Download - Math Biostatistics Boot Camp 1
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Mathematical Biostatistics Boot Camp 1
Quiz
Week 1
1. (1 point) What is P (A B) always equal to?a. 1 P (AC BC)b. 1 P (AC BC)c. 1 P (AC)P (BC)d. P (AC BC)
1.
a.
Solution:
P (A B) = 1 P ((A B)C) = 1 P (AC BC)2. (1 point) Which of the following are always true about P (
ni=1Ei)? (Check all that apply.)It is smaller than or equal to
ni=1 P (Ei).
It is equal to ni=1 P (Ei). It is smaller than maxi P (Ei).It is larger than or equal to mini P (Ei).It is larger than or equal to maxi P (Ei).
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
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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 75thpercentile? Express your answer to two decimal places.
a. 0.25
b. 0.50
c. 0.75
d. 0.10
e. 0.90
4.
c.
Solution:
The cumulative distribution function for f(x) is:
F (x) =
x0
f(t) dt =
x0
1 dt = [t]x0 = x 0 = xWe 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
1x2 for 1 < x
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6. (1 point) What is the quantile p from the density ex (1 + ex)2?
a. log ((1 p) /p)b. p/ (1 p)c. (1 p) /pd. 1/ (1 + ex)
e. log (p/ (1 p))
6.
e.
Solution:
First, we need to nd cumulative distribution function:
F (x) =
x
et
(1 + et)2dt,
using substitution u(t) = 1 + et and du = et dt we get:
F (x) =
u(x)u()
1u2
du =
1+ex
1u2
du =
[1
u
]1+ex
=1
1 + ex=
exex
ex (ex + 1)=
ex
ex + 1
Now we want to nd a value of xp such that F (xp) = p.
F (xp) = p
exp
exp + 1= p
exp = p (exp + 1)
exp = pexp + p
exp pexp = pexp (1 p) = p
exp =p
1 pxp = log
(p
1 p)
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7. (1 point) Suppose that a density is of the form cxk for some constant c > 1 and 0 < x < 1. What is the valueof c?
a. k + 2
b.
1k
c. 2
d.
1k+1
e. k 1f. k
g. k + 1
7.
g.
Solution: 10
cxkdx = c
[xk+1
k + 1
]10
=c
k + 1
[xk+1
]10=
c
k + 1(1 0) = c
k + 1must be equal to 1, so:
c
k + 1= 1 c = k + 1.
8. (1 point) Suppose that the time in days until hospital discharge for a certain patient population follows a
density f(x) = 12ex/2for 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):
F (x) =
x0
1
2et/2 dt =
[et/2
]x0=[et/2
]0x= e0 ex/2 = 1 ex/2
Now we need to nd the value of x50 such that F (x50) = 0.5:
F (x50) = 0.5
1 ex50/2 = 0.5ex50/2 = 0.5
x50/2 = log(1
2
)x50/2 = log
(1
2
)x50/2 = log (2)
x50 = 2 log (2) 1.386
R code:
qexp ( 0 . 5 , 1/2)
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9. (1 point) Consider the density given by 2xex2
for x > 0. What is the median?
a. 1.03
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):
F (x) =
x0
2tet2
dt,
using substitution u(t) = t2 and du = 2tdt we get:
F (x) = u(x)u(0)
eu du = x20
eu du = [eu]0x2 = e
0 ex2 = 1 ex2
Median is the value of x50 such that F (x50) = 0.5.
F (x50) = 0.5
1 ex250 = 0.5ex
250 = 0.5
x250 = log(1
2
)x250 = log (2)
x50 =log (2) 0.833
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- 10. (1 point) Suppose h(x) is such that 0 < h(x)