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```Mathematical Biostatistics Boot Camp 1
Week 1
Quiz
1. (1 point) What is P (A ∪ B) always equal to?
a. 1 − P (A
C
∩ BC )
b. 1 − P (AC ∪ B C )
c. 1 − P (AC )P (B C )
d. P (AC ∩ B C )
1.
a.
Solution:
P (A ∪ B) = 1 − P (A ∪ B)C = 1 − P (AC ∩ B C )
2. (1 point) Which of the following are always true about P (
√
It is smallerPthan or equal to
Pn
i=1
P (Ei )
It is equal to
P (Ei ).
It is smaller than maxi P (Ei ).
n
i=1
√
√
.
Sn
i=1
Ei )? (Check all that apply.)
It is larger than or equal to min P (E ).
It is larger than or equal to max P (E ).
i
i
i
i
It is smaller than mini P (Ei ).
3. (1 point) Consider inuenza epidemics for two parent heterosexual families. Suppose that the probability is
17% that at least one of the parents has contracted the disease. The probability that the father has contracted
inuenza is 12% while the probability that both the mother and father have contracted the disease is 6%. What
is the probability that the mother has contracted inuenza?
a. 12%
b. 6%
c. 5%
d. 25%
e. 11%
f. 17%
3.
e.
Solution:
Let's have following events:
A : father has the u, and B : mother has the u
We know the following:
P (A ∪ B) = 0.17, P (A) = 0.12, and P (A ∩ B) = 0.06.
From the following formula:
P (A ∪ B) = P (A) + P (B) − P (A ∩ B)
we can derive:
P (B) = P (A ∪ B) + P (A ∩ B) − P (A) = 0.17 + 0.06 − 0.12 = 0.11
Page 1 of 6
4. (1 point) A random variable, X is uniform, so that it's density is f (x) = 1 for 0 ≤ x ≤ 1. What is it's 75th
a. 0.25
b. 0.50
c. 0.75
d. 0.10
e. 0.90
c.
4.
Solution:
The cumulative distribution function for f (x) is:
x
Z
F (x) =
Z
x
f (t) dt =
0
0
x
1 dt = [t]0 = x − 0 = x
We want to nd value of x75 such that F (x75 ) = 0.75. We know that F (x75 ) = x75 , so x75 = 0.75.
5. (1 point) A Pareto density is
for 1 < x < ∞?
a. x1 − 1
b. 1 −
c. 1 −
d. x13
e. x1
1
x2
for 1 < x < ∞. What is the distribution function associated with this density
1
x
1
x3
5.
Solution:
Z
x
F (x) =
1
b.
x 1
1
1
1 1
1
1
dt
=
−
=
= − =1−
t2
t 1
t x
1 x
x
Page 2 of 6
6. (1 point)
a.
b.
c.
d.
What is the quantile p from the density e−x (1 + e−x )−2 ?
log ((1 − p) /p)
p/ (1 − p)
(1 − p) /p
1/ (1 + ex )
e. log (p/ (1 − p))
6.
e.
Solution:
First, we need to nd cumulative distribution function:
x
e−t
−∞
(1 + e−t )
Z
F (x) =
2
dt,
using substitution u(t) = 1 + e−t and du = −e−t dt we get:
Z
u(x)
F (x) =
u(−∞)
−1
du =
u2
Z
1+e−x
∞
1+e−x
−1
1
1
ex e−x
ex
du
=
=
=
=
u2
u ∞
1 + e−x
e−x (ex + 1)
ex + 1
Now we want to nd a value of xp such that F (xp ) = p.
F (xp ) = p
exp
=p
exp + 1
exp = p (exp + 1)
exp = pexp + p
exp − pexp = p
exp (1 − p) = p
p
1−p
p
xp = log
1−p
exp =
Page 3 of 6
7. (1 point)
of c?
a.
b.
c.
d.
e.
f.
g.
Suppose that a density is of the form cxk for some constant c > 1 and 0 < x < 1. What is the value
k+2
1
k
2
1
k+1
k−1
k
k+1
7.
Solution:
Z
1
xk+1
cx dx = c
k+1
k
0
1
=
0
g.
c k+1 1
c
c
x
=
(1 − 0) =
must be equal to 1, so:
0
k+1
k+1
k+1
c
= 1 ⇒ c = k + 1.
k+1
8. (1 point) Suppose that the time in days until hospital discharge for a certain patient population follows a
density f (x) = 12 e−x/2 for x > 0. What is the median discharge time in days?
a. 1.0
b. 1.4
c. 1.8
d. 2.2
e. 2.6
8.
b.
Solution:
First, we need to nd cumulative distribution function F (x):
Z
x
h
ix h
i0
1 −t/2
e
dt = −e−t/2 = e−t/2 = e0 − e−x/2 = 1 − e−x/2
0
x
0 2
Now we need to nd the value of x50 such that F (x50 ) = 0.5:
F (x) =
F (x50 ) = 0.5
1 − e−x50 /2 = 0.5
e−x50 /2 = 0.5
1
2
1
x50 /2 = − log
2
−x50 /2 = log
x50 /2 = log (2)
x50 = 2 log (2) ≈ 1.386
R code:
qexp ( 0 . 5 , 1 / 2)
Page 4 of 6
9. (1 point) Consider the density given by 2xe−x for x > 0. What is the median?
a. 1.03
2
b. 0.83
c. 0.79
d. 0.24
e. 0.15
9.
b.
Solution:
First, we need to nd cumulative distribution function F (x):
x
Z
2
2te−t dt,
F (x) =
0
using substitution u(t) = −t2 and du = −2tdt we get:
Z
u(x)
u
Z
e du = −
F (x) = −
u(0)
0
−x2
2
0
eu du = [eu ]−x2 = e0 − e−x = 1 − e−x
2
Median is the value of x50 such that F (x50 ) = 0.5.
F (x50 ) = 0.5
2
1 − e−x50 = 0.5
2
e−x50 = 0.5
−x250
1
= log
2
x250 = log (2)
p
x50 = log (2) ≈ 0.833
Page 5 of 6
10. (1 point) Suppose h(x) is such that 0 < h(x) < ∞ for x = 1, 2, . . . , I . Then c · h(x) is a valid PMF (probability
mass function) when c is equal to what?
a.
b.
c.
1
PI
1+
PI
x=1
x=1
h(x)
h(x)
P
I
x=1
−1
h(x)
d. h(I)
c.
10.
Solution:
We know that 0 < c·h(x) < ∞. The second condition to function to be a valid PMF is, that
I
X
P
x
c·h(x) = 1.
c · h(x) = 1
x=1
c
I
X
h(x) = 1
x=1
c=
I
X
!−1
h(x)
x=1
Question:
1
2
3
4
5
6
7
8
9
10
Total
Points:
1
1
1
1
1
1
1
1
1
1
10
Score:
Page 6 of 6
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