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Course: AP®︎/College Statistics > Unit 9
Lesson 4: Sampling distributions for sample proportions- Sampling distribution of sample proportion part 1
- Sampling distribution of sample proportion part 2
- Normal conditions for sampling distributions of sample proportions
- The normal condition for sample proportions
- Mean and standard deviation of sample proportions
- Probability of sample proportions example
- Finding probabilities with sample proportions
- Sampling distribution of a sample proportion example
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Normal conditions for sampling distributions of sample proportions
Conditions for roughly normal sampling distribution of sample proportions.
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What is the intuition for the rule np>=10 , n(1-p) >=10 ?(17 votes)
Think about the edge case ofp = 0.1, thenn * 0.1 >= 10i.e.n >= 100- so, if your probability parameter is around10%then you would need at least 100 samples before the SD is so tight that it (i.e. the left side) gets mostly captured in the[0-0.1]interval(3 votes)
So as n increases, even very low or very high values of p start to produce normal sampling distributions?
Are normal distributions impossible if n < 10?(3 votes)
As for your first question, you might be interested in this video by jbstatistics on YT: https://www.youtube.com/watch?v=fuGwbG9_W1c.
The part6:45~7:57graphs the sample distribution of sample proportion with p=0.04 and increasing n, and the part8:15~8:52does the same thing except with p=0.96. That might answer your question.
As for the second question:
First, I think the 10 is more of an arbitrary value than an actual rule. That's because it's quite subjective whether a graph with lower (np) and/or n(1-p) is normal or not (i.e., the video I linked for your first question gives 15 instead of 10 as in this video.)
Secondly though, as you can see in the video on youtube, if your np or n(1-p) takes on small values, a part of your otherwise quite normal histogram gets cut off at zero. Higher np or n(1-p) values kind of makes the distribution skinnier and therefore prevents the cutoff.
I know your question was posted six months earlier, but hopefully this answers your question if you are still confused with it.(7 votes)
- In the first example, how could we tell which way it was going to be skewed?(5 votes)
Someone correct me if I'm wrong.
The mean of the sampling dist is p (population proportion). If your sampling dist is indeed skewed, then when p is closer to 0 than 1, the top of the distribution "hump" will be closer to 0 than to 1, so it will be skewed to the left, and vice versa.(2 votes)
- what does np represent?(3 votes)
That's a good question! np is the population mean of a binomial distribution. Where n is the sample size, and p is the probability of success. Since it's a binomial variable, the probability of success is constant.(4 votes)
- Why do the conditions need to be n*p > 10 and n * q > 10?
I thought that n > 30 is what it needs to be in order for any sample distribution of sample statistic to be normally distributed according the central limit theorem.
So,
The sample distribution of the sample means will be normally distributed if n > 30.
The sample distribution of the sample proportions will be normally distributed if n > 30?
Am I not correct about this?
What am I missing?(1 vote)
The conditions n*p > 10 and n*q > 10 ensure that p is not too close to 0 or 1.
For any given value of n, if p is too close to 0 or 1, then the distribution of the number of successes in a binomial distribution with n trials and success probability p would be significantly asymmetric about its mean (and so significantly non-normal).(6 votes)
- for the first problem, 'a shipment of 50 tangerines everyday' is it means the 'population'? if yes , then how can she sampled 50 tangerines out of 50 population?(2 votes)
- In the example, the shipment of 50 tangerines every day represents the sample size (n), not the population. The population would be the larger pool from which these tangerines are drawn, potentially encompassing all tangerines supplied by the distributor. Emiliana's daily sample of 50 tangerines is considered a random sample from this larger population. The confusion might arise from the wording, but in the context of statistical sampling, the population refers to the total set of observations that could be made, not just the number in a specific shipment.(1 vote)
- If we already know the true population proportion, why are we interested in calculating a sample proportion? We are using the "true" population proportion to validate the normal distribution of the sample, but then why not just work with the (known to be accurate) population data? Why bother with sampling at all in this case?(1 vote)
In short, if the sampling distribution is approximately normal, then we can calculate how likely it is for a sample proportion to deviate from the population proportion by a certain number of standard deviations.
In later lessons we will use this to figure out how likely it is that the population proportion is what it is said to be.(3 votes)
For these examples, we know that the sample size is 50 and 125 respectively, but what about the number of samples (each of which consists of 50 or 125 in this case) taken? Obviously, the more the number of samples, the smoother the curve, so it's implicitly assumed that they take many samples (of 50 or 125)?(1 vote)- It doesn't really matter how many samples we take, the proportion of each sample is still deviating from the population proportion with a probability that resembles that of a normal distribution.
It does however take quite a few samples before we can actually see this in a graph.(2 votes)
How do you see if the sampling distribution will turn out to be uniform?(1 vote)
Can the mean and standard deviation for the sample distribution of sample proportions still be determined even if the sample distribution is not a normal distribution (ex: skewed left or skewed right)?(1 vote)- Yes, the mean (μ sub p hat) and standard deviation (σ sub p hat) of the sampling distribution of sample proportions can still be determined and are meaningful even if the distribution is not normal (e.g., skewed left or right). The mean of the sampling distribution is always equal to the population proportion (p), and the standard deviation is calculated as sqrt(p(1 − p) / n), where n is the sample size. These measures are useful for understanding the distribution's center and spread, respectively, regardless of its shape. However, the normal approximation (and thus the use of z-scores for probability calculations) is more accurate when the distribution is approximately normal, fulfilling the rule of thumb criteria.(1 vote)
6:45Video transcript
- [Instructor] What we're
going to do in this video is think about under which conditions does the sampling distribution
of the sample portions, in which situations does
it look roughly normal, and under which situations
does it look skewed right so does it look something like this, and under which situations
does it look skewed left maybe something like that. And the conditions that
we're going to talk about, and this is a rough rule of thumb, that if we take our sample size and we multiply it by
the population proportion that we care about and that
is greater than or equal to 10 and if we take the sample size and we multiply it times one
minus the population proportion and that also is greater
than or equal to 10, if both of these are true, the rule of thumb tells us that this is going to be
approximately normal in shape, the sampling distribution
of the sample proportions. So with that in our minds,
let's do some examples here. So this first example says, Emiliana runs a restaurant that receives a shipment
of 50 tangerines every day. According to the supplier, approximately 12% of the population of these tangerines is overripe. Suppose that Emiliana calculates the daily proportion
of overripe tangerines in her sample of 50. We can assume the supplier's claim is true and that the tangerines each
day represent a random sample. What will be the shape of
the sampling distribution, what will be the shape of
the sampling distribution of the daily proportions
of overripe tangerines? Pause this video, think about
what we just talked about and see if you can answer this. All right, so right over here, we're getting daily
samples of 50 tangerines. So for this particular example, our n is equal to 50 and our population proportion, the proportion that is
overripe is 12% so p is 0.12. So if we take n times p, what do we get? NP is equal to 50 times 0.12, well 100 times this would be 12 so 50 times this is
going to be equal to six and this is less than or equal to 10. So this immediately violates
this first condition and so we know that we're
not going to be dealing with a normal distribution. And so the question is, how
is it going to be skewed? And the key realization is remember, the mean of the sample proportions or the sampling distribution
of the sample proportions or the mean of the sampling distribution of the daily proportions that that's going to be the same thing as our population proportion
so the mean is going to be 12%. So if I were to draw it, let me see if I were to
draw it right over here where this is 50% and this is 100%, our mean is gonna be
right over here at 12% and so you're gonna have
it really high over there and then it's gonna be
skewed to the right. You're gonna have a big long tail. So this is going to be
skewed to the right. Let's do another example. So here we're told, according
to a Nielsen survey, radio reaches 88% of children each week. Suppose we took weekly random samples of n equals 125 children
from this population and computed the proportion of children in each sample whom radio reaches. What will be the shape of
the sampling distribution of the proportions of
children the radio reaches? Once again, pause this video and see if you can figure it out. All right, well let's just
figure out what n and p are. Our sample size here n is equal to 125 and our population proportion of the proportion of children that are reached each week
by radio is 88% so p is 0.88. So now let's calculate np
so n is 125 times p is 0.88 and is this going to be
greater than or equal to 10. Well, we don't even have
to calculate this exactly. This is almost 90% of 125. This is actually going to be over 100 so it for sure is going
to be greater than 10 so we meet this first condition. But what about the second condition? We could take n 125 times one minus p so this is times 0.12 so this is 12% of 125. Well, even 10% of 125 would be 12.5 so 12% is for sure going
to be greater than that so this too is going
to be greater than 10. I didn't even have to calculate it. I could just estimate it and so we meet that second condition. So even though our population
proportion is quite high, it's quite close to one here, because our sample size is so large, it still will be roughly normal and one way to get the
intuition for that is so this is a proportion of zero, let's say this is 50% and this is 100%, so our mean right over
here is gonna be 0.88 for our sampling distribution
of the sample proportions. If we had a low sample size, then our standard deviation
would be quite large and so then you would end up with a left skewed distribution. But we saw before the
higher your sample size, the smaller your standard deviation for the sampling distribution and so what that does is it tightens up, it tightens up the standard deviation and so it's going to look more normal. It's gonna look closer to being normal. So we'll say approximately normal because it met our conditions
for this rule of thumb. Is it gonna be perfectly normal? No. In fact, if we didn't
have this rule of thumb to draw the line, some might even argue that we still have a
longer tail to the left than we do to the right,
maybe it's skewed to the left, but using this threshold,
using this rule of thumb which is the standard in statistics, this would be viewed as
approximately normal.