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STAB22 section 5.1
ual observations, but the approximation will be
better if the distribution of the observations is
closer to normal. If you take a simple random
sample of size 144 from this population, there
is an approximate 95% chance that the sample
mean will be between 240 − 2(1.5) = 237 and
240 + 2(1.5) = 243.
5.2 The sample mean has a sampling distribution
with
√ mean µ (the population mean) and SD
σ/ n (smaller than the population SD). Here
the √
sampling distribution has mean 240 and SD
18/ 36 = 3. That is, if you take a sample of
36 observations, the sample mean is likely to be
close to 240 (because the SD of the sampling dis- 5.5 Replace 144 by 1296 in 5.3. The sample mean
again has an approximate normal distribution
tribution is a relatively small 3), even though the
(the approximation will be better because the
individual observations could be quite a bit larger
sample √
size is bigger) with mean 240 and now
or smaller than 240. (The high and low values will
SD 18/ 1296 = 0.5. Now there is about a 95%
tend to cancel each other out. If you only take
chance
that the sample mean will be between
a small sample, you may be unlucky enough to
240 − 2(0.5) = 239 and 240 + 2(0.5) = 241. Beget, say, mostly large values, but the larger the
cause of the larger sample, we can make a more
sample you take, the more likely the large and
precise statement about what the sample mean
small values are to be more or less 50-50.)
might be. (With a sample size this large, it pretty
5.3 The sampling distribution
still
has
mean
240,
but
much doesn’t matter what the distribution of the
√
now has SD 18/ 144 = 1.5, only half as big.
original observations looks like; the sampling disThat is, if you take a bigger sample, the sample
tribution of the sample mean will be pretty much
mean is more likely to be even closer to the popnormal. See example 5.6 for an example of this.
ulation mean (here 240) than it would be if you
5.6 The sampling distribution of the√sample mean
took a smaller sample.
has mean 25 mins and SD 25/ 50 = 3.5355
5.4 According to the central limit theorem, the samminutes (as in Example 5.7). It also has apple mean has an approximate normal
distribution
proximately a normal shape, so we can go ahead
√
with mean µ = 240 and SD σ/ n = 18/12 = 1.5.
and use the normal distribution to figure out the
chance that the mean time between text mes(These are the same numbers as 5.3.) This is
sages is less than 28 minutes, even though we
true regardless of the distribution of the individ1
know that times between individual messages are
not at all normal (see Figure 5.3 in the text).
z = (28 − 25)/3.5355 = 0.85, so the probability that the mean time between text messages is
less than 28 minutes is (Table A) approximately
0.8023. Note that times between individual text
messages are all over the place, but the mean time
between 50 of them is very likely to be less than
28 minutes, which is only a little bigger than the
mean. This is because the mean time between 28
is quite predictable; a sample size of 28 is on the
large side for this kind of calculation.
mation required, regardless of the size of the sample.
If you’re thinking about one can, it has a SD of
3 ml. If you look at 6 cans
√ the sam√ and take
ple mean x̄, x̄ has SD σ/ n = 3/ 6 = 1.225,
smaller. The mean in each case is 250 ml, from
the last sentence of 5.13.
So your picture should look like Figure 1, with
the solid line showing the curve for a single measurement and the dashed line showing the curve
for the mean of 6. (Note that the dashed curve is
less than half as wide, so it goes more than twice
as high, roughly speaking.)
5.7 (a) would be correct if “variance” were replaced by “SD”. (“Sampling distribution of sample mean” would be clearer than just “mean”,
too.) In (b), larger sample sizes will have sampling distributions for the sample mean with
smaller SDs (which is a bit clearer than the statement given in the question). For (c), the mean
of the sampling distribution of the sample mean
stays where it is (it is equal to the population
mean µ no matter how big the sample size is); it is
the SD of the sampling distribution that changes
when the sample size changes (gets smaller as the
sample size gets bigger).
In (b), the can has to contain less than 250 −
1 = 249 or more than 250 + 1. These two values
correspond to z = (249 − 250)/3 = −0.33 and
z = 0.33. The chance of either of those is 0.3707
(Table A), so 0.7414 of the time a single can will
be outside these limits.
In (c), because the SD is now less than half as
big, the z values are now more than twice as big.
z = ±0.82; the chance of being “outside” either
of those is 0.2061 (Table A), so 0.4122 of the time,
the mean of 6 cans will be outside those limits.
The results of (b) and (c) make sense according to
the picture. The moral of the story is that means
of several measurements are more accurate than
values of individual measurements, so the mean
5.15 First off, if individual observations have a normal distribution, the sampling distribution of a
sample mean is exactly normal, with no approxi2
of 6 measurements is more likely to be inside a
“tolerance” than an individual measurement is.
This is why, in science class, you made several
measurements and averaged them.
5.17 The sample mean that you actually observe
might be smaller or bigger than the population
mean, but on average (that is, over the many possible samples you might take), the sample mean
will be the same as the population mean. (The
sample mean is neither systematically too small,
nor systematically too high. The distribution of
the sample mean might look like Figure 3.14(b)
on page 207 of the text, but at least it doesn’t
look like (a) or (c).) The other side of the issue
is of how close the sample mean might be to the
population mean; this is summarized by the SD
of the sampling distribution of the sample mean
(in other words, how much sampling variability
there is), and a larger sample will have less sampling variability, so it will produce a picture more
like 3.14(d) than 3.14(b).
5.18 (a) z = (23 − 19.2)/5.1 = 0.75, so the chance of
a value higher than this is 0.2266. (b)
√ The mean
25 = 0.96.
is still 20.8, but the SD is now 5.1/
√
(c) z is now (23 − 19.2)/(5.1/ 25) = (23 −
19.2)/0.96 = 3.73, and the probability of exceeding this is close to zero (since 3.73 is off the end
of Table A). My software gives 0.000096. (d) The
Figure 1: Normal curves for volume and mean volume
of cans
3
0.5)/0.099 = 1.01 and is 1 − 0.8438 = 0.1562.
Because the number of moths trapped is highly
variable, the sample mean could easily be bigger
than 0.6 even though the population mean is only
0.5.
scores were only “roughly normal”, so the answer
to (a) may not be very accurate. The answer
to (c), however, is based on the mean of a number of observations, and (by the Central Limit
Theorem) the sampling distribution of the sample
mean is approximately normal even if the individual observations are not. So (c) is more accurate
than (a). (25 is probably a large enough sample
size to make the Central Limit Theorem work if
the population is “roughly normal”, though of
course this is guesswork.)
A hint that that the normal approximation is
going to be OK here is to look at whether the
normal approximation to the sampling distribution of the sample mean could go down below
zero (which we know to be impossible for these
data). The 68–95–99.7 rule says that almost all
of the sampling distribution should be between
0.5 − 3(0.099) ' 0.2 and 0.5 + 3(0.099) ' 0.8.
This doesn’t get too close to zero, so we are all
right. (0 is actually five SDs below the mean of
the sampling distribution, so the normal approximation giving us a value below zero is as good
as impossible.
5.19 When your observations have to be zero or
positive (as here, they are numbers of moths
trapped), and the SD is bigger than the mean,
that’s a sign that the distribution is skewed to
the right. (Try this for yourself with your software: make some lists of numbers that are zero
or positive, and see what it takes to get the SD
bigger than the mean.) My guess here is that
5.20 To find the mean of this discrete distribution,
there are a lot of traps with no moths at all. But
use “value times probability added up”, to get
because of the central limit theorem, we can do
a mean of µ = 4(0.33) + 3(0.24) + 2(0.18) +
calculations for mean numbers of moths trapped
1(0.16) + 0(0.09) = 2.56. Then find the variance
in a sample of traps with some confidence that
as σ 2 = (4 − 2.56)2 (0.33) + (3 − 2.56)2 (0.24) +
the answers will be accurate enough.
· · · +√(0 − 2.56)2 (0.09) = 1.7664, and so the SD is
The sampling distribution of the mean number
σ = 1.7664 = 1.329.
of moths trapped
in
50
traps
has
mean
µ
=
0.5
√
√
The mean √
and SD of the sample mean are 2.56
and SD σ/ n = 0.7/ 50 = 0.099, so that the
and 1.329/ 50 = 0.1880.
chance of finding a sample mean number of moths
trapped that is bigger than 0.6 uses z = (0.6 −
For a randomly chosen student, just add up the
4
√
probabilities for “B or better”: 0.24+0.33 = 0.57.
125 and SD 10/ 3 = 5.773. The chance of this
The average grade for the class as a whole was
sample mean coming out above 140 now uses
around B- (a grade point value of 2.56), and
z = (140 − 125)/5.773 = 2.60, so the probabilso the chance that the 50 students will have a
ity is 1 − 0.9953 = 0.0047, which is much smaller.
mean grade point value of 3 or more ought to
The chance of a wrong diagnosis has been much
be less than 0.57. Using the answers for (b),
reduced by using 3 measurements rather than 1.
z = (3 − 2.56)/0.1880 = 2.34, and the chance
of being higher than this is 1 − 0.9904 = 0.0096, 5.22 (a) is the same idea as 5.20: mean is 0(0.999) +
2
500(0.001) = 0.5, variance is (0 − 0.5)√
(0.999) +
which is indeed (a lot) smaller than 0.57. (The
2
(500 − 0.5) (0.001) = 249.75, so SD is 249.75 =
grade point mean for a sample of 50 students is
15.80. The law of large numbers says that Joe’s
more likely to be close to 2.56 than the grade
average winnings per play will be very close to
point score for a single student.)
0.5 over a large number of plays (such as a year’s
The mean, by the way, is less than 3 even though
worth). The Central Limit Theorem says that
the median is greater than 3 (more than 50% of
the distribution of average winnings per play will
students got a B or better), because the distribube close to normal
√ over the year, with mean 0.5
tion is strongly left-skewed: you can’t do better
and SD 15.80/ 104 = 1.549. (Actually, the Law
than A, but there were some students who got a
of Large Numbers is kind of optimistic here: beD or F and they pulled the mean downwards.
cause the SD of winnings if you play once is so
large, Joe’s average winnings over the year is still
5.21 Sheila ought not to get diagnosed with gestaquite variable even though he’s played 104 times.
tional diabetes because her average glucose level
This is saying that 104 isn’t really “large” in this
is 125, a lot less than 140. But if only a single
case. We’ll have to keep a close eye on the normal
measurement is made, she might get diagnosed as
approximation, too.)
having gestational diabetes by chance. The probUsing the normal approx., find the chance that
ability of this comes from z = (140 − 125)/10 =
Joe comes out ahead for the year as P (x̄ ≥ 1) =
1.5, and is 1 − 0.9332 = 0.0668.
P (Z ≥ (1 − 0.5)/1.549) = P (z ≥ 0.32). This,
Measurements made on 3 separate days ought
from Table A, is 1 − 0.6255 = 0.3745.
to be independent. Sheila’s sample mean meaAlternatively, Joe will be ahead for the year if he
surement has a sampling distribution with mean
5
wins once or more, because the $500 (or more)
should be very close to 1.3. The 99.7 part of the
he wins will pay for the whole year’s tickets, with
rule says that the mean number of flaws is almost
a profit left over. The probability that he fails
certainly going to be between 1.0 and 1.6, and 2
104
to win at all over the year is 0.999 = 0.9012,
is way above that.) This time, the Central Limit
so the probability that he wins at least once is
Theorem should work all right because the to1 − 0.9012 = 0.0988. This answer is quite a bit
tal number of flaws found should be around 300,
different from the one obtained using the norand there are many possible different numbers of
mal approximation, which means that the norflaws that might be observed.
mal approximation isn’t very good, even though
the sample size (104) appears to be large. This is 5.25 If the weights of individual passengers are “not
very non-normal”, the mean weight of 25 pasbecause the distribution of single observations is
sengers should be pretty close√to normal, with
very skewed, and n = 104 is not yet big enough
mean 195 pounds and SD 35/ 25 = 7 pounds.
for the central limit theorem to take hold. So
If
the total weight of passengers is going to exbeware: the normal approximation is only an apceed 5200 pounds, the mean weight has to exproximation, and sometimes it’s not very good!
ceed 5200/25 = 208 pounds. The chance of this
The clue here is that Joe’s number of wins is
is figured using z = (208 − 195)/7 = 1.86, and
likely to be very small, 0,1 or 2 say, whereas if
is 1 − 0.9686 = 0.0314. (This kind of calculation
the normal approx. is going to work well there
is routinely done by airlines when deciding how
needs to be a large number of possible outcomes.
much fuel to put in: if the plane (and passen5.24 This is like a sample of 200 individual square
gers) are heavier, it will need more fuel to get to
yards of carpet, so the mean number of flaws has
its destination, but if the airline puts in too much
approx. a√normal distribution with mean 1.3 and
fuel, it will be more costly because the fuel has
SD 1.5/ 200 = 0.1061 (by the Central Limit
weight too (and it takes more fuel to carry more
Theorem). The chance that the mean number
fuel).
of flaws is more than 2 per square yard uses
z = (2 − 1.3)/0.1061 = 6.60; the chance of ex- 5.26 According to the law of large numbers, the averceeding this is more or less 0. (2 isn’t much bigage loss to the company will be close to $250 per
ger than 1.3, but the law of large numbers says
house, if the number of policies sold is large. 12
that the mean number of flaws per square yard
is probably not large enough; if one of those 12
6
mean 57 and variance 2.22 /6 = 0.8067; the sample mean of 6 treated specimens has a normal
distribution with mean 30 and variance 1.62 /6 =
0.4267. The difference also has a normal distribution. The mean is the difference of means,
57 − 30 = 27; the variance is the sum of variances (remember the rules for means and variances?), which is 0.8067 + 0.4267 = 1.2333. For
the chance that this difference
is 25 or bigger,
√
find z = (25 − 27)/ 1.2333 = −1.80; table
A gives probability 0.0359 of being below, so
1 − 0.0359 = 0.9641 of being above.
houses burns down, the insurance company will
have to pay out a lot more than they took in in
premiums. But with a large number of policies
sold, the amount taken in in premiums will be
enough to cover even a few houses burning down,
almost certainly.
If n = 10000, the average
√ loss per policy sold
has mean 250, SD 1000/ 10000 = 10 and approx. a normal distribution. Thus the chance
that the average loss per year exceeds $275 uses
z = (275 − 250)/10 = 2.5 and is approx. 1 −
0.9938 = 0.0062. Insurance, in sufficient volume,
is a very safe business (from the point of view
of an insurance company), just as owning a big
casino is a safe business (from the point of view of
the casino). Probabilistically speaking, insurance
and gambling are really the same thing.
If the individual values have a normal distribution, the sample mean has a normal distribution
exactly; there’s no need to rely on the Central
Limit Theorem. In this case, that’s probably a
good thing, since the samples aren’t very big.
5.27 The sample mean strength of 6 untreated spec- 5.28 (a) The Central Limit Theorem says that samimens has a √
normal distribution with mean 57
ple means have approximately a normal distriand SD 2.2/ 6 = 0.8981 pounds. To find the
bution in large samples. The individual scores
probability that this exceeds 50 pounds, calcuwon’t be too non-normal (they are scores between
late z = (50 − 57)/0.8981 = −7.79. The chance
1 and 7, so there can’t be any extreme outliers),
of getting a value smaller than this is basically 0,
so a sample of 28 scores will probably be large
so the chance of exceeding it is basically 1. (The
enough.
sample mean will be close to 57; specifically, alThe sample mean
of Journal scores has mean 4.8
√
most certainly closer than 50 is.)
and SD 1.5/ 28 = 0.2835; the sample mean
√ of
Part (b) is tricky. The sample mean of 6 unEnquirer scores has mean 2.4 and SD 1.6/ 28 =
treated specimens has a normal distribution with
0.3024.
7
The difference ȳ − x̄ has a normal distribution
with mean the difference of means 4.8 − 2.4 =
2.4 and variance the sum of variances (like part
(b) of the previous question): the variance of the
difference √
is 0.28352 + 0.30242 = 0.1718 and so
the SD is 0.1718 = 0.4145.
the wall’s height differs from the design height by
more than half an inch is 1−0.9876 = 0.0124: not
very likely.
If you want to, you can figure out the mean and
SD of the total height of the wall by taking the
mean and SD for the mean height of a row and
multiplying them by 4. Then you can work with
these values and find the chance that the total height is between 31.5 and 32.5. You’ll get
the same z-values (±2.5) that we got above, and
therefore the same answers.
For the chance that the difference in sample
means is greater than 1, calculate z = (1 −
2.4)/0.4145 = −3.38, so the chance of the difference being less than 1 is (Table A) 0.0004, and
the chance of the difference being 1 or bigger is
1 − 0.0004 = 0.9996: almost a sure thing.
5.31 Like the question about the airline passengers,
this question about totals can be translated into
a question about means (by dividing by 4, there
being 4 rows of blocks). We can figure out the
probability that the wall does not differ from the
design height by more than half an inch, and then
subtract the answer from 1. We are going, therefore, to figure out the probability that the total height is between 31.5 and 32.5 inches, which
(dividing by 4) is the same as saying that the
mean height of a row is between 7.875 and 8.125
inches. The mean height of 4 rows has a sampling√distribution with mean 32/4 = 8 and SD
0.1/ 4 = 0.05 inches. So we need the chance of
being between z = (7.875 − 8)/0.05 = −2.5 and
z = (8.125 − 8)/0.05 = 2.5. This probability is
0.9938 − 0.0062 = 0.9876, so the probability that
8