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CS 3341 HW # 6 (SOLUTION)
1. Travelbyus is an internet-based travel agency wherein the customers can see videos of the
cities they plan to visit. The number of hits daily is normally distributed with a mean of 1,000
and a standard deviation of 240.
(a) What is the probability of getting fewer than 900 hits?
(b) The website of this company has a limited bandwidth, which is measured in terms of the
number of hits the site can handle. How large a bandwidth should Travelbyus have in
order to handle 99% of the daily traffic?
X = # hits at the website. Given that, X ~ Normal (µ = 1000, σ2 = 2402). Let Z = (X – µ)/σ
denote a N(0, 1) random variable.
(a) P (X < 900) = P (Z < (900 – 1000)/240)
= P( Z < -0.42) = 0.34 [using normal table and symmetry].
(b) We want k such that, 0.99 = P(X ≤ k ) = P(Z ≤ (k – 1000)/240).
Now, the normal table suggests, (k – 1000)/240 = 2.33, which means
k = 2.33(240) + 1000 = 1559.
2. Suppose that the CPU time for an execution of a particular software package has gamma
distribution with mean 5 seconds and standard deviation of 2.5 seconds. Find the probability
that it will take more than 10 seconds for an execution of this software.
Let X = execution time.
Given that X ~ Gamma (r, λ), and E(X) = r/λ = 5, Var(X) = r/λ2 = 2.52.
Solving these two equations, we get r = 4, and λ = 0.8.
Now, P( X > 10) = P[ Gamma (r = 4, λ = 0.8) > 10]
= P[Poisson (0.8*10 = 8) < 4]
(using the Gamma-Poisson formula)
= 0.042
(using Poisson PMF)
3. A program consists of 3 blocks. Compilation time for each block is exponentially distributed
with the same parameter
sec-1, and it is independent of the compilation of other blocks.
Compute the probability that the whole program compiles in less than 2.5 seconds.
Let T = total compilation time = sum of 3 independent Exponential (λ = 1). Hence, T ~
Gamma (r = 3, λ =1).
P(T < 2.5) = 1 – P[Gamma(r =3, λ =1) ≥ 2.5]
= 1 – P[Poisson(2.5*1 = 2.5) < 3]
= 1 – 0.544 = 0.456
(using Gamma-Poisson formula)
(using Poisson PMF)
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4. Recall the pick-six lottery example from the class. In this Lottery, a person picks six numbers
from the numbers 1 through 50 with no repetitions and pays $1.00. On Wednesday evenings,
the Texas State Lottery Commission televises one of their employees randomly picking six
balls without replacement, each with a number from 1 to 50 on it, from a large hopper. The
player is paid if his/her number matches with the selected balls for three or more numbers. In
particular, for matching three balls one receives $3, for matching four balls one receives $89,
for matching five balls one receives $1,268 and for matching all six balls one received
$4,000,000. Let X = # of matches a customer has with the selected balls. The PMF of X is
given in the following table:
X
0
1
2
3
4
5
6
Total
P(X = x)
0.44422536452
0.41005418264
0.12814193207
0.01666886921
0.00089297514
0.00001661349
0.00000006293
1.00000000000
How much do you expect to win for the purchase of a $1 lottery ticket? Would you want
to purchase this ticket?
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Let g(X) = Amount won if X balls are matched. We want to find Eg(X) = ∑ g ( x) P( X = x)
x =0
X
0
1
2
3
4
5
6
Total
P(X = x)
0.44422536452
0.41005418264
0.12814193207
0.01666886921
0.00089297514
0.00001661349
0.00000006293
1.00000000000
g(x)
$0
$0
$0
$3
$89
$1,268
$4,000,000
g(x)*P(X = x)
$0
$0
$0
$0.05
$0.08
$0.02
$0.25
0.40 = Eg(X)
Thus, for every $1 bet we can expect to get a return of only 40 cents. I will not want to bet.
Would you?
5. The number of errors in each of 200 files has a Poisson distribution with 1.4 errors per file on
average. What is the probability that the total number of errors exceeds 500? Assume the
errors in different files are independent.
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Let X = # errors in a file.
Given that, X ~ Poisson (λ = 1.4). So that E(X) = 1.4 and Var(X) = 1.4.
T = total # errors in n = 200 files. From CLT, approximately T ~Normal (µ = n E(X) = 280,
σ2 = n Var(X) = 280).
Let Z = (X - µ)/σ denote a N(0, 1) random variable.
P(T > 500) = P(Z > (500 – 280)/16.73) = P(Z > 13.15) = 0.
6. The homework consists of 50 independent problems. On the average, it takes 5 minutes to
solve a problem, with a standard deviation of 2 minutes. Find the probability that the
homework will be completed in less than 3 hours.
X = amount of time (in minutes) it takes solve a HW problem. Given that E(X) = 5 and
Var(X) = 4.
T = total time to complete the HW with n = 50 independent problems. From CLT, T ~
Normal (µ = n E(X) = 250, σ2 = n Var(X) = 200).
P(T < 180) = P(Z < (180-250)/14.14) = P(Z < - 4.95) = 0.
7. A keyword search program lists the files that contain a given keyword. If it runs through 200
files, and each file contains the keyword with probability 0.25, independently of other files,
compute the probability that least 50 files will be listed.
T = total # files listed. Since this is a binomial experiment, we have that T ~ Binomial (n =
200, p = 0.25).
Also since n is large and p is not too small or too big, approximately, T ~ Normal (µ = n p =
50, σ2 = n p (1 –p) = 37.5).
As before, P(T ≥ 50) = P(Z ≥ 0) = 0.5 (using symmetry).
8. A company has two manufacturing plants for producing integrated circuits. Let X and Y
denote the daily number of defective circuits produced at plants 1 and 2, respectively.
Assume that X and Y are independent. Suppose E(X) = 23, SD(X) = 5.5, E(Y) = 27 and
SD(Y) = 4.7. Find the expected value and the SD of the total number of defective circuits
produced by this company in a day. What is the likely range for the daily total number of
defectives?
Let T = X+ Y = daily total of defectives.
E(T) = E(X + Y) = E(X) + E(Y) = 23 + 27 = 50.
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Var(T) = Var(X+Y) = Var(X) + Var(Y) + 2 Cov(X, Y).
Here, Cov(X, Y) = 0 since X and Y are independent.
So, Var(T) = 5.52 + 4.72 = 52.75, which means SD(T) = 7.26
Likely range = [50 – 2*7.26, 50 + 2*7.26] = 35.48 to 64.52
9. The time (in minutes) it takes to install a certain software package is a continuous random
variable with the density
(a) Find the probability that it takes less than 30 seconds to install the package
(b) Find the expected installation time.
(a) 30 sec = 0.5 min
(b)
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