Download 1 PRACTICE MIDTERM I (CS 3341) 1. Weights of elephants are

PRACTICE MIDTERM I
(CS 3341)
1. Weights of elephants are approximately distributed as Exponential random variables with a
mean of 3 tons. One hundred elephants are being transported on a ship that has a cargo limit
of 250 tons. What is the chance that this ship will sink?
Let X = # weight of an elephant. Given that, X ~ Exponential (λ = 1/3). So that E(X) =3 and
Var(X) = 9.
T = total weight of n = 100 elephants. Now, since n is large, using CLT, approximately T ~
Normal (µ = n E(X) = 300, σ2 = n Var(X) = 900).
Let Z = (T – µ)/σ denote a N(0, 1) random variable.
P(Ship will sink) = P(T > 250) = P[Z > (250 – µ)/ σ]
= P(Z > (250 – 300)/30)
= P(Z > – 1.67) = 0.953 (using the normal table and the symmetry).
2. Because of the relatively high interest rates, most consumers attempt to pay off their credit
card bills promptly. However, this is not always possible. An analysis of the amount of
interest paid monthly by a bank’s Visa cardholders reveals that the amount is normally
distributed with a mean of $27 and standard deviation of $7.
(a) What proportion of the bank’s Visa cardholders pay more than $30 in interest?
(b) What interest payment is exceeded by only 20% of the bank’s Visa cardholders?
X = amount of interest paid. Given that, X ~ Normal (µ = 27, σ2 = 72). Let Z = (X – µ)/σ
denote a N(0, 1) random variable.
(a) P(X > 30) = P(Z > (30 – 27)/7) = P( Z > 0.43) = 1 – 0.67 = 0.33 (using the normal
table)
(b) We want k such that, 0.20 = P(X > k), or equivalently k such that
0.80 = P(X ≤ k) = P(Z ≤ (k-27)/7).
Now, the normal table suggests, (k – 27)/7 = 0.84, which means k = 7(0.84) + 27 =
32.88.
3. Two vendors supply a large company with computers. Employees of this company are asked
to rate them. Vendor A supplied 60% of the computers, and vendor B supplied the rest.
Overall, 10% of all the computers received a Poor rating, 40% received a Good rating, and
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the rest received an Excellent rating. Also, 5% of B's computers were rated Poor, 45% of B's
computers were rated Good, and the rest of B's computers were rated Excellent.
(a) What proportion of A’s computers were rated Poor?
(b) Are the events “a computer came from vendor B” and “a computer was rated
Excellent” independent? Why or why not?
We can solve this problem by thinking in terms of probability. But sometimes, as we saw
in the class, it is may be easier to think in terms of numbers.
So, let us suppose that total # of computers = 1000.
Given that, # A = 600, # B = 1000 – 600 = 400,
Overall, # Poor = 10% of 1000 = 100, # Good = 40% of 1000 = 400, # Excellent = 500
Also, # Poor |B = 5 % of 400 = 20, # Good |B = 45% of 400 = 180, # Excellent |B = 200
(a) # Poor | A = Overall 10 poor – 20 B’s poor = 80. So, P(poor | A) = 80/600 =
13.33%
(b) Yes, because P(Excellent | B) = 0.50 = P(Excellent).
4. About twenty percent of users do not close Windows-NT properly. Suppose that WindowsNT is installed on a public PC that is used by random people in a random order.
(a) On the average, how many people use this PC until someone closes Windows-NT
properly?
(b) What is the probability that at least 6 of the next 9 users will close Windows
properly?
(a) Let X = # users until someone closes properly. This represents the number of Bernoulli
trials to get the first success, where success = “close properly”.
Hence, X ~ Geometric (p = 0.80), which implies that E[X] = 1/p = 1/0.80 = 1.25
(b) Let Y = # users among the 9 users who close properly. This represents the number of
success in 9 trials.
Hence Y ~ Binomial (n = 9, p = 0.80), which implies that
P[Y ≥ 6] = 1 – P[Y ≤ 5] = 1 – 0.0856 = 0.9144 [from Table C.1].
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5. In Thunder-city, the probability of a thunderstorm on any day is 0.6. During a thunderstorm,
the number of traffic accidents has Poisson distribution with parameter 10. Otherwise, the
number of traffic accidents has Poisson distribution with parameter 4. If there were 7
accidents yesterday, what is the probability that there was a thunderstorm?
Let T = Thunderstorm and X = Poisson # of accidents. From the Bayes’ theorem,
P(T | X = 7) =
=
P(T ∩ X = 7)
P( X = 7 | T ) P(T )
=
P ( X = 7)
P( X = 7 | T ) P(T ) + P( X = 7 | no T ) P(no T )
(e
−10
(e
−10
)
10 7 / 7% 0.6
)
(
)
10 7 / 7% 0.6 + e −4 4 7 / 7% 0.4
= 0.6941
6. Amount of load of a network, X, is a continuous random variable that takes values between 0
(no load) and 1 (full load). Let its PDF be
f(x) = 3 x2, when 0 < x <1; and f(x) = 0, otherwise.
(a) Find P(X ≤ 0.2)
(b) What is the average load?
0.2
(a) P( X ≤ 2) = 3 ∫ x 2 dx =0.2 3 = 0.008
0
1
1
0
0
(b) E ( X ) = 3∫ x.x 2 dx = 3∫ x 3 dx =3 / 4 = 0.75
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