Description

CENG222
Statistical Methods for Computer Engineering
Homework 2

Student Information
Full Name : Yavuz Selim Yesilyurt
Id Number : 2259166
Answer 3.10
P(X = 1) × P(Y = 2) = 0.2 × 0.2 = 0.04
P(X = 0) × P(Y = 2) = 0.2 × 0.6 = 0.12
P(X = 0) × P(Y = 1) = 0.2 × 0.6 = 0.12
Adding them up we get = 0.28
Answer 3.15
a)
To compute the probability of at least one hardware failure happening in any lab we will subtract the possibility of not happening of any failure of the hardware from total probability (1) and then we will reach to the required probability. We have from table P(0,0) = 0.52 and subtracting, 1 − 0.52 = 0.48
b)
To determine if X and Y are independent we need to make sure that PXY (x,y) = PX(x).PY (y) for each marginal/joint distribution values. If we bump into any inequality in this calculations then we can say X and Y are dependent. So we start computing with P(X = 0) and P(Y = 0), P(0,0) = 0.52, the marginal distribution of X at 0 is 0.76 and the marginal distribution of Y at 0 is 0.72, 0.76×0.72 = 0.54, but 0.54 6= 0.52 therefore X and Y are not independent, they are infact dependent.
Answer 3.19
a)
Let’s compute the Expectation and Variance values of X (a profit made on 1 share of A) and Y (a profit made on 1 share of B):
E(X) = (−2)(0.5) + (2)(0.5) = 0 and V ar(X) = (−2)2(0.5) + (2)2(0.5) − 02 = 4 E(Y ) = (−1)(0.8) + (4)(0.2) = 0 and V ar(Y ) = (−1)2(.8) + (4)2(0.2) − 02 = 4
Considering X and Y are independent, Expectation and variance values for 100 shares of A;
E(100X) = 100E(X) = 0 and V ar(100X) = 1002V ar(X) = 40.000 b)
Expectation and variance values for 100 shares of B;
E(100Y ) = 100E(Y ) = 0 and V ar(100Y ) = 1002V ar(Y ) = 40.000 c)
Expectation and variance values for 50 shares of A and 50 shares of B;
E(50X + 50Y ) = 50E(X) + 50E(Y ) = 0
V ar(50X + 50Y ) = 502V ar(X) + 502V ar(Y ) = 20.000
Answer 3.26
a)
We need to find the probability P(X ≥ 5), where X is the number of damaged files, out of 20 files that is being checked. This is the number of successes in 20 Bernoulli trials, therefore, X has Binomial distribution with parameters n = 20 and p = 0.2. From Table A2,
P(X ≥ 5) = 1 − F(4) = 1 − 0.630 = 0.370
b)
Let X be the number of files tested until 3 undamaged files are found. It is a number of trials needed to see 3 successes, hence X has Negative Binomial distribution with k = 3 and p = 0.8. We need to find:
P(X > 5) = 1 − F(5)
However, there is no table of Negative Binomial distribution in the Appendix on the book. We can convert X into a new random variable Y with Binomial Distribution and solve it as such. We will convert X into Y as the following:
P(X > 5) = P(more than 5 trials needed to get 3 successes)
= P(5 trials are not sufficient)
= P(there are fewer than 3 successes in 5 trials)
= P(Y < 3)
where Y is the number of successes (undamaged files) in 5 trials, which is a Binomial variable with parameters n = 5 and p = 0.8. From Table A2,
P(X > 5) = P(Y < 3) = P(Y ≤ 2) = F(2) = 0.058