# Practice Final Solutions

advertisement ```CS3130: Probability and Statistics for Engineers
Practice Final Solutions
1. (a) Probability of being male, given that you have autism spectrum disorder.
P (M |A) = 0.8 (given in problem).
(b) Probability of having autism spectrum disorder, given that you are a male. Using
Bayes’ Rule:
P (A|M ) =
0.8 &times; 0.006
P (M |A)P (A)
=
= 0.0096
P (M )
0.5
(c)
P (M c ∩ A) = P (A − M )
= P (A) − P (A ∩ M )
= P (A) − P (M |A)P (A)
= 0.006 − 0.8 &times; 0.006
= 0.0012
definition of set minus
difference rule
multiplication rule
2. (a) Probability of success on first attepmt: P (X = 1) = p. So, probability that it isn’t
on the first attempt is: P (X &gt; 1) = 1 − P (X = 1) = 1 − p.
(b) Using the pmf for the Geometric distribution and the formula given in the problem,
this is:
n
n
X
X
P (X ≤ n) =
P (X = i) =
(1 − p)i−1 p = 1 − (1 − p)n
i=1
i=1
(c) This is just the complement of (b), so we get
P (X &gt; n) = 1 − P (X ≤ n) = 1 − (1 − (1 − p)n ) = (1 − p)n
P (A ∩ B)
,
P (B)
with A = {X = k + n} and B = {X &gt; n}. Notice that if X = k + n is true, then
X &gt; n is also true, so A ∩ B = A in this case. We get:
(d) Here we apply the standard formula for conditional probability: P (A|B) =
P (X = k+n | X &gt; n) =
P (X = k + n)
(1 − p)k+n−1 p
=
= (1−p)k−1 p = P (X = k)
P (X &gt; n)
(1 − p)n
3. (a) The cdf must be equal to one at the end of its range (at x = π). That is,
1 = FX (π) = k(π − sin(π)) = kπ
Solving for k gives us k = π1 .
(b) For x in the range 0 ≤ x ≤ π, we get
d
FX (x)
dx
d 1
=
(x + sin(x))
dx π
1
= (1 + cos(x))
π
For all other values of x, fX (x) = 0.
fX (x) =
(c)
π π
= FX
P X≤
2
2
π 1 π
=
+ sin
π 2
2
1 1
= +
2 π
(d)
1
E[X] =
π
Z
π
x(1 + cos(x))dx
0
4. (a) Because X and Y are uncorrelated Var(X + Y ) = Var(X) + Var(Y ) = 2σ 2 .
(b) Note from the definition of correlation that
p
Cov(X, Y ) = Var(X)Var(Y )ρ(X, Y ) = −σ 2 .
This gives
Var(X + Y ) = Var(X) + Var(Y ) + 2Cov(X, Y )
= 2σ 2 − 2σ 2
=0
(c) For part (a) it is “not enough information”. Uncorrelated does not imply independent. For part (b) they are “dependent”. Correlated (either positive or negative)
variables are always dependent.
5. First, for a 99% confidence interval, we are using the level α = 0.01. This means we
want to use zα/2 = z0.005 = 2.58. Now, the confidence interval is (ln , un ):
0.25
σ
= 0.3695
ln = x̄n − z0.005 √ = 0.305 − 2.58 &times;
10
n
σ
0.25
= 0.2405
un = x̄n + z0.005 √ = 0.305 + 2.58 &times;
10
n
6. (a) Assuming the monthly sales come from a Gaussian distribution, N (&micro;, σ 2 ).
H0 : &micro; = 800
H1 : &micro; &gt; 800
(average monthly sales are equal to 800 units)
(average monthly sales are greater than 800)
(b) You should use the critical value F −1 (0.95) = 2.35 because you are doing a onesided test to the right.
(c) Using the T statistic formula:
t=
x̄n − &micro;0
805 − 800
5
√ =
√
= =1
5
sn / n
10/ 4
(d) No, the t statistic is not greater than our critical value.
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