Transcript Unit 4 Starters

Unit 4 Starters
Starter 7.1.1
• Suppose a fair coin is tossed 4 times.
Find the probability that heads comes up
exactly two times.
Answer
• A tree diagram will show that there are 16
possible outcomes.
• The tree diagram also shows that 6
branches contain exactly two heads.
• Therefore P(two H) = 6/16
• Another approach:
– There are 2x2x2x2=16 outcomes from 4 flips
– Choose any two flips to be heads: 4C2=6
– So P(two H) = 6/16
Starter 7.1.2
• A 9-sided die has three faces that show 1,
two faces that show 2, and one face each
showing 3, 4, 5, 6.
• Let X be the number that shows face-up.
• Draw the PDF histogram of X.
Answer
• What is the total area under the histogram?
• (3/9) + (2/9) + 4 x (1/9) = 9/9 = 1
Starter 7.1.3
•
What is the difference between a
discrete random variable and a
continuous random variable?
– There are two important ideas here
•
Assume a certain continuous random
variable X is known to be N(55, 1.2). If
you randomly choose one observation of
X, what is the probability that:
a) X < 55
b) X < 53
c) X > 53
Hint: normalcdf(LB, UB, mean, s.d.)
Answer
• A discrete random variable has a finite number of
outcomes that can only take on specific (usually
integer) values. It can be represented by a table or
histogram.
• A continuous random variable has an infinite
number of outcomes. We can never list them all,
and they typically include rational and irrational
decimals. It can be represented by a density curve
(only).
• P(X<55) = .5 by definition: 55 is the median
• P(X<53) = normalcdf(0, 53, 55, 1.2) = .0478
• P(X>53) = 1 – .0478 = .9521
Starter 7.2.1
• An unfair six-sided die comes up 1 on half
of its rolls. The other half of its rolls are
evenly spread through the other 5
outcomes.
• If X is the number that comes up, write
PDF for X.
Answer
X
1
2
3
4
5
6
P(X) .5
.1
.1
.1
.1
.1
Starter 7.2.2
• Calculate from formula the variance of
these data:
1, 2, 3, 4, 5
Answer
1 2  3  4  5
x
3
5
2
2
2
2
2
(1  3)  (2  3)  (3  3)  (4  3)  (5  3)
2
X 
2
5
Starter 7.2.3
• Suppose I measure the heights of a class
of fourth-graders and find the distribution
of heights to be N(100, 2 cm).
• Then I have them all stand on top of a 10
cm high step and I measure again.
• What change would you expect in the
mean?
• What change would you expect in the
standard deviation?
Answer
• Since we added 10 to all values, the mean
should increase by 10 cm.
• Adding 10 to all values did not change the
spread; it just changed the location if we
plotted on an axis. So we expect the
standard deviation to remain unchanged at
2.
Starter 7.2.4
• A bag contains 4 red marbles and one
white marble. Two marbles are chosen
without replacement. What is the
probability that they are both red?
• A worker is paid each week in the
following manner: He draws two bills from
a bag which contains four $20 bills and
one $100 bill. What is the expected
outcome of his average weekly pay over
the long run?
Answer
• Let X = the amount of weekly pay
– Notice that X can only be $40 or $120
• There are 5C2 (= 10) ways to draw two bills
from the five in the bag
• There are 4C2 (= 6) ways to draw two $20
bills from the four in the bag
– So P(40) = 6/10
• There are 4 ways to draw one $100 and
one $20
– So P(120) = 4/10
• E(X) = $40 (6/10) + $120 (4/10) = $72
Starter 8.1.1
• Here’s a game you will like: Let’s bet a
dollar on this proposition: I will roll a fair
die once. If it comes up 1 or 2, I win. If it
comes up 3, 4, 5, or 6, you win!
– What is the probability that I win?
– What is the probability that I lose?
– How much should we each bet to make this a
fair game?
Answer
• P(win) = 2/6 or 1/3
• P(lose) = 4/6 or 2/3
• Odds that I win are 2:4 (or 1:2) against, so
you should bet $2 and I bet $1
– Note: “fair” in this context means E(x) =0,
where x represents my net winnings.
– E(x) = (+2)(1/3) + (-1)(2/3) = 0, so it’s fair
Starter 8.1.2
• Five white marbles are on a table. Two of
them are to be painted with a “W” and the
rest will be painted with a “L”.
• How many ways are there to choose the
two marbles to be painted “W”?
Starter 8.1.3
• A baseball player bats .325 over a full
season. If he bats 5 times today, find the
probability he gets at least 3 hits.
Answer
• This is a binomial setting, so use the
binomial CDF of at most 2 hits, then
subtract from 1
• 1 – binomcdf(5, .325, 2) = .197
Starter 8.1.4
• A manufacturer produces a large number of
toasters. From past experience, he knows that
about 2% are defective. In a quality control
procedure, we randomly select 20 toasters for
testing. We want to determine the probability
that no more than one of these toasters is
defective.
–
–
–
–
Is this a binomial setting? Justify your answer.
Find the probability that exactly one toaster is defective.
Find the probability that at most one toaster is defective.
Find the mean and standard deviation for the problem.
Answer
• Two outcomes (good / defective), a fixed
number of trials (20), independent trials,
probability is fixed (2%), so it is binomial.
• P(X=1) = binompdf(20, .02, 1) = 27%
• P(X1) = binomcdf(20, .02, 1) = 94%
• μ = np = (20)(.02) = .4
– So we expect on average .4 defectives in
each group of 20
• σ = √npq = √(20)(.02)(.98) = .626
Starter 8.2.1
• Fred Funk hits his tee shots straight most
of the time. In fact, last year he put 78%
of his tee shots in the fairway.
• In yesterday’s round, he hit only 7 of 14
shots in the fairway.
• What is the likelihood that he would have
that bad a day (or worse) from the tee?
Answer
• This is a binomial setting
– There are 14 trials where he succeeds or fails
at hitting the fairway
– All trials are independent with fixed probability
• To do this poorly (or worse) means he hits
at most 7 fairways.
– He could hit 7 or 6 or 5 or …
• Binomcdf(14, .78, 7) = .02
– So there is only a 2% chance he will do that
poorly.
Starter 8.2.2
• The SAT Math and Verbal sections are
both designed to be approximately normal
with a mean of 500 and standard deviation
of 100.
• If we defined a new measure (TOTAL) by
adding the scores on the two sections,
what would you expect the mean and
standard deviation of TOTAL to be?
(Assume math and verbal are independent)
Answer
TOTAL  Math  Verbal  500  500  1000
 2TOTAL   2 Math   2Verbal

2
TOTAL
 10000  10000  20000
 TOTAL  20000  141.4
Starter
• Write the characteristics of the binomial
setting.
• What is the difference between the binomial
setting and the geometric setting?
Answer
• The binomial setting has 4 characteristics:
–
–
–
–
There are only two possible outcomes
Each trial has fixed probability of success
Each trial is independent of all other trials
There are a fixed number of trials; the variable of
interest is the number of successes
• The geometric setting has the same first three
characteristics. The difference is in the fourth:
there are an unknown number of trials; trials
stop after the first success; the variable of
interest is the number of trials until the first
success