 49.5?

3) Given a normal distribution with a µ = 50 and σ = 5, if you select a random sample of n = 100, there is a 35% chance that the sample mean is above what value?

I'm really struggling here with the Z's being greater than 3. working on this for three days. Not just trolling for answers and being lazy. I desperately want to know the techniques and steps to calculate situations like this. Thank you very very much.
The key to all of these questions is using the standard error of the mean which is described in one of the next videos in the section.

Briefly, the SE (standard error) = standard deviation / sqr (sample size).

For 1) the SE = 10 (standard deviation) / 5 (sqr of 25) = 2. If you use a z table, we are looking for the probability of z between -5 {(90-100)/2} and -1.25 {(97.5-100) / 2}. This is .1056 using this online table (

The other problems are solved similarly
I need help putting together the formula to anser the question, "A population is bimodal with a variance of 5.77. One hundred samples of size 30 are randomly selected and the 100 sample means are calculated. The standard deviation of the sample means is approximately:
If the distdistribution for a histrogram showing weight is normal what does it mean for the mean median and mode
Thanks for info! Just one question, what about the distribution of the means of the means? So if I take 10 samples from a population find the averages and do this 10 times and average all the means again, what is the standard deviation of the means of the means?
You'd apply the same ideas. Say we have the following:

Original population has `µ = 1000` and `σ = 24`

Step 1:
Take samples of size n=16 and record the sample mean. If we do this over and over, we'd get the sampling distribution of the sample mean, which is a new population with `µ* = 100` and `σ* = 24 / √16 = 6`

Step 2:
If take samples of size m=9 from the population resulting from Step 1. That is, in order to get 1 observation, we'd:
a. Draw samples of size n=16 from the original population, record the mean.
b. Do this 9 times, and record the mean of the mean.
If we did this over and over, then we'd have a new sampling distribution with `µ** = 100` and `σ** = 6 / √9 = 2`. This can also be expressed as `σ** = 24 / (√14 * √9) = 24 / √(14 *9) `. Which is `σ / √(n*m)`.

In other words: Each value from the sampling distribution in Step 2 used 144 (`16*9`) draws from the original population. Doing this in "stages" of size 16, replicated 9 times, doesn't gain us anything but more bookkeeping, the resulting sampling distribution is exactly the same as if we had just taken samples of size 144 from the original distribution.

Hence, we generally don't bother with this two-stage sampling, we just take one sample, and use the theoretical idea of the sampling distribution to determine how the sample mean will behave.

There are some instances where the ideas of this two-stage sampling could conceivably come in handy. One such possibility would be meta-analysis, where the results of several studies (which implies several samples) are combined. Meta-analysis is a bit of a specialty (and importantly, not my specialty), so I don't know that these ideas are used, it's just my speculation that they could be used.
Thanks a ton Sal. This video really helped me in understanding the concept very well. Can you explain how variance of this distribution is sigma square by n and not just sigma square
You are given the distribution of ages of students enrolled at College of the Redwoods from a simple random sample of size 53. You plot the data and it appears to be strongly skewed to the right. You consult the college admissions office and they inform you that the population mean age of students is 21.3 years with a population standard deviation of 10.7 years. You calculate the mean from your sample of 53. What would be the sampling distribution that this value belongs to?
Is kurtosis independent of narrowness (standard deviation)? What kinds of factors influence kurtosis?
Well, the standard deviation is in the formula for kurtosis. So they are _related_ to one another, if that's what you mean.

Though "independence" has a special meaning in Statistics, and I do not know if kurtosis and the variance are independent in the Statistical sense. To explain what I mean: the sample mean is a part of the formula for the sample variance, yet these two statistics are independent.
Can it be demonstrated that a sampling mean is likely to yield more accuracy than one "large" sample mean, for the same amount of work?
To use what you have, data wise, to make an inference or to find out information about something you do not have data about. So if I know somthing about the sick people in a study for example, I may want to use that knowledge to make an inference about the people who are not sick...or I might want to make an inference about the population in general from the data I have.
Given that a larger sample would result in more variance in opinions- particularly when it comes to qualitative questions- shouldn't he range be wider with a larger population?
In Stats we generally make the assumption that the phenomenon we're investigating has certain parameters. For instance, if we're measuring the number of calories consumed per day, there will be some mean (μ) and some standard deviation (σ). For the most part, the caloric intake will cluster around, say, μ +/- σ (within 1 standard deviation of the mean). There will be some folks a bit more on the extreme sides, say μ +/- 2σ. But the further out we go, the less likely a value becomes.

If we collect a larger sample, we don't really expect to see many more values that a further out - we would expect to see more values clustered around the mean. Sure, there would be some values further out on the tails, but by and large the new values would "fill in" the same area. If you think of a histogram of the observed sample values, think of it getting more and more dense, not wider and wider.
I'm liking these videos. Nice change from scrolling over lecture notes. Thanks Sal.
can you explain the theory of sampling distributions of sample variance?
My guess:
We need to know how get a reliable mean, because for most populations (unless we want to grade a class test on a curve, or do something like that) it's not practical to get the population mean.
I understand that all samples of data are going to vary from time to time and some of them maybe the same, if the population is higher in one distribution than the other. Do you think that sample of means would still normal or do you think that the sample of means would have to be adjusted to fit within the parameters?
I think he was saying that the distribution of the sample means will be approach a normal distribution, regardless of the population the samples come from, as the number of samples increases. I wish he would have discussed some more extreme examples - some even smaller populations, and talked about the limits of use of this statistic.
I have a statistics midterm tomorrow and we only need to understand what the sample mean difference is and what the standard error of the mean is. Can you help me understand those two terms?
In terms of platykuric and leptokuric, which is considered the "positive kurtosis" and "negative kurtosis" as discussed in the video?
From Wikipedia:
"A high-kurtosis distribution has a sharper peak and fatter tails, while a low-kurtosis distribution has a more rounded peak and thinner tails.
Distributions with zero excess kurtosis ["excess" is the difference from kurtosis of a normal distribution, which is +3], are called mesokurtic, or mesokurtotic. The most prominent example of a mesokurtic distribution is the normal [Gaussian] distribution family, regardless of the values of its parameters. ...
A distribution with positive excess kurtosis is called leptokurtic, or leptokurtotic. "Lepto-" means "slender" [as in "lepton"]. In terms of shape, a leptokurtic distribution has a more acute peak around the mean and fatter tails. ... Sometimes called 'super-Gaussian'.
A distribution with negative excess kurtosis is called platykurtic, or platykurtotic. "Platy-" means "broad" [as in "platypus"]. In terms of shape, a platykurtic distribution has a lower, wider peak around the mean and thinner tails. ... Sometimes termed 'sub-Gaussian'."
THIS ISN'T HELPING. I still don't know how to find the probability of getting a specific sample mean given a normal bell curve. He's not even using the bell curve! I'm extremely confused, please help.
I have a question m failing to solve. ' A population has a mean of 200 and a standard deviation of 50. A simple random sample of size 100 will be taken and the sample mean x will be used to estimate the population mean. Show the sampling distribution of the sample mean
I think it means "distribution of the sample means" - the distribution of all of the mean values of the (same sized) samples taken from the original population.
what is the minimum number of sample means that should be used to get a reasonable accurate Sampling distribution of the sample mean since thousands of sample means may not be practical in a gage repeatability and reliability study.
In practice we get one, and use the _theoretical_ sampling distribution to let us draw conclusions.
at 8:45, it has been said that even for single samples the central limit theorem is true. It is not so, central limit theorem is applicable only for sample MEANS. For example, out of a population of 5000 if I have taken the sample of n=50, central limit theorem does NOT apply to that. It applies only when I have taken (e.g.)40 samples of n=50. However, this is as per my understanding. Please correct me if I am wrong.
What would be a real world application for this? If anyone has any examples in the manufacturing world that would be very helpful. Thanks!
What I don't understand is when you have a large Binary distribution for example, and you approximate it using Normal distribution.. If you only have one sample consisting of x values, you haven't got a standard deviation really.. we always have those kinds of questions on the exam but i always get the formula wrong then..
As long as you know all the values in the sample, you can do the series of calculations described under "basic examples" here to figure out what the sample's standard deviation is. Of course, you have to divide by N-1 with samples like the wikipedia article (as well as Sal's video on standard deviation) explains, otherwise it's exactly the same. Perhaps you are limiting your definition of "standard deviation" to "standard deviation of population", which you of course can't figure out with just one sample of values? If it's not specified that the population's SD is asked for in the exam question you're describing, it's safe to assume that they are asking for the sample's SD.
I'm trying to picture skew and kurtosis, but I have no idea how much the numbers actually mean. Is there a video that gives a good idea of how much skew is, say: 0.1, 0.5, 1, and 10? Same thing with kurtosis. I like having a feel for what the value means in my brain.
What would be the difference between the distribution of a sample variable and the sampling distribution of the mean?..? I'm so confused between these two terms
Sal repeats "well defined mean" and "well defined variance" a couple of times at the very beginning of the video. When are these quantities not well defined?
This seems to be a simple question to answer, but I'm actually not 100% certain about it:

say there's a population that's normally distributed with mean u and standard deviation s. An independent sample of N observations is drawn from the population. What is the distribution of the sample mean? I think it's still a normal distribution, but I'm not sure if this is correct and sufficient, because I'm still in the process of getting comfortable with all this stat lingo.

Yes. If X is normally distributed, then the sample mean xbar will also be normally distributed regardless of the sample size. If X is _not_ normally distributed, then we have to make sure the sample size is large enough for the Central Limit Theorem to kick in.
There were two cases talked about; n=5 and n=25. It was said that after 10,000 samples the n=25 was a closer fit to the normal distribution than the n=5 case. What I want to know is, if there were infinite samples, would the n=5 and the n=25 cases both be a perfect normal distribution?

If this is so: As the number of samples tends to infinity, does the n=25 case converge to the normal distribution faster than the n=5 case?
2:20 Can you really do it with the mode? It seems like there would be some distributions in which no matter how many samples you take, the mode would not be normally distributed.
If you can do it with any population, then all the possible modes of all the possible n-samples of a population also constitute a population, no?
In this example Sal took 10,000 samples of 5 for a total of 50,000 samples in the first example. Why not just take 50,000 samples of the original distribution and calculate the mean and SD?
In addition to what Casey wrote, I think you get a different sense of how confident you can be that the sample mean x-bar is close to the population mean mu if you take several samples and compare them.
I might take one sample of size 400, get x-bar = 11 and std. dev=1.5 and decide that mu is probably between 10 & 12. If I take 4 samples of size 100 and get ```
x-bar | std dev
10 | 2.1
8 | 1.4
12 | 1.6
5 | 1.8
``` I will notice that many of the sample means are more than one standard deviation away from another, so I am no longer confident that any of them are close to the population mean.
are sample mean and population mean the same? while solving ques for confidence intervals why do we always subtract the sample mean from the value when the formula includes population mean?
how do distributions provide a link between probabilities and statistical tests
Statistical tests are generally trying to compute the probability of something. Most often, there is an assumption (hypothesis), and we find the probability of the observed results assuming that hypothesis is true.

The probabilities can be calculated in a few different ways, but a very common method is through a distribution. So, we think that the data or a function of it, like a test statistic, has a particular distribution (this is generally _proven_, so it's not just a guess), and we can use that distribution to calculate probabilities.
What's an example of a random variable with a symmetrical PDF which is not normal? Would this happen in the real world?
There are also the Laplace and Cauchy distributions both have a similar shape to the Normal - symmetric with a peak - but are very much not Normal.

The Uniform distribution is symmetric, but has no peak. This most certainly comes up in the real world, the first I can think of is die rolling and board games that use spinners. There is also wait times for something that occurs at regular intervals: Say a bus arrives at a given bus stop every 8 minutes. You arrive at a random time. The length of time you're waiting for the bus will follow a uniform distribution.

Depending on the parameters, the Beta distribution can be symmetric, and is certainly non-normal, since it can only be between 0 and 1. One useful thing is to model probabilities, for instance in Baseball, we can model batting averages. Full disclaimer, I stole this example from the StackOverflow answer here:
See the first answer.
Can someone help explain the relationship among population, sampling frame, and sample? They all are so interlaced to me.
According to wikipedia, "a sampling frame is the source material or device from which a sample is drawn. It is a list of all those within a population who can be sampled."
Let's take an example. Suppose someone wanted to find out how many houses in the city of Yelm were painted white.
The *population* is everything being considered: all houses in the city of Yelm, WA.
There are many choices of *sampling frames*. I might choose the property tax rolls of the city, since that probably lists every house in the city (so it covers the population) and probably assigns each house an ID # (so it would be easy to choose a random sample by using a random number generator).
The *sample* is which houses I actually draw from the population to see what color they are. Maybe I decide to sample 70 houses. I use a random number generator to give me random numbers across the range of ID #'s in Yelm's property tax rolls, take the first 70 valid ID #'s, go to those houses and record what color they are.
A manufacturer knows that their items have a normally distributed lifespan, with a mean of 2.6 years, and standard deviation of 0.5 years.
If you randomly purchase 25 items, what is the probability that their mean life will be longer than 3 years?
The average number of defective hard disks made by a certain manufacturer is 3. What is the probability of seeing no more than 10 defective hard disks in a large sample?
You would need at least the standard deviation of the number of defective disks in order to calculate that. With the information presented it's impossible to give an answer.