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12/1/04
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Quiz on §6.1, especially LOLN. Also, the Post
health stories will be covered as usual.
HW due: Write (1) the methodology
for a new simulation to find the answer to the Car Talk problem as posed on the
show, and (2) perform at least 24 iterations, recording both your raw
data and your outcomes. You may use the “chain” shorthand illustrated in
yesterday’s class if you wish. If you are willing to perform more than 24
trials, please feel free. If you are willing to program your TI-83
calculator, you can easily execute hundreds or even thousands of trials, and
in that case you would be exempt from recording your raw data. (Of course,
you would need to write out your program and make your source code available
for inspection.)
Results: Martin M.’s calculator simulation (several thousand trials) was
excellent, a model of clarity in programming. With or without his data, however,
we learned that there was insufficient evidence to reject the claim that .5
is the true probability that passenger #100 is able to sit in the correct
seat. Note: That is not the same as
proving that .5 is the correct answer. We are saying that even if .5 is
wrong, our test does not have enough power to detect a statistically
significant difference.
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W
12/15/04
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Washington
Post Quiz (4 points). This week, the quiz will be multiple choice only (no
fill-in-the-blank).
Also: Quiz (20 points) on random
variables and expected value. The following five facts will be covered,
even though we did not have time to get to them yesterday:
1. Let X be a r.v., and let Y = mX + b be another r.v.
that is related to X, where m and b are constants. (Obviously, the relationship is linear. This
describes a wide variety of situations, including the classic example of
random temperature readings where one r.v. is measured in °C and the other is measured in °F.) Then the
expected value of Y is given by the
formula mY = mmX + b = mmX + b. In words: The expected value of a linearly transformed
r.v. is exactly what common sense would suggest: Take the old mean (i.e., mX), multiply by the slope, and add the intercept.
Or, to put it another way, the expected value of a linearly transformed
variable can be found simply by plugging the mean into the linear function.
Example: Let X = temperature
reading (°C) at Walrus Point on a
randomly selected day, and let Y =
temperature reading (°F) at Walrus Point on a randomly selected day. If the expected value
of X is 21.2, what is the expected
value of Y?
Answer: Since the linear conversion from Celsius to Fahrenheit is given by F = 1.8C + 32, we can simply plug 21.2 into the equation to get
1.8(21.2) + 32 = 70.16°F.
Here is a version that shows more work. Make sure that you understand each
step in this method:
mY = mmX + b = m1.8X + 32 = 1.8mX + 32 = 1.8(21.2) + 32 = 70.16°F.
2. As in #1, let let X be a r.v.,
and let Y = mX + b be another r.v.
that is linearly related to X. Then the standard deviation of Y equals msX. (Note that b plays no role
at all here.) In words: The s.d. of a linearly transformed r.v. is again what
your common sense would tell you: Take the old s.d. (i.e., sX) and multiply by the slope. Ignore the intercept,
since shifting values up or down by a constant has no effect on their
standard deviation.
Example: If the Walrus Point Celsius readings have a s.d. of 2.3°C, what is the s.d. of the Walrus Point Fahrenheit
readings?
Answer: Since the linear conversion from Celsius to Fahrenheit is given by F = 1.8C + 32, with a slope of 1.8, the s.d. must be multiplied by 1.8.
We get sY = msX = 1.8sX = 1.8(2.3) = 4.14°F.
3. Let X and Y be r.v.’s, and let Z
be a new r.v. formed by adding X + Y. Then the expected value of Z
equals the expected value of X plus
the expected value of Y. In
words: The mean of a sum equals the sum of the means. In formula: mX + Y = mX + mY.
Example: Male residents of Globarnia have a mean net worth of $44,300, and
female residents of Globarnia have a mean net worth of $38,180. Compute the
mean net worth of a couple formed by pairing a randomly chosen male resident
with a randomly chosen female resident.
Answer: Since the mean of the sum equals the sum of the means, the expected
value of the couple’s net worth is $82,480. In all of my years of teaching,
I have never had a student who had trouble with this concept.
4. Let X and Y be independent
r.v.’s, and let Z = X + Y. Then the variance of Z equals the variance of X plus the variance of Y, provided X and Y are
independent. In words: The variance of a sum of independent variables
equals the sum of the variances.
In formula: s2X + Y
= s2X + s2Y.
Since the s.d. of Z is simply the
square root, we also have the useful formula sX + Y = sqrt(s2X + s2Y). Important: These results are valid only if X and Y are
independent. Independence means that knowing the value of X does not change the probability
distribution of Y.
Example: Boots are manufactured by two independent machines, one for left
feet and one for right feet. Although the boots should weigh the same, they
vary because of normal manufacturing fluctuations. Let X = the weight (lbs.) of a randomly selected left boot, and let Y = the weight (lbs.) of a randomly
selected right boot. If X and Y follow the N(1.5, 0.18) and N(1.5,
0.16) distributions, respectively, compute the weight distribution of a pair
of randomly selected boots.
Answer: Obviously, the expected value of X
+ Y is 3 lbs., using either common
sense or rule #3 above. The s.d. of X
+ Y is harder. By rule #4, the variance
of X + Y is 0.182 + 0.162 = 0.058. Therefore, s.d.
of X + Y = sqrt(0.058) = 0.241 lb. The sum of boot weights therefore
obeys the N(3, 0.241)
distribution. Note that independence of the boot-making machines is required
for the second part. (Without that, there would be insufficient information
to solve the problem.) Independence is not required in order to find the mean
of the sum; the mean of a sum is always equal to the sum of means.
5. This is the trickiest one of them all. Let X and Y be independent r.v.’s, and let W = X – Y. Then the variance of W equals the variance of X
plus the variance of Y,
provided X and Y are independent. In words: The variance of a difference of
independent variables equals the sum of the variances. Important: This may be different from
what you expected. Read this paragraph very carefully. Also note that
independence is required in order for the result to be true.
Example: A machine that bores holes into a pegboard is adjusted to make holes
whose diameters (in inches) follow N(0.250,
0.002). A completely independent machine that manufactures pegs is adjusted
so that the pegs have mean diameter 0.248" and s.d. 0.003",
normally distributed. What is the probability that a randomly selected peg
will fit into a randomly selected hole?
Answer: Let X = diameter of a
random hole, Y = diameter of a
random peg. The peg will fit iff X
> Y, which is another way of
saying X – Y > 0. For convenience, we will let Z = X – Y, so that our problem reduces to that
of finding P(Z > 0). Note that Z
is normally distributed, with an expected value of 0.002" (either by
common sense or by rule #3) and a variance of 0.0022 + 0.00032
= 0.000013 (by rule #5). Note that we add
variances even though variables X
and Y are involved in a subtraction
process. Thus the s.d. of Z equals
sqrt(0.000013) = 0.0036055513", and Z
follows the N(0.002, 0.0036055513)
distribution. By calculator (normalcdf), we get P(Z > 0) = 0.710.
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