Chapter 17 Answers to questions 15, 16, 17, 18 and 20

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Transcript Chapter 17 Answers to questions 15, 16, 17, 18 and 20

Chapter 17
Answers to questions 15, 16, 17, 18
and 20
AP Statistics B
• This is the introduction to using the binomial
• I will do some of the calculations by hand
• Others, we will use the binomial function on your
calculator (“about $#%^@! time, “I hear you say,
though it was in the text which you could have
• I will show you how the calculator works by doing
some examples that we first will work through by
Chapter 17, Exercise 15
• This is the first time we’ve been using the
binomial theorem to calculate mean and
standard deviation
• Set-ups and 15 and 16 are straightforward
• Remember Exercise 13 on lefties (lefthandedness, not political orientation):
p=0.13, q=0.87, n=5
Chapter 17, Exercise 15(a) and (b)
• “How many lefties do you expect?” in (a) asks
for the mean. (b) is expressly the standard
• Formulas:
Chapter 17, Exercise 15(c)
• “If we keep picking people until we find a lefty,
how long do you expect it will take?”
• TERRIBLE wording of the question: where
have we been dealing with TIME? (“how
• What they REALLY want is the expected value,
which is what for the binomial distribution?
(next slide has answer, but quiz yourselves
first and come to a consensus)
Chapter 17, Exercise 15(c)
μ = 1/p = 1/.13 = 7.69 people
Chapter 17, Exercise 16
Virtually the identical set-up as Exercise 15. I am
therefore not going to repeat the workout, but
just give you the answers.
a. μ= 4.8
b. σ=0.98
c. 1.25 shots
Confused? Email me!
Chapter 17, Exercise 17
• Same set-up as 15, except we increase n from 5
to 12
• p still is 0.13, q is 0.87, n=12.
• Is this correct?
• Well, not exactly….because the questions asks for
mu and sigma (mean and standard deviation) for
the number of RIGHT-HANDERS in the group
• So p becomes q and q become p, since success is
Chapter 17, Exercise 17(a)
Right-handed mean and standard deviation
Chapter 17, Exercise 17(b)
• What’s the probability that they’re not all
• Trick: find the percentage of all right-handers,
and subtract from 1.
• First calculate:
• Then subtract from 1: 1-0.188=0.812
Chapter 17, Exercise 17(c)
• “What’s the probability that there are no more than 10
• Think before calculating.
• Do you really want to calculate all the following binomial
coefficients? (The answer is “no”, in case you’re wondering.)
Discuss among yourselves alternatives to such massive
calculations, and when you reach a consensus, go on to the
next slide.
Chapter 17, Exercise 17(c)
Alternatives to brute-force calculation
• We do what we did before: find the
percentage MORE than 10, and calculate that
• It’s considerably easier, since we only have
two binomial coefficients to calculate:
• The second is just 1, so let’s calculate the first:
Chapter 17, Exercise 17(c)
Putting everything together
• We have the binomial coefficients, so all we
have to do is connect them to the terms they
belong with.
• Don’t forget to subtract this from 1! (Why?)
Chapter 17, Exercise 17(d)
• The probability of exactly 6 of each?
• So n=12, k=16, and we calculate:
Chapter 17, Exercise 17(e):
Time to throw in the towel on computation
• OK, I surrender to technology
• Yes, we COULD calculate the answer to this. But we’d
have to calculate each of the following binomial
coefficients (or generate Pascal’s triangle for n=12),
and pair them with the appropriate exponentiated
variables, then add them up:
Chapter 17: calculating binomials with
the TI 83+/84+ calculator
• As Monsanto used to say, “better living
through chemistry”
• We’re doing better living through calculators
(for once)
• You need to learn about the binomial
• On the next page are pictures of your
calculators showing you what button to push
Calculating binomials with TI 83+/84+
Using the 2ND-DISTR key combination
Location of keys on TI 83+
Location of keys on TI 84+
Getting to the binomial distributions
Keys to press
• Push the 2nd-Distr keys
together (Distr is on the
VARS key, just below the
cursor keys)
• You should get a screen that
looks like this.
• Move to the binom
distributions by pushing the
top cursor key to get this
Screen of the TI
binompdf and binomcdf
• Here are the two functions
we’ll be working with
• binompdf is “binomial
probability density function”
which we use to determine
probabilities of a single
parameter (e.g., 5 out of 12)
• binomcdf=“binomial
cumulative density function”,
used to find things like “find 4
or fewer hits out of 12”
How to use binompdf
• Let’s redo Exercise 17(d)
• n=12, p=0.87, and x=6 (i.e., a population of 12,
p of 0.87, and 6 out of 12 are right-handed
• Answer: 0.00193
• On the next slide, I’ll take you through how to
do it on the calculator.
n=6, p=0.87, x=6
• Choosing binompdf
gives you a screen like
• Fill it in like this:
• Press ENTER and….
• Pretty easy, huh?
• MUCH easier than
doing it by hand.
• This is how we’ll do the
rest of them
Recap on binompdf
• Use to solve questions like “exactly 6 of 13
have are left-handed”
• Access through VARS key, but be sure you
press 2ND-VARS (actually 2ND-DISTR)
• Syntax is binompdf(n, p, x)
Introduction to binomcdf
• binomcdf=binomial cumulative density function
• Think of this as being like what we did with the normal
function, except easier
• Remember how we’d say on the normal model that the
percentage of the population below Z was x%?
• It will be similar here: this function answers questions
like “what’s the probability of finding 5 left-handers or
less in a population of 9?”
• Good news: we don’t have to calculate all the binary
coefficients and hook them up with the probability
products of success and failure
• Bad news: well, see the next slide
binomcdf: good news and bad
• Good news: easy to calculate IF set up right
• Bad news: not always set up right.
• For example, Exercise 17(e): “What is the
probability that the MAJORITY is right handed?”
• This is the equivalent of asking for 1-(Probability
of 6 or fewer right-handers)
• Good news (redux): we can set up the equation
like this: answer=1-binomcdf(12, 0.87, 6)
binomcdf: Exercise 17(e)
• Choose binomcdf
• Enter values for n, p, and x and
press enter
• Not so hard, is it? Except that
we forgot to subtract this from
binomcdf: subtracting
• Here’s the way you
SHOULD set it up
• Set up, as usual, is
harder and more
challenging than the
Redoing Exercises 17(b) and (c)
using the TI
• Look back on 17(b) and (c)
• See if you can write out ways of calculating
those answers by using either binompdf or
• Take 5-6 minutes to practice.
• Share at your table, then share out to the class
• Somebody write the collective wisdom on the
board, then go on to the next couple of slides
for the answers
Redoing Exercises 17(b) and (c)
using the TI (answer to set-up)
b. probability that not all are right-handers =
1-binompdf(12, .87, 12)
OR binomcdf(12, .87, 11)
c. probability that no more than 10 righties =
binomcdf(12, .87, 10)
d. we’ve already redone (d), right?
b. 1-binompdf(12, .87, 12)
or binomcdf(12, .87, 11)
c. probability that no more
than 10 righties =
binomcdf(12, .87, 10)
Using the TI for Exercise 18
• Enough of working by hand. Technology is a tool. You
shouldn’t try to hammer a nail in with your hand or
your neighbor’s head.
• Data for Exercise 18 on our archer Diana:
p=.80 (would Jonathan Zuniga or Jose
Fuentes be this successful? Don’t believe
Do the problems together using your calculators. Answers
with screen shots are on next slides
Answers, Exercise 18(a)-(c)
a. μ= 8
b. Probability that never
misses=10 of 10
hits=use binompdf
c. No more than 8 bull’s
eyes: use binomcdf
Answers, Exercise 18(d)-(e)
d. Exactly 8 bull’s
eyes=use binompdf
e. Hits bull’s-eye more
often than she misses:
use binomcdf, but
subtract from 1:
1-binomcdf (5
Exercise 20
Set-up on the TI
• Success=presence of the trait; failure=lack of trait (or
vice versa, as we’ve seen)
• Success=1 out of 8. What’s that in percentages? 12.5%
or 0.125 for success, 0.875 for failure
• n= 12
• (a), (b), and (d) are fairly straightforward, though you
can do (a) a couple of different ways.
• (c) is something we haven’t done yet, using the TI,
anyway, so you should stop at that slide and watch
• Except for(c), I’m just going to show you what to use
and how to set it up, and give you screens shots of the
Exercise 20(a)
Two options, for once
(a) No frogs with traits…
use either binomcdf or
Why can you use either to
get the same answer?
Exercise 20(b)
Subtraction and reframing again…
• “at least 2 frogs” means we
could have 2, 3, 4, 5,6,7, 8,
9, 10, 11 or 12 frogs with
the unusual trait
• We already have the figure
from 20(a) for 0 frogs with
condition (=.201). Add that
to the probability of 1 frog
with condition, and you
have the probability of
exactly 1 frog having the
• Subtract this sum from 1 to
get your answer
Exercise 20(d)
Straightforward “plug and chug”
(d): “no more than 4 frogs
with the (1 out of 8) trait”
Obvious and straightforward
application of binomcdf
Finally, 20(c)
Sum of two probabilities
• “3 or 4 frogs” with the
• Easiest approach here is
to use binompdf twice:
one for exactly three
frogs and one for
exactly 4 frogs
• Answer: binompdf(3
frogs with
trait)+binompdf(4 frogs
with trait)