Investigation 7
a) Answers will vary because it's a simulation, but everyone should find that
11, 22, 33, 44, 55, and 66 occur a little less than 3% of the time, while 21,
31, 32, 41, 42, 43, 51, 52, 53, 54, 61, 62, 63, 64, 65 occur about 5.5% of the
time.
b) Answers will vary. Take the number of cases where the condition holds
and divide by 10,000.
c) The histogram should reveal that most prices occur about equally often, and
some (the repeat digit ones) occur about half as often. The histogram
also reveals many gaps, because lots of values are not possible prices.
d) Answers will vary but should be in the neighborhood of 47.25.
e) The graph reveals that the average price varies a lot at first but gradually
starts to converge as the number of repetitions increases.
f) The probability is 1/36 for 11, 22, 33, 44, 55, and 66; the probability is
2/36 for 21, 31, 32, 41, 42, 43, 51, 52, 53, 54, 61, 62, 63, 64, 65.
g) The simulation results should come close to the
theoretical results.
h) The expected value
i) 20/36
j) 21/36
Investigation 8
a) The answer is that X = 1 with probability (.9)^10 =
.349 and X=11 with probability 1-(.9^10) = .651.
b) The answer is that E(X) = 1(.349) + 11(.651) = 7.51. This is less than
the 10 tests required if each indivisual is tested
individually.
c) No, it does not guarantee this. In fact, the prob. is only .348 that
fewer than 10 tests will be needed.
d) Now E(X) = (1-p)^10 + 11[1-(1-p)^10], which reduces
to 11-10(1-p)^10.
e) Setting E(X) < 10 and solving gives 11-10(1-p)^10
< 10 and so (1-p)^10 > 1/10 and so (1-p) > (.1)^(.1) and so p <
1-(.1)^(.1) or about .206.
f) E(X) = (1-p)^n + (n+1)[1-(1-p)^n] = n+1-n(1-p)^n
g) Composite testing makes sense whenever n+1-n(1-p)^n < n, which is when n(1-p)^n > 1.
Investigation 9
a) X has a binomial distribution with n=10 and p=.15.
b) The possible values are integers from 0 through 10, inclusive. The probs (to 3 decimal places) are: .197, .347, .276, .130,
.040, .008, .001, .000, .000, .000, .000.
c) The most likely value is 1, its prob is .347.
d) p-value = P(X>=8) = .00001.
e) The answer is 4, because P(X>=4) = 1-P(X<=3) = 1-.95003 = .04997.
f) p-value = P(X>=79) = 1-P(X<=78) = 1-.99926= .00074.
g) The p-values are very small, so we have strong evidence that the mortality
rate in this hospital is actually higher than the national rate of .15.
Why? Because the p-values show that if this hospital's underlying
mortality rate were .15, then it's very unlikely to have observed as many
deaths as they had.
Investigation 10
a) Because independence is violated: the outcome of the
first draw affects the outcome of the next one.
b) P(X=0)=.467, P(X=1)=.467, P(X=2)=.067
c) P(X=0)=.488, P(X=1)=.424, P(X=2)=.088
d) P(X=0)=.490, P(X=1)=.420, P(X=2)=.090 (to three decimal places)
e) P(X=0)=.490, P(X=1)=.420, P(X=2)=.090 (to three decimal places)
f) P(X=0)=.49, P(X=1)=.42, P(X=2)=.09 (exactly)
g) Yes, the binomial approximation is very accurate in the N=1000 and N=10,000
cases. It's pretty close in the N=100 case. It's not so close in
the n=10 case.
h) The exact distribution of X is hypergeometric with
N=200 million, N=105 million, n=1000.
X could be approximated by a binomial distribution with n=1000 and p=.525.
The approximation should be very accurate because the population size (200
million) is very much larger than the sample size (1000).