Review of Exercises from Chapter 17:

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Review of Exercises from Chapter 17
Statistics, Spring 2012
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Outline for today:
1. Outline for next two weeks
2. Review homework assignment
3. Reading and notes on Chapter 18
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Outline for next two weeks
• We need to move really quickly
• You need to do the reading ahead of time
• I will provide you with exercises to do in class to
cement what we’re studying
• After you take notes, I will present a QUICK
review
• You MUST take notes and be ahead on the
reading or you won’t make it
• We will be finished with Chapter 20 by next
Friday
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Chapter 17, Exercise 24
Apples (set-up)
• Data:
6% of apples have to go for cider because of
bruises or blemishes
300 trees
What information are we missing or need to
calculate? (actually the latter)
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Chapter 17, Exercise 24
Apples (finding missing variables)
• What are the variables in a binomial
distribution? n, p, and q
• Which do we have? n and p.
• What’s q? Always, always, always 1-p.
• So q = 1-p =1.00-0.06 which equals 0.94
• What can we calculate with these?
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Chapter 17, Exercise 24
Apples (choosing the model)
• So which model do we use?
• We could try the binomial model….OK since
events are independent, so Binom (300,0.06)
would work
• Since np>=10 and nq>=10, normal model OK,
too.
• ?????WHERE DID THE NORMAL MODEL COME
FROM????? Well, let’s see…..
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Chapter 17, Exercise 24
Apples (calculating μ and σ)
• Nothing new here, just plug-and-chug with
our n, p, and q:
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Chapter 17, Exercise 24
Apples (normal model)
• Since the numbers are big, we can also do the
Normal model: N(18, 4.11)
• Your answer for (a) is therefore that either the
binomial model of Binom (300, 0.06) or the
normal model of N(18, 4.11) would be
appropriate.
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Chapter 17, Exercise 24(b)
Apples (Probability that no more than a dozen cider
apples)
• Do it first on your calculator by calculating
Binom (300, 0.06). I’ll post the answer in a
couple of slides (go back over your notes from
before break; it was the last lecture we had)
• Let’s do the Normal model to remind
ourselves how to do it (you can do the
binomial by plugging it into your calculator)
• Our formula is our old friend, the z-equation:
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Chapter 17, Exercise 24(b)
Apples (Probability that no more than a dozen cider
apples)
• What is y for the equation? 12.
• Can we take it directly from the z-table, or do
we need to subtract our result from 1? (talk
about it; answer on next slide)
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Chapter 17, Exercise 24(b)
Apples (Probability that no more than a dozen cider
apples)
• You can read it off directly, since the problem
says “no more than a dozen cider apples.”
• So let’s calculate:
• The z-value of -1.46 is about 0.072.
• Answer for the binomial is a little different:
0.085. Remember, the normal model IS an
approximation!
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Chapter 17, Exercise 24(c)
Apples (how likely are more than 50 cider apples)
• Before you start plugging and chugging, think for
a moment.
• We’ve seen this type of problem before.
• Actually, we don’t have to do any more
calculations.
• When you see a problem like this, it does NOT
require an exact answer.
• You can answer it from N(18, 4.11).
• Talk at your table and come up with an answer.
Share out, and let’s see how yours compares to
mine. (¡Cállate, Jason!)
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Chapter 17, Exercise 24(c)
Apples (estimating)
• Whenever you see a question like this, just
measure off the value they give you—here,
50—and find out how many standard
deviations it is from the mean. If it’s more
than 3σ, no way is it probable.
• Here you should get something like 7.8
standard deviation, so no way, José (there,
you got mentioned!).
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Chapter 17, Exercise 26
Airline no-shows
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•
•
•
•
p=0.05 (success is a no-show!)
q therefore is 0.95
n = number of tickets sold =275.
Now, how do we set this up?
We need the probability that FEWER than 9
people will cancel. (If 10 people cancel, we
have 265 seats and 265 passengers, but if only
9 cancel, we have 265 seats and 266 (angry)
pasajeros.
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Chapter 17, Exercise 26
Airline no-shows (set-up)
• Let’s calculate mean and standard deviation,
since we’re going to use a normal model (also
calculate the binomial model using your
calculators; let Ms. Thien know if you’re
having problems and I’ll review this with you
tomorrow)
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Chapter 17, Exercise 26
Setting up the normal model
• So we have N(13.75, 3.61)
• Calculating our z-value:
• The area under the normal curve that
corresponds with -1.32 is 0.0934.
• Binom should give you 0.116, which is the exact
answer (your answers may be a bit different if
you use the calculator)
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Chapter 17, Exercise 27
Annoying phone calls
• Like, there’s another kind of phone call? <rimshot, please>
• This problem is actually very much like 24(c). If
you didn’t get it, take about 5 minutes and
rework it, using 24(c) as a model. Or go on to
the next page, and we’ll work it together.
(Better if you try on your own….you’ll
remember it longer!)
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Chapter 17, Exercise 27
Annoying phone calls (set up)
• n=200; p=0.12, so q = 0.88
• Calculate mean and standard deviations for
normal model:
• What does this mean? That we expect sales of
24, with a standard deviation of 4.6
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Chapter 17, Exercise 27
Annoying phone calls (solution)
• Once again, we want to evaluate 10 sales as being
how for from the mean (i.e., expected value) of
24.
• If the standard deviation is 4.6, 10 is more than 3
standard deviations away. NOT a good sign.
• We can therefore conclude that he was probably
misled. From a telemarketer? I’m shocked,
SHOCKED to find that there’s gambling in
here….now where are my winnings?
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Chapter 17, Exercise 34
A skilled archer, a new bow
• This is a simple question to answer, if you
press the right buttons on your calculator
• Use the binompdf function. Instructions are
on the next page, which you can skip if you
know how to use the function (it’s discussed
on p. 393 of the text)
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Chapter 17, Exercise 34
A skilled archer, a new bow..and TIs 83 and 84+
• Access by pressing the
2nd-Vars button to get
distributions
• Scroll up until you get to
binompdf
• WARNING! Binompdf is
at the “A” label in the TI
83+ and at “0” in the TI
84+
• Press enter
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Chapter 17, Exercise 34
A skilled archer, a new bow..and TIs 83 and 84+
• You should get
something that looks
like the first screen on
the right
• The syntax is
binompdf(n, p, X),
where X is what we
want (here, 6 in a row)
• Answer is about 0.26
• What does that mean?
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Chapter 17, Exercise 34
Analysis of the numerical answer from binompdf
• 0.26 means that a little more than 1 time out
of 4 (26% of the time), you can expect a skilled
archer like Diana to get 6 bull’s-eyes in a row.
• The result is therefore not really surprising,
and doesn’t suggest that there’s anything
special about the new bow.
• Note: why did we use the binomial
distribution here instead of the normal
distribution?
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Chapter 17, Exercise 36
More archery
• Same problem as 34, except the archer shoots
50 shots and gets 45 of them. Are you NOW
convinced?
• Well, let’s see if you are.
• Let’s do our standard calculations first. Since
we have 50 shots, let’s use a normal model.
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Chapter 17, Exercise 36
More archery
• Standard calculations for normal model:
n=50, p=.8, q therefore is 0.2
• What’s your take on this? Good, great, and
unlikely?
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Chapter 17, Exercise 36
More archery
• For sure, the shooting is good, and above the
80% mark. But 45 is less than 2 standard
deviations above the mean.
• Good…indeed, MUCH better than just 6 in a
row, as in Exercise 34, but not SO spectacular
that we’re going to give it credit.
• Close, but no cigar…..
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Where we go from here
• It’s really important for you guys to be reading
ahead.
• I want you to take notes on Chapter 18 now. I
expect full notes by tomorrow. That’s your
homework, but Ms. Thien is going to check
and grade your progress before you leave
today, so you’ll have two grades entered.
• See the next slide for the conclusion of the
sermon.
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End of sermon (almost)
• Tomorrow I’ll give you as good a lecture as I
can about Chapter 18, and we’ll spend a
couple of days doing problems.
• Read ahead, because we’ll be repeating this
on Thursday for Chapter 19.
• Welcome to college and adulthood….you
heard it here, first.
• Send me feedback on how we’re doing.
--Hartley.
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