5.5) Pr(1.5<z<2.0) = PHI(2.0)-PHI(1.5) = .9772 - .9332 = .0440 (from a table)
So about 4.4% of the people have borderline cholesterol.
5.6) Take Z=(X-124)/20. We want Pr(X>140) = Pr (Z>16/20) = 1-PHI(0.8) = .2119
5.15) We use a Poisson distribution with u=24 (lambda = 2.4 deaths
per year). Then
sigma = sqrt(24) = 4.90; since 24+2(4.9) = 33.8 > 33; so 33 is not
quite excessive using
the 2 sigma rule, but still quite high.
YOU ARE NOT RESPONSIBLE FOR THIS FOR OUR FINAL:
Since u is big (u > 10), we can actually use
a normal approximation to calculate Pr(X>=33) in
this case. We take (32.5-24)/4.9 = 1.84 and then
Pr(X>=33) ~ Pr(Z > 1.84) = 1 - PHI(1.84) ~ 3.29%
(very low, but not quite exceptional if you are using the 2.5% threshold).
5.19) We use a binomial distribution with p=0.002 and n=100,000.
Then u=200 and
sigma ~ sqrt(200) = 14.14.
Since
100 < 200-2(14.14) ~ 171, this proportion is not consistent with the
previous census,
i.e., if the study was correctly done, this is good evidence that the
disease rate decreased.
YOU ARE NOT RESPONSIBLE FOR THIS FOR OUR FINAL:
Since npq = 200 > 5, we can actually use a normal
approximation to calculate Pr(X<=100)
in this case. We take (100.5-200)/14.14 = -7.03 and then
Pr(X<=100) ~ Pr(Z<-7.03) = PHI(-7.03) < .0001 !
5.27) For 1-14 years old, we take Z=(X-105)/5. We are looking
for
Pr(Z>(115-105)/5) = Pr (Z>2) = 1 - PHI(2) ~ 2.28%
5.28) For 15-44 years old, we take Z=(X-125)/10. We are looking
for
Pr(Z>(140-125)/10) = Pr (Z>1.5) = 1 - PHI(1.5) ~ 6.68%
5.29) The probability of one family being hypersensitive is approximately
p = 1 - (0.9772)^2 (0.9332)^2 = 0.1684
(i.e., 1 - Pr(no one is hyper), note how we used independence)
5.30) If we are looking at 200 families, this is a binomial distribution with n=200 and p=0.1684. Then
Pr (10 <= X <= 25) = SUM (k=10 to k=25) (200,k) p^k (1-p)^(200-k)
YOU ARE NOT RESPONSIBLE FOR THIS FOR OUR FINAL:
We can approximate this SUM by a normal
approximation. The expected value of this
binomial is u = np = 33.68 and the variance is s2 = npq = (33.68)(0.8316)
= 28.01, so
sigma = 5.292 (note npq > 5, so the approximation is good).
We want Z=(X-33.68)/5.292 and then
Pr(10<=X<=25) ~ Pr((9.5-33.68)/5.292 < Z < ((25.5-33.68)/5.292))
=
Pr (-4.57 < Z < -1.55) = PHI(-1.55) - PHI(-4.57) ~ 0.0606
5.35) Since p = 1.5% = 0.015 is the probability that a random
cell is eosinophils, the
probability that k of them are eosinophils follows a binomial distribution:
Pr(X=k) ~ (100,k) p^k (1-p)^(100-k)
So Pr(X>=5) = 1 - Pr(X=0) - Pr(X=1) - Pr(X=2) - Pr(X=3)
- Pr(X=4) =
= 1 - 0.2206 - 0.3360 - 0.2532 - 0.1260 - 0.0465 =
= 0.0177 = 1.77%
YOU ARE NOT RESPONSIBLE FOR THIS FOR THE FINAL:
Since u=1.5 is TOO SMALL, the normal approximation
will not be very good.
Indeed, use u = np = 1.5 and sigma = sqrt(npq) = sqrt((1.5)(0.985))
= 1.216.
Then Z = (X-1.5)/1.216 and
Pr(X>=5) ~ Pr(Z>(4.5-1.5)/1.216) = Pr (Z>2.46) ~ 0.0069 = 0.69%
(DOH!)
5.36) Same as above with p=0.34 and 40 instead of 5, so now the
mean is
u = np = 34 and the st.dev. is sigma = sqrt(34*(0.66)) = 4.74.
Pr (X>=40) = 1 - SUM (k=0 to k=39) (100,k) p^k (1-p)^(100-k)
YOU ARE NOT RESPONSIBLE FOR THIS FOR THE FINAL:
Since u > 10 here, the normal approximation
ought to work.
We take Z=(X-34)/4.74 and
Pr(X>=40) ~ Pr(Z>(39.5-34)/4.74) = Pr(Z>1.16) = 1-PHI(1.16) = 12.3%
5.37) Same as above
Pr(X>=50) = 1 - SUM (k=0 to k=49) (100,k) p^k (1-p)^(100-k)
YOU ARE NOT RESPONSIBLE FOR THIS FOR THE FINAL:
Pr(X>=50) ~ Pr(Z>(49.5-34)/4.74) = Pr(Z>3.27) = 1-PHI(3.27) = 0.05%
5.38) We look for a value k such that Pr(X>=k)~2.5%. We use the 2 sigma interval, so
u+2sigma = 34 + 2 (4.74) = 43.48
Then we will use 44 and above as exceptional cell counts.
YOU ARE NOT RESPONSIBLE FOR THIS FOR THE FINAL:
As we know, this "2sigma" idea is just an approximation. More exactly,
we should use
1.96sigma since PHI(1.96)=0.975, that is
u+1.96sigma = 43.29
Then we look at a normal approximation: Pr(X>=44) = Pr(Z >= s) =
1-PHI(s)
where s=(43.5-u)/sigma is slightly above 1.96, i.e., PHI(s)>0.975
Pr(X>=44) is slightly below 2.5%.
OBS: The book seems to use a different idea for "exceptional"; they
seem to be looking for
Pr(X>=?)<5%. Indeed Pr(X>=44)~2.25%; Pr(X>=43)~3.65%; Pr(X>=42)~5.68%
5.39) YOU ARE NOT RESPONSIBLE FOR THIS FOR THE FINAL:
Using a normal approximation, we see that PHI(1.645) =.95. That
means we
want about u + 1.645 sigma. Now we use u=60 and sigma=sqrt(60*0.4)=4.89
(neutrophils parameters), that is, 60 + 1.645(4.89) = 68.04. Since
X at 68
means Z at X=67.5, we really need X=69 or more.
5.40) YOU ARE NOT RESPONSIBLE FOR THIS FOR THE FINAL:
Using a normal approximation, we see that PHI(2.33) =.99. That means
we
want about u + 2.33 sigma, that is, 60 + 2.33(4.89) = 71.39. Since
X at 71
means Z at X=70.5, we really need X=72 or more.
5.46) For E takers, we use Z=(X-0.8)/0.48, so
Pr(X>0.30) = Pr(Z>-1.04) = 1-PHI(-1.04) = PHI(1.04) = 85.08%
5.47) This is Pr((test +) given (took E)) => SENSITIVITY
5.48) For placebo takers, we use Z=(X-0.05)/0.16, so
Pr(X<0.30) = Pr(Z<1.5625) = PHI(1.5625) = 94.06%
5.49) This is Pr((test -) given (no compliance)) => SPECIFICITY
5.50) If Z1 = (X-0.8)/0.48 and Z2 = (X-0.05)/0.16, we want
1 - PHI(Z1) = PHI(Z2) for the particular X we are looking for, i.e.,
Z1 = -Z2 => (X-0.8)/0.48 = -(X-0.05)/0.16 =>
=> X-0.8 = -3X+0.15 => 4X = 0.95 => X= 0.2375
Then Z2 = 1.172 and PHI(Z2) = 87.9% is both the sensitivity and specificity
of the test.