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Current time:0:00Total duration:5:57

It never hurts to get
a bit more practice. So this is problem number five
from the normal distribution chapter from ck12.org's
AP statistics FlexBook. So they're saying, the 2007 AP
statistics examination scores were not normally distributed
with a mean of 2.8 and a standard
deviation of 1.34. They cite some College
Board stuff here. I didn't copy and paste that. What is the approximate z-score? Remember, z-score
is just how many standard deviations you
are away from the mean. What is the approximate
z-score that corresponds to an
exam score of 5? So we really just
have to figure out-- this is a pretty
straightforward problem. We just need to figure out how
many standard deviations is 5 from the mean? Well, you just take
5 minus 2.8, right? The mean is 2.8. Let me be very
clear, mean is 2.8. They give us that. Didn't even have
to calculate it. So the mean is 2.8. So 5 minus 2.8 is equal to 2.2. So we're 2.2 above the mean. And if we want that in terms
of standard deviations, we just divide by our
standard deviation. You divide by 1.34. Divide by 1.34. I'll take out the
calculator for this. So we have 2.2 divided
by 1.34 is equal to 1.64. So this is equal to 1.64. And that's choice C. So this was
actually very straightforward. We just have to see how far
away we are from the mean if we get a score of
5-- which hopefully you will get if you're
taking the AP statistics exam after watching
these videos. And then you divide by the
standard deviation to say, how many standard deviations
away from the mean is the score of 5? It's 1.64. I think the only tricky
thing here might have been, you might have been tempted
to pick choice E, which says, the z-score cannot be calculated
because the distribution is not normal. And I think the reason why you
might have had that temptation is because we've
been using z-scores within the context of
a normal distribution. But a z-score literally
just means how many standard deviations you
are away from the mean. It could apply to
any distribution that you could calculate a mean
and a standard deviation for. So E is not the correct answer. A z-score can apply to a
non-normal distribution. So the answer is C. And I
guess that's a good point of clarification to
get out of the way. And I thought I would do
two problems in this video, just because that
one was pretty short. So problem number six. The height of fifth grade
boys in the United States is approximately
normally distributed-- that's good to know-- with
a mean height of 143.5 centimeters. So it's a mean of
143.5 centimeters and a standard deviation
of about 7.1 centimeters. What is the probability that
a randomly chosen fifth grade boy would be taller
than 157.7 centimeters? So let's just draw
out this distribution like we've done in a
bunch of problems so far. They're just asking
us one question, so we can mark this
distribution up a good bit. Let's say that's
our distribution. And the mean here, the
mean they told us is 143.5. They're asking us
taller than 157.7. So we're going in the
upwards direction. So one standard
deviation above the mean will take us right there. And we just have to add 7.1
to this number right here. We're going up by 7.1. So 143.5 plus 7.1 is what? 150.6. That's one standard deviation. If we were to go another
standard deviation, we'd go 7.1 more. What's 7.1 plus 150.6? It's 157.7, which
just happens to be the exact number they ask for. They're asking for
the probability of getting a height
higher than that. So they want to know, what's
the probability that we fall under this area right here? Or essentially more than
two standard deviations from the mean. Or above two
standard deviations. We can't count this
left tail right there. So we can use the
empirical rule. If we do our standard
deviations to the left, that's one standard deviation,
two standard deviations. We know what this whole area is. Let me pick a different
color so that I don't. So we know what this
area is, the area within two standard deviations. The empirical rule tells us. Or even better, the
68, 95, 99.7 rule tells us that this
area-- because it's within two standard
deviations-- is 95%, or 0.95. Or it's 95% of the area under
the normal distribution. Which tells us that what's
left over-- this tail that we care about and
this left tail right here-- has to make up the
rest of it, or 5%. So those two combined
have to be 5%. And these are symmetrical. We've done this before. This is actually a little
redundant from other problems we've done. But if these are added, combined
5%, and they're the same, then each of these are 2.5%. Each of these are 2.5%. So the answer to
the question, what is the probability that a
randomly chosen fifth grade boy would be taller then
157.7 centimeters. Well, that's literally
just the area under this right green part. Maybe I'll do it in
a different color. This magenta part that
I'm coloring right now. That's just that area. We just figured out it's 2.5%. So there's a 2.5% chance we'd
randomly find a fifth grade boy who's taller than
157.7 centimeters, assuming this is the mean,
the standard deviation, and we are dealing with
a normal distribution.