If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains ***.kastatic.org** and ***.kasandbox.org** are unblocked.

Main content

Current time:0:00Total duration:7:35

AP Stats: UNC‑3 (EU), UNC‑3.M (LO), UNC‑3.M.1 (EK)

- [Instructor] We're told
suppose that 15% of the 1,750 students at a school have
experienced extreme levels of stress during the past month. A high school newspaper doesn't
know this figure, but they are curious what it is, so
they decide to ask a simple random sample of 160 students
if they have experienced extreme levels of stress
during the past month. Subsequently, they find that
10% of the sample replied "yes" to the question. Assuming the true proportion
is 15%, which they tell us up here, they say 15% of
the population of the 1,750 students actually have experienced
extreme levels of stress during the past month, so
that is the true proportion, so let me just write that,
the true proportion for our population is 0.15, what is
the approximate probability that more than 10% of the
sample would report that they experienced extreme levels of
stress during the past month. So pause this video and see
if you can answer it on your own and there are four
choices here I'll scroll down a little bit and see if you
can answer this on your own. So the way that we're going
to tackle this is we're going to think about the
sampling distribution of our sample proportions and first
we're going to say well is this sampling distribution
approximately normal, is it approximately normal
and if it is, then we can use its mean and the standard
deviation and create a normal distribution that has that same
mean and standard deviation in order to approximate the
probability that they're asking for. So first this first part,
how do we decide this? Well the rule of thumb we
have here and it is a rule of thumb, is that if we take
our sample size times our population proportion and that
is greater than or equal to ten and our sample size times
one minus our population proportion is greater than
or equal to ten, then if both of these are true then our
sampling distribution of our sample proportions is going
to be approximately normal. So in this case the newspaper
is asking 160 students, that's the sample size, so
160, the true population proportion is 0.15 and that
needs to be greater than or equal to ten and so let's
see this is going to be 16 plus eight which is 24 and
24 is indeed greater than or equal to ten so that
checks out and then if I take our sample size times one
minus P well one minus 15 hundredths is going to be
85 hundredths and this is definitely going to be
greater than or equal to ten. Let's see this is going to
be 24 less than 160 so this is going to be 136 which is
way larger than ten so that checks out and so the sampling
distribution of our sample proportions is approximately
going to be normal. And so what is the mean and
standard deviation of our sampling distribution? So the mean of our sampling
distribution is just going to be our population proportion,
we've seen that in other videos, which is equal to 0.15. And our standard deviation
of our sampling distribution of our sample proportions is
going to be equal to the square root of P times one minus
P over N which is equal to the square root of 0.15
times 0.85 all of that over our sample size 160, so now
let's get our calculator out. So I'm going to take the
square root of .15 times .85 divided by 160 and we
close those parentheses and so what is this going to give me? So it's going to give me
approximately 0.028 and I'll go to the thousandths place here. So this is approximately 0.028. This is going to be approximately
a normal distribution so you could draw your classic
bell curve for a normal distribution, so something like this. And our normal distribution
is going to have a mean, it's going to have a mean
right over here of, so this is the mean, of our sampling
distribution, so this is going to be equal to the
same thing as our population proportion 0.15 and we
also know that our standard deviation here is going to
be approximately equal to 0.028 and what we want to know
is what is the approximate probability that more than
10% of the sample would report that they experienced
extreme levels of stress during the past month,
so we could say that 10% would be right over here,
I'll say 0.10 and so the probability that in
a sample of 160 you get a proportion for that sample,
a sample proportion that is larger than 10% would be
this area right over here. So this right over here
would be the probability that your sample proportion
is greater than, they say is more than 10%, is more
than 0.10 just like that. And then to calculate it, I can
get out our calculator again so here I'm going to go to
my distribution menu right over there and then I'm going
to do a normal cumulative distribution function, so
let me click Enter there. And so what is my lower bound? Well my lower bound is 10%
0.1, what is my upper bound? Well we'll just make this one
'cause that is the highest proportion you could have
for a sampling distribution of sample proportions. Now what is our mean well
we already know that's 0.15. What is the standard deviation
of our sampling distribution? Well it's approximately 0.028
and then I can click Enter and if you're taking an AP
exam you actually should write this you should say, you
should tell the graders what you're actually typing
in in your normal CDF function, but if we click
Enter right over here, and then Enter, there we have
it, it's approximately 96%. So this is approximately 0.96
and then out of our choices it would be this one right over here. If you were taking this on
the AP exam you would say that called, called normal normal
CDF where you have your lower bound, lower bound, and
you would put in your 0.10 you would say they use an
upper bound, upper bound of one, you would say that
you gave a mean of 0.15 and then you gave a
standard deviation of 0.028, just so people know that you
knew what you were doing. But hopefully this is helpful.