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ck12.org: More Empirical Rule and Z-score practice

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

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Video Language:
English
Team:
Khan Academy
Duration:
05:57

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