Note: Required work is shown for each
problem. Additional commentary is shown in [square brackets].
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1.
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The sampling
distribution of a statistic
is the distribution of all possible values that that statistic could have when samples of a fixed size are drawn from a fixed population. The two statistics for which we are
most interested in constructing these types of distributions are (1) xbar (the sample
mean) and (2) phat (the sample proportion). However,
IQR, s, and median are also examples of statistics for which we could compute sampling distributions if we were so inclined. [This
question will not be covered on the 2/11/2003
test.]
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2.
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The Central Limit Theorem
states that the sampling
distribution of xbar approaches N(m , s/Ön) as n grows large. This is due,
in part, to the fact that xbar is
a(n) unbiased estimator of m,
the population mean. In fact, regardless of the sample size, E(xbar) = m.
[This question will not be covered on the 2/11/2003 test.]
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3.
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The expected value of X + Y, where X and Y are
random variables, always
equals E(X)
+ E(Y) . (The alternate notation
for E(X) is mX .) However, we can
compute the variance of a sum or difference a priori only if X
and Y are independent,
in which case Var(X
+ Y) = Var(X)
+ Var(Y)
and Var(X
– Y) = Var(X)
+ Var(Y)
also.
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4.
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Let X = r.v.
denoting temp. in °F.
Let Y = r.v.
denoting temp. in °C.
[You need to be very clear about this. Or, perhaps you would like to use
subscripts F and C instead of X and Y to denote the
random variables.]
Then mY = E(Y)
= E(5/9 · (X – 32)) [by substitution]
= 5/9 · E(X – 32) [since 5/9 is a constant]
= 5/9 E(X) – 5/9 (32) [since the mean of a
difference is the difference of the means]
= 5/9 (54) – 5/9 (32) [by substitution]
= 12.222°C.
Var(Y) = Var(5/9 · (X –
32)) [by substitution]
= (5/9)2 · Var(X – 32) [by rules for variances, since
Y is a linear function of X]
= 25/81 · Var(X) [since variance is not affected by
a horizontal shift]
= 25/81 · (11.82)
= 25/81 · 139.24
= 42.975
sY = Övariance = Ö42.975
= 6.556°C.
[Note that units are generally meaningless for variance. Shorter solutions
that demonstrate that you understand the principles involved are acceptable.
The bracketed reasons are for your information and are not required if you
show your steps clearly and systematically. However, unsupported answers or
answers that do not use proper notation would be acceptable only in the
multiple-choice section.]
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5.
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Let X = # of HIV-pos. subjects
in SRS of 216
Let Y = position # of the first
HIV-pos. subject found
[You need to be very clear about these definitions. Full credit cannot be
awarded if you start slinging X and
Y around without explanation.]
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(a)
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E(X) = np
= 216(.018) = 3.888
[You should show formula, plug-ins, and answer. Remember that E(X)
means exactly the same as mX
since the terms “expected value” and “mean” are synonymous.]
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(b)
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Since X is approx. B(216,.018),
P(3 £ X £
5) = P(X = 3) + P(X = 4) + P(X = 5)
= .2017… + .196879… + .153…
= .552 by calc.
[You should state that X is approximately
binomially distributed. The most compact way of doing this is as shown
above.]
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(c)
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Since Y is approx. geometrically
distrib. w/ p = .018,
P(73 £ Y £
144) = P(Y £
144) – P(Y £
72)
= (1 – P(Y > 144)) – (1 – P(Y
> 72))
= (1 – q144)
– (1 – q72)
= (1 – .982144) – (1 – .98272)
= .197
Alternate method:
Since Y is approx. geometrically distrib. w/ p =
.018,
P(73 £ Y £
144) = P(Y £
144) – P(Y £
72)
= .926… – .729… by calc.
= .197
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(d)
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binomial
·
X counts
the # of successes in a fixed number of trials (n = 216)
·
p =
.018, essentially a constant
·
only 2 outcomes are poss. on each trial
Independence is slightly
violated since we are using an SRS instead of sampling w/ replacement; but
since Buffalonia is a “large metropolis,” the
discrepancy is negligible.
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(e)
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geometric
·
Y
counts the # of trials needed to obtain success (n is undetermined)
·
p =
.018, essentially a constant
·
only 2 outcomes are poss. on each trial
Independence is slightly
violated since we are using an SRS instead of sampling w/ replacement; but
since Buffalonia is a “large metropolis,” the
discrepancy is negligible.
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6.(a)
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Let X = # of sixes when die is
rolled 600 times
X is exactly B(600, 1/6).
mX = np = 600(1/6) = 100
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(b)
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P(X = 100) = .04366 by calc.
[If you are required to show work, the work would be as follows, using the
formula provided on the AP formula sheet:
P(X = 100) = nCk pkqn – k
= 600C100
(1/6)100(5/6)500
= .04366
But note that this is really asking a lot, since your calculator may not be
able to handle the calculations precisely as shown anyway! You may have
overflow on the large numbers and underflow on the small ones. The binompdf function is a safe alternative, provided you
show your notation properly. A sketch is always a good idea, too.]
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(c)
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Since equal numbers of bettors choose “higher” or “lower,” the problem can
be viewed strictly from the casino’s point of view as consisting of only 2
cases:
(1) We win. (This happens 4.366% of the time, and the net profit is $1.00.)
(2) They win. (This happens 95.634% of the time, and the net profit is zero.)
Let X = casino winnings on a $1.00
bet.
E(X) = Spixi
= .04366…(1) + .95633….(0)
= $0.04366
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7.
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[This is almost exactly like the challenge problem posed 1/29/03, which in
turn was based on an AP exam problem a few years ago. I don’t believe anyone
ever asked about that problem. Anyway, here’s how to do it. The trick is to
consider the difference between the random variables.]
Let C = r.v.
denoting a random cylinder’s diam.
Let P = r.v.
denoting a random piston’s diam.
Let X = C – P = r.v. denoting the difference betw.
cyl. & piston diam.
We want to know P(.08 < X < .13). Since C and P are both normal r.v.’s,
we can assume that so is their difference. (This can be proved
mathematically, but we will take this as an article of faith. It is a true
statement.) If we additionally can assume that C and P are indep. [a key point!], Var(X) = Var(C) + Var(P) = .012
+ .022 = .0005 Þ sX
= Ö.0005
= .02236…
The assumption of independence is plausible if different machines and
different work crews are used to manufacture the cylinders and pistons.
Also, E(X) = E(C – P) = E(C) – E(P) = 4.3 – 4.2 = .1.
Therefore, since X follows N(.1, .02236…), P(.08 < X < .13)
= .725 by calc. Thus the scrap percentage is 1 – .725 = 27.5%.
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