Download Lung Cancer - Introductory Statistics - Lab Solutions and more Study notes Mathematical Statistics in PDF only on Docsity!
TA: Yury Petrachenko, CAB 484, [email protected], http://www.ualberta.ca/∼yuryp/
Review Questions, Chapters 2, 3, 4, 7, 8
2.105 A study of the residents of region showed that 20% were smokers. The probability of death
due to lung cancer, given that a person smoked, was ten times the probability of death due to
lung cancer, given the person did not smoke. If the probability of death due to lung cancer in
the region is .006, what is the probability of death due to lung cancer given that the person
is a smoker?
Solution. Consider an experiment consisting of a random selection of a late resident in the
area. Some of those who died were smokers. Let’s denote S the event that the selected person
smoked. Now, some of the deaths were due to lung cancer. Denote D the event that the
person died because of lung cancer.
It is stated in the problem that P (S) = 0.2 and P (D) = 0.006. We are also given that
P (D|S) = 10 · P (D|S), or P (D|S) = 0. 1 · P (D|S), where S is the event that the person didn’t
smoke.
Let’s apply the law of total probability to P (D) conditioning on S and S:
P (D) = P (D|S)P (S) + P (D|S)P (S).
Substitute all available information into this equation:
0 .006 = P (D|S) · 0 .2 + 0. 1 · P (D|S) · (1 − 0 .2).
Now, P (D|S) = 0.0021. §
3.42 A particular sale involves four items randomly selected from a large lot that is known to
contain 10% of defectives. Let Y denote the number of defectives among the four sold. The
purchaser of the items will return the defectives for repair, and the repair cost is given by
C = 3Y
2
- Y + 2. Find the expected repair cost.
Solution. A binomial model is depicted here. There are n = 4 trials of selecting an item from
a large lot. There are two outcomes each time: an item is either defective or not. Assuming
the lot is large enough, the trials are independent. Since Y is the number of defectives, it
makes sense to consider finding a defective success. Then, p = 0.1 and Y ∼ Binomial(n, p).
To find the expected repair cost C, let’s use the linearity of expected values:
E[C] = E[3Y
2
2
] + E[Y ] + E[2].
The expectation of 2 is 2. The expectation of Y is np (since we know the distribution of this
random variable), so E[Y ] = 4 · 0 .1 = 0.4. To find E[Y
2 ] recall the formula
V [Y ] = σ
2
= E[Y
2
] − μ
2
= E[Y
2
] −
E[Y ]
2
We have E[Y
2 ] = V [Y ] +
E[Y ]
2
= npq + (0.4)
2 = 0.36 + 0.16 = 0.52. Let’s finally plug
everything in: E[C] = 3 · 0 .52 + 0.4 + 2 = 3.96. §
3.107 A salesperson has found that the probability of a sale on a single contract is approximately .03.
If the salesperson contacts 100 prospects, what is the approximate probability of making at
least one sale?
Solution. If we ignore the word “approximate”, we can approach this problem with a binomial
distribution. There are n = 100 prospects, seemingly independent, with the salesperson
making a single sale with the probability p = 0.03 in each of the 100 cases. The question can be
reformulated in terms of probabilities as follows. Find P (Y ≥ 1) if Y is distributed binomially
with parameters n and p. To answer this question let’s use the fact that
P (Y = y) = 1,
and Y takes values from 0, 1, 2,... , 100. So,
P (Y ≥ 1) =
100 ∑
y=
P (Y = y) = 1 − P (Y = 0) = 1 −
0
(0.97)
100
≈ 0. 9524.
To find a truly approximate value, we can use the Poisson distribution. In this case, since p
is close to zero, the variance and expected value doesn’t differ much. So, let λ = np = 3 and
assume that Y follows the Poisson distribution:
P (Y = y) =
λ
y
y!
e
−λ
, y = 0, 1 , 2 ,...
With this assumption,
P (Y ≥ 1) = 1 − P (Y = 0) = 1 −
0
e
−λ
= 1 − e
− 3
≈ 0. 9502.
The second method is faster and implements a model with one parameter only. §
We want this function minimized. Denote
f (a) = a
2
σ
2
1
2
σ
2
2
, then
f
′
(a) = 2aσ
2
1
− 2(1 − a)σ
2
2
and solving for a:
a =
σ
2
2
σ
2
1
2
2
This value of a minimizes the variance of the third estimator. §
8.11 Let Y 1
, Y
2
,... , Y
n
denote a random sample of size n from a population whose density is given
by
f (y) =
3 β
3
y
− 4
, β ≤ y,
0 , elsewhere
where β is unknown. Consider the estimator
β = min(Y 1
, Y
2
,... , Y
n
(a) Derive the bias of the estimator
β.
(b) Derive MSE(
β).
Solution. See Prof. Prasad’s notes.
