Homework Solution: Probability and Statistics - Detailed Analysis

This document provides a detailed solution to a statistics and probability homework assignment. The solution covers various concepts including the analysis of critical failures using different beta values and time intervals, calculating the expected number of critical failures using the intensity function, and determining probabilities using the Poisson distribution. The assignment further delves into queuing systems, exploring birth and death processes, arrival and service rates, and the probabilities associated with customer waiting times. Additionally, the solution examines Markov chains, calculating transition probabilities, limiting distributions, and time reversibility. The analysis includes calculations related to shale layers, demonstrating the application of statistical methods in different contexts. Finally, the solution provides insights into the thickness of various layers, illustrating how these concepts can be used to solve real-world problems.

Question 1
Part a
Using the formula given and substituting in for different values of beta and t; we are able to see
that the number of critical failures is the same across the time interval (5, 15) for beta equivalent
to 1. The expected number of critical failures decrease as beta tends to 0 and time increases from
5 to 15. Also, the expected number of critical failures increase for beta increase for B>1, as time
increases from 5 to 15.
Part b
To get the expected number critical failure we find the sum of the intensity function for beta=0.5
and time=5, 6, 7,..., 15. Hence:
E ( t )=∑
t=6
15
(0.5 β t ( β−1 ) ), forβ=0.5
E ( t )= ( 0.5 β 5 ( β−1 ) ) + ( 0.5 β 6 ( β −1 ) ) +…+(0.5 β 15 ( β −1 )) , forβ=0.5
E(t)=0.91
The probability is given by
Exactly 2 hence x=2
Interval (5, 15)
Lambda= 0.5
p ( x ; λ )=¿ λx e− λ
x !
p ( 2; 0.5 ) =¿ 0.52 e−0.5
2 !
p ( 2; 0.5 ) =¿0.075816
Part C

The cumulative distribution function is given by
F ( x ; λ )=∑
x=0
∞ λx e− λ
x !
And the probability density function
p ( x ; λ )=¿ λx e− λ
x !
They both have elements of the exponential distribution and it's important to note that Poisson
distribution is a subset of binomial distribution.
Question 2
Part a
(i)
Birth takes place when the process moves from state t to t+1 (the customers enter the store and
join the line) and dead takes place when the process moves from state t to t-1(the customer exists
the store because he finds a given number of people in the line). The process is subject to birth
rate (rate of customer entry into store), and death rate (rate of customer leaving).
(ii)
The customers arrive at rate 20 per hour hence 1 person per 3 minutes
The service time per customer is 3 minutes
Hence, the probability that the first customer would have left the queuing system before the
second customer arrives is one or 100%.
Part b
(i)

The first person to arrive at the store will take 3 minutes (and has service time of 3 minutes) and
the second person to arrive will be the first person in the queuing system and this individual will
take 6 minutes to arrive.
(ii)
Due to the fact that service time is equivalent to arrival time means that there is really no queuing
indicating there will never be a situation where there are 4 people in the queuing system.
Part c
The waiting time for customer two is zero hence the first and the second have the same
probability distribution however it decreases from person three as the number of individual in the
queue increase i.e. the exit probability increase as P tends to 4 (P=number of people in queue)
Part d
(i)
The probability is one or 100% because the service time is equivalent to the service time hence
customers are serviced almost as soon as they arrive; hence, there is no one in the queue when a
customer arrives.
(ii)
Average customers in queue system=0.0003
(iii)
Mean time customers are served= 0.008 minutes
Mean time customers spend in the system=0.0508 minutes
Question 3
1
1.54.0

Part a
(i)
Probability for moving to
state
from to 0 to 1 to 2
State 0 0 0.8 0.2
State 1 0.25 0 0.75
State 2 0 1 0
(ii)
Probability that the shale layer will be greater than 1 meter is 0.25
Part b
The limiting distribution is given by
f(x)= {X ( t ) =t∗Qi , for Q 0= 25
100 Q1=180
100 Q2= 95
100
0 , otherwise
Part C
Let's check the forward probability
0 to 2= 1.55
Checking the reverse probability
2 to 0=1.25
0 20.5 0.5
1.0
0

Hence the Markov chain is not time reversible because the forwards and reverse probabilities are
not equivalent.
Part D
For the system to move from shale to shaly sandstone to purely sandstone and back to shale it
will have moved in terms of "t" t=3 were the three represents 3 meters. Hence we deduct 1 meter
for the first shale layer and we are left with 2 meters that cover the distance between two shale
layers.
Hence, the thickness of layer of pure sandstone and shaly sandstone together is 2 meters