Probability and Statistics Assignment Solutions

This assignment covers various topics in probability and statistics, including mutually exclusive events, normal distribution, reliability of manufacturing processes, and more. The solutions are provided with step-by-step calculations and explanations.

Running head: ASSIGNMENT 1
Assignment
Student Name
Institution

ASSIGNMENT 2
Question 11
Consider a simple experiment that is replicated a number of times. Consider four events A,
B, C, D. In this experiment events B and D are not independent and the only mutually
exclusive events are events A and B and events C and D. If the probability of event D
occurring is 0.69, the probability of event B occurring is 0.35 and the probability of event B
given D occurring together is 0.58, then what is the probability of event B AND D
occurring, P(B AND D)? Provide your answer to 3 decimal places.
8 points
A and B are mutually exclusive
Thus
P (A and B) = 0
C and D are mutually exclusive
Thus
P(C and D) = 0
B and D are independent
P (D) =0.69
P (B) = 0.35
P (B|D) =0.58
Using conditional probability formula;
P (B|D) P (B∧D)
P(D)
0.58 = P (B∧D)
0.69

ASSIGNMENT 3
P( B∧D) = 0.69*0.58 = 0.4
Question 12
A full deck of playing cards (excluding Jokers) is well shuffled and a card drawn at
random from the pack. Its characteristic (e.g. colour and number) is noted and the card is
put back into the pack and the pack re-shuffled. This process is repeated a further seven
times. Expressing probability as a number between 0 and 1, calculate to 3 decimal places
(and never specify units or enter text of any kind):
The probability that there are exactly two cards with the number 9 or 10 on it is
P (number 9) or P (number 10) = 4
52 + 4
52 = 0.154
The probability that there are exactly five cards with the Jack, a Queen, an Ace or King on
it is
P (Jack or Queen or Ace or King) = 4
52 + 4
52+ 4
52 + 4
52 = 0.308
The probability that there are neither four nor more cards showing neither an ace nor a
two on it is
P (ace) = 4
52
P (a two) = 4
52
P (neither) = 1- ( 4
52+ 4
52 ¿ = 0.846
the mean number of cards showing neither an ace nor a queen is
P (ace) = 4
52
P(queen ) = 4
52

ASSIGNMENT 4
P (neither ace nor queen) = 1- ( 4
52+ 4
52 ¿ = 0.846
with standard deviation
X P(x) X2P(X)
1 0.0769 0.0769
2 0.0769 0.3076
sum 0.3845
Variance = 0.3845
The probability that there are two or less cards with the numbers 5 or 6 or 7 on it is
P(5 or 6 or 7) = ( 4
52+ 4
52 + 4
52 ¿ *2 = 0.462
20 points
Question 13
The density of a chemical solution is normally distributed with a median of 0.04. Suppose
also that 80% of all chemical solutions have a density of 0.05 or less.
Where a probability is asked for, do not express this as a percentage. Use fully un-rounded
values in all calculations, but provide answers to the following questions to 3 decimal places
and do not specify units or enter text of any kind:
The normal distribution parameter ? is equal to
probability , P = 0.8
Density = 0.05
For normal distribution, median = mean = 0.04
Z0.20 = 0.842
Z = x−μ
σ = x−μ
σ

ASSIGNMENT 5
0.842= 0.05−0.04
σ
Thus
σ = 0.012
N( μ , σ) = N(0.04,0.012)
The probability that the density is less than or equal to 0.03 is
P(x≤ 0.03) = P(Z≤ x−μ
σ ) = 0.03−0.04
0.012 = -0.833
Using standard normal tables, the associated p-value at z = -.0833
P(density ≤ 0.03)= 0.203
The probability that the density is between 0.03 and 0.05 is
P(x¿0.03) = P(Z¿ x−μ
σ ) = 0.03−0.04
0.012 = -0.833
Using standard normal tables, the associated p-value at z = -.0833
P(density=0.03) = 0.203
P(x¿0.05) = P(Z¿ x−μ
σ ) = 0.05−0.04
0.012 = 0.833
Using standard normal tables, the associated p-value at z = 0.833
P(density= 0.05)= 0.80
P(density is between 0.03 and 0.05) = 0.80-0.203 = 0.597
A density level of 0.055 has a standardised value equal to
P(x¿0.055) = P(Z ¿ x−μ
σ ) = 0.055−0.04
0.012 = 1.25
Using standard normal tables, the associated p-value at z = 1.25
P(density = 0.055) = 0.894
The probability that the density is more than 0.055 is

ASSIGNMENT 6
P(density > 0.055) = 1- P(density =0.055)
= 1- 0.894
= 0.106
A company needs to be able to guarantee to its customers that no more than 5 in 100 of its
solutions will have a density above 0.045. With a mean of 0.04, the most the standard
deviation can be to achieve this is
Probability = 5/100 = 0.05
Mean = 0.04
X = 0.045
P(x¿0.045) = 1- P(Z ¿ x−μ
σ ) = 0.045−0.04
0.012 = 0.416
P(Z¿ x−μ
σ ) = 0.045−0.04
0.012 = 0.416
Using standard normal tables, the associated p-value at z = 0.416
P(density < 0.045) = 0.663
P(density > 0.045 ) =1- 0.663 = 0.337
Question 14
Consider a manufacturing process composed of three stages - (R1, R2 and R3). Stage one
consists of 2 machines arranged in a doubly redundant sub system. Stage two has 3
machines that work in series, meaning that this stage operates successfully only if all the
machines operate. Stage three is made up of 4 machines and only operates successfully if at
least 2 machines operate. The three manufacturing stages are connected in series to
produce the final product, meaning that they must all work properly to produce a final
product.

ASSIGNMENT 7
Between two scheduled maintenances, each machine may fail independently of the others,
with the following failure probabilities:
R1
0.11 0.18
R2
0.30 0.14 0.21
R3
0.38 0.21 0.29 0.22
Use this information to answer the following questions to 3 decimal places (expressing
probability as a number between 0 and 1 and do not specify units or enter text of any
kind ):
1. The reliability of sub system stage 1 is
Reliability
P( machine 1 or machine 2 ) = 0.11 + 0.18 = 0.29
reliability = 1- P(failure) = 1- 0.29 = 0.71
2. The reliability of sub system stage 2 is
The machines in stage 2 are dependent
P (machine 1 and machine 2 and machine 3 failing) = 0.30+0.14+0.21 = 0.65
Reliability = 1- P (failure) = 1- 0.65 = 0.35
3. The reliability of sub system stage 3 is
P(production occurs| 2 machines work) =
The machines in stage 2 are dependent
P (Failure ) = 0.38 *0.21 *0.29 * 0.22

ASSIGNMENT 8
Reliability = 1- P (failure) = 1- 0.00509 = 0.99491
4. At stage 3 the probability of 1 machines breaking down between scheduled maintenances
is
P( one machine breaking ) = P(machine 1) or P(machine 2) or P(machine 3) or P(machine 4)
= 0.38+0.21+0.29+0.22
= 1
5. The probability of the manufacturing process as a whole failing between scheduled
maintenances is
P (Whole process failing) = 0.29+0.35 +0.00509 = 0.6451