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The final exam for a university-level course in probability and statistics. It includes 9 problems covering topics such as poisson processes, cumulative distribution functions, probability density functions, conditional probability, bayes' rule, and expected values. Students are required to solve each problem and show their work.
Typology: Exams
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July 1, 2011
Show all your work. Answers out of the blue and without any supporting work may receive no credit even if they are right! Write clearly. You may use a calculator.
Box your final answers. No notes allowed.
(a) No arrivals are simultaneous. (b) The number of arrivals in disjoint time intervals are independent. (c) The number of arrivals in an interval is Poisson distributed with mean proportional to the length of the interval.
(a) cumulative distribution function The cdf of the random variable X is F (x) = P (X ≤ x).
(b) probability density function A pdf is a function f such that f (x) ≥ 0 for all x and
−∞ f^ (x)dx^ = 1.
(c) conditional probability of A given B The conditional probability of A given B is P (A|B) =
P (B) if^ P^ (B)^ >^ 0.
P (A) = P (X ≥ 2) =^48 =^12 ,
P (B) = P (X 6 = 0, X 6 = 3) =^68 =^34 , and
P (AB) = P (X = 2) =
So, P (AB) = 38 = 12 · 34 = P (A)P (B), so that the events A and B are independent.
P (A) = P (A|B 1 )P (B 1 ) + P (A|B 2 )P (B 2 ) + P (A|B 3 )P (B 3 ) =
P (X ≤ x) =
volume of sphere of radius x volume of sphere of radius r =
4 3 πx^3 4 3 πr^3
x^3 r^3.
So, the probability density function of X is f (x) = (^) ddx [P (X ≤ x)] = 3 x
2 r^3 for 0^ < x < r, and
f (x) =
{ (^3) x 2 r^3 if 0^ < x < r 0 otherwise
Then, E(X) =
∫ (^) r 0 xf^ (x)dx^ =^
∫ (^) r 0 x^ ·^
3 x^2 r^3 dx^ =^
3 r^4 4 r^3 =^
3 4 r.
k=
k(0.99)k−^1 (0.01) = 0.^01 (1 − 0 .99)^2
k=
k^2 (0.99)k−^1 (0.01) =
(1 − 0 .99)^3 = 19900,^ and so var(X) = 19900 − (100)^2 = 9900.
Alternatively, you can assume that X has an exponential distribution. (This will only be approx- imately true, but I gave credit for solving the problem by this method.) Then E(X) = 100 and var(X) = 10000.
6
)k− 2 , where we have k − 2 instead of k − 1 since the first roll can be anything) and that the kth roll is the same as the previous roll (with probability 16 ), so that P (X = k) =
k− (^2 ) 6 for^ k^ = 2,^3 ,.... Then
E(X) =
k=
k
)k− 2 1 6
=^1 6
k=
(k + 1)
)k− 1
k=
k
)k− 1 +^16
k=
)k− 1
