Statistics Homework Assignment: Probability Solutions

This homework assignment solution addresses several statistical problems. It begins with calculating the expected number of arrivals and standard deviation based on provided frequency data. The solution then analyzes stock investment strategies, computing expected returns and standard deviations for two stocks under different economic conditions, recommending an investment choice based on the analysis. Further, the assignment covers probability calculations for binomial distributions related to stock price increases. Finally, it explores Poisson distribution problems, determining probabilities for the number of website visitors, providing comprehensive solutions for each question.

HOMEWORK ASSIGNMENT 3 1
Homework Assignment 3
By
(Name of student)
(Institutional Affiliation)
(Date of Submission)
Question

HOMEWORK ASSIGNMENT 3 2
Solution
X P(X) P(X) E[X]=Ʃ
X*P(X)
Arrivals Frequency Freq
. X*P(X) x^
2 x^2*P(X) Varianc
e
0 14 0.07 0 0 0 3.14
1 31 0.15
5 0.155 1 0.155
2 47 0.23
5 0.47 4 0.94
3 41 0.20
5 0.615 9 1.845
4 29 0.14
5 0.58 16 2.32
5 21 0.10
5 0.525 25 2.625
6 10 0.05 0.3 36 1.8
7 5 0.02
5 0.175 49 1.225
8 2 0.01 0.08 64 0.64
200 1
Mean
= 2.9 11.55
a. Compute the expected number of
arrivals per minute. = 2.9
b. Compute the standard deviation =1.77
2
Question
Solution
a. Expected return for Stock X and Stock Y
E(x) = Mx = (0.1) (-50) + (0.2) (20) + (0.45) (100) + (0.25) (150)
= 81.5
E(y) = My = (0.1) (-100) + (0.2) (50) + (0.45) (130) + (0.25) (200)
= 108.8
Mx + My = 81.5 + 108.8
= 190
b. Standard Deviation for Stock X and Stock Y
σ x2 = (-50 – 81.5)2 (0.1) + (20 – 81.5)2 (0.2) + (100 – 81.5)2
(0.45) + (150 – 81.5)2 (0.25)
= 3829
σ x = √ 3,829 = 61.88

HOMEWORK ASSIGNMENT 3 3
c. Would you invest in Stock X or Stock Y? Explain.
I would invest in stock y, because of having fast growth with the 0.2
probability
Question
Solution
Given that: n = 5; x = 5 and p = 0.5
The probability that a stock will show an increase in its closing price on five
consecutive days is,
P (X = 5) = ( 5
5 )(0.5)5(1-0.5)5-5
= 0.0312
Question
Solution
P (X = 6) = ( 5
6 )(0.2)6(1-0.2)6-5
= 0.0016
a. At least four questions correct?
P (X = 4) = ( 5
4 )(0.2)4(1-0.2)5-4
= 0.0272
b. No questions correct?
c. No more than two questions correct?

HOMEWORK ASSIGNMENT 3 4
Question
Solution
Mean of Expected number of events of interest: 4
Probabilities Table for Poisson
Distribution
X P(X) P(<=X) P(<X) P(>X) P(>=X)
0 0.0183 0.0183 0 0.9817 1
1 0.0733 0.0916 0.0183 0.9084 0.9817
2 0.1465 0.2381 0.0916 0.7619 0.9084
3 0.1954 0.4335 0.2381 0.5665 0.7619
a. Zero new visitors will arrive at the website?
Thus P(X = 0) = 0.0183
b. Exactly one new visitor will arrive at the website?
Thus P(X = 1) = 0.0733
c. Two or more new visitors will arrive at the website?
Thus P(X ≥ 2) = 0.9084
d. Fewer than three new visitors will arrive at the website?
Thus P(X < 3) = 0.2381