This formatting template has been developed to provide guidance to students when preparing this assignment. It is difficult to mark the assignment if Excel components are inserted as appendices in the assignment. This template should be used to ensure the Excel components are included in the correct places throughout the assignment.
Question 1
 Typed or handwritten answers to be inserted here.
 Copy table and complete with typed or handwritten answers.
 Typed or handwritten answers to be inserted here.
 Cut and paste the pivot table from Excel here.
 and f. Typed or handwritten answers to be inserted here.
 i. Cut and paste the pivot table from Excel here.
 Cut and paste the bar chart from Excel after the pivot table.
 Typed or handwritten answer for this part to be inserted after the bar chart.
 i. Cut and paste the descriptive statistics table from Excel here.
Typed or handwritten answer for parts ii., iii. and iv. to be inserted after the descriptive statistics table.
 Cut and paste the frequency distribution from Excel here.
Cut and paste the histogram from Excel after the frequency distribution.
 Typed or handwritten answers to be inserted here.
Question 2
Handwritten answers for all parts to be inserted here.
Question 3
Handwritten answers for all parts to be inserted here.
Question 4
Handwritten answers for all parts to be inserted here.
Question 5
 Handwritten answer to be inserted here.
Question 6
 Handwritten answer to be inserted here.
Question 7
Handwritten answers for all parts to be inserted here.
Part a
 Variables from the dataset, which are Qualitative and Normal, are Suburb, Address, Type, Result and Agent.
 There are no variables in the dataset, which are both Qualitative and Normal.
 Variable from the dataset, which is Quantitative and ratio is price.
 The variable from the dataset, which is both Quantitative and Ordinal, is bedroom.
Part b
Result Code 
Overall Outcome 
PI, NB, VB 
Property not sold 
SP, PN 
Property sold prior to auction 
S, SN 
Property sold at auction 
SA, SS 
Property sold after auction 
W 
Withdrawn from auction 
Part c
 By Suburb and Address, the four properties where the number of bedrooms is missing is shown by the following table:
Suburb 
Address 
Darlington 
9/299 Abercrombie St. 
Kirribilli 
49/20 Carabella St. 
Manly 
1/1923 Pittwater Rd. 
North Sydney 
307/54 High St. 
 The properties which are errors, from the above identified properties are given as follows:
Suburb 
Address 
Bedrooms 
Type 
Price 
Result 
Agent 
Darlington 
9/299 Abercrombie St 
u 
N/A 
PN 
Blues Point Real Estate 

North Sydney 
307/54 High St 
u 
N/A 
PN 
Blues Point Real Estate 
These two properties are the errors because the price of the property is missing in both the cases and both of the properties were sold before auction. Thus, these two properties do not have any records during the auction.
Part d
Part e
 It can be seen from the pivot table that there were 467 properties, which was listed for auction in that day.
 The total number of properties that were sold that day (at auction, prior or after) is 244 + 31 + 121 = 396.
 The percentage of the number of properties sold at auction is = 396/467 = 84.8%.
Part f
The number of four bedroom houses listed for the auction that day is 106. These include 102 houses, 4 duplex and 1 townhouse.
Out of these 106 houses, 91 were sold at the auction, prior or after.
The percentage of the number of four bedroom houses sold (at the auction, prior or after) is 91/106 = 85.8%
Of all the listed properties, the clearance rate is 84.8% and for the four bedroom houses, the clearance rate is 85.8%. Thus, it can be said the rate of clearance for the four bedroom houses is better than the rate of clearance of all the properties overall for that week.
Part g
 The two way pivot table for type by result is given below:
 The single horizontal component chart is given below:
 The two types of properties which had approximately the same proportion of properties passed in that week are township types of properties and duplex types of properties as identified from the graph above.
Part h
 The table of descriptive statistics for price is
Price 

Mean 
1790575 
Standard Error 
63516.72 
Median 
1566250 
Mode 
1150000 
Standard Deviation 
963278.8 
Sample Variance 
9.28E+11 
Kurtosis 
16.14936 
Skewness 
2.850252 
Range 
8871500 
Minimum 
428500 
Maximum 
9300000 
Sum 
4.12E+08 
Count 
230 
From the above table, the median selling price is $1566250.
The standard deviation of these selling prices (expressed to the nearest thousands of dollars) is $963279.
The cheapest house sold that week was sold at a selling price of $428500.
It can be seen from the table that the house with the cheapest selling price was a three bedroom house located in San Remo.
 The required sample variance and its scientific notation is given by the following table:
Sample Variance 

Actual Number Value 
927906035169.89 
Scientific Notation 
9.27906E+11 
 The frequency distribution table and histogram as obtained using Excel are given below:
Household Selling Prices 
Frequency 
700000 
6 
1500000 
100 
2300000 
78 
3100000 
29 
3900000 
8 
4700000 
7 
5500000 
1 
6300000 
0 
7100000 
0 
7900000 
0 
8700000 
0 
9500000 
1 
More 
0 
In order to quote the prices of houses in Sydney, the median of the prices are considered and not the mean. The mean of a set of numbers is usually determined by adding all the numbers divided by the number of values. Thus, it involves all the values of the dataset. The median is denoted by the middlemost value of the data. It does not involve all the values of the data. Thus, it can be said that the median is more reliable than the mean. In this case, the end class is open. Thus, mean will not be accurate. Hence, median is a more appropriate measure.
Part a
Let A be the event that the number of need to be repaired.
Therefore,
 P (A = 1) = 0.17
P (A= 2) = 0.08
P (A > 2) = 0.06
Thus, P (no repairs) = P (A = 0) = 1 – [P (A = 1) + P (A= 2) + P (A > 2)]
Question 2
= 1 – (0.17 + 0.08 + 0.06)
= 0.69
 P (No more than one repair) = P (A ≤1) = P (A = 0) + P (A = 1)
= 0.69 + 0.17
=0.86
 P (Some repairs) = P (A ≥1) = 1 P (A = 0)
= 1 – 0.69
= 0.31
Part b
The probability distribution of X is given in the following table where X is the number of cars repaired by a mechanic in a day.
No. of cars (X) 
6 
7 
8 
9 
10 
Probability 
0.15 
0.25 
0.3 
0.23 
0.07 
The mean number of cars repaired in a day is given by;
= (6*0.15) + (7*0.25) + (8*0.3) + (9*0.23) + (10*0.07)
= 7.82
= 8 [approx.]
The formula for calculating the standard deviation of the number of cars repaired in a day is given by
The necessary calculations to find out the standard deviation is given as follows:
No. cars X 
Probability 
X_{i}μ 
(X_{i}μ)^2 
P_{i}*(X_{i}μ)^2 
6 
0.15 
1.82 
3.3124 
19.8744 
7 
0.25 
0.82 
0.6724 
4.7068 
8 
0.3 
0.18 
0.0324 
0.2592 
9 
0.23 
1.18 
1.3924 
12.5316 
10 
0.07 
2.18 
4.7524 
47.524 
Total 
1 
0.9 
10.162 
84.896 
The required standard deviation is .
Part c
The sample that Paul selected is bias because
 The sample was drawn on the basis of the participation of the people of one particular day. The result can vary a lot in the next day or the day after that. The number of people watching the particular cinema today might not be the same on any other day. Thus, the sample could have been collected for a week and not a day.
 Only female participants were considered in the sample. Considering only female participants will not give a clear analysis of the frequency of people visiting the movies.
Part d
The table showing the members of Sydney golf club is given below:
Age Range 
Male 
Female 
Total 
Under 18 
55 
23 
78 
18 to 30 
165 
45 
201 
31 to 50 
300 
90 
390 
Over 50 
180 
42 
222 
Total 


900 
The number of male members to be sampled from the age group of 31 to 50 years is (300/900)*90 = 30.
The number of female members to be sampled = (200/900)*90 = 20.
Part a
From the records of a company, it has been found out that four types of incidents are described in the information. The events are, day shift workers, night shift workers, the workers who turned up and workers who did not turn up.
Let the day shift workers be denoted as A and the workers who did not turn up be denoted as B. Then, the night shift workers will be denoted as and the workers who turned up is denoted by . The following diagram can give the probability tree containing the above information.
The tree clearly states the following information:
P (A) = 0.7, P () = 1 – 0.7 = 0.3
Again, P (BD) = 0.2 P ( = 1  0.02 = 0.98.
Further, P (B/) = 0.04, P (/ = 1 – 0.04 = 0.96
Therefore, the percentage of dayshift workers who are not present on any given day = (0.7 * 0.2) * 100 = 1.4%
The percentage of night shift workers are absent on any given day = (0.3 * 0.4) * 100 = 1.2%
Therefore, the total percentage of workers who are absent on any given day = 1.4% + 1.2% = 2.6%
Part b
Absenteeism will be independent of the shift worked if P (BA) = P (B). Here, this is not the case. Therefore, absenteeism depends on the working shifts.
Part a
 P (Z > 0.4)
= 1 – P (Z < 0.4)
= 1 – 0.6554
= 0.3466.
 P (1.35 ≤Z ≤ 1.25)
= P (Z ≤ 1.25) – P (Z ≤ 1.35)
= 0.8944 – 0.0885
=0.8059.
Part b
 The probability of making 30 sales = BINOM.DIST (30, 300, 0.12, FALSE) = 0.042101.
 The probability of making more than 30 sales = 1  BINOM.DIST (31, 300, 0.12, FALSE) = 0.949998.
Part c
 The probability of less than two collisions in a six month period = POISSON.DIST(2, 1.8, TRUE) = 0.730621086.
 The probability of one collision in a two month period = POISSON.DIST(1, 1.8, FALSE) ÷ 3 = 0.09918.
Part a
In the given problem, the random variable described follows a poisson distribution defined as . The mean is denoted as λ, which is equal to 13 in this problem.
Part b
The probability of the company receiving 13 emergency calls in a specified month is
P (X ≥ 13) = 1 – P (X ≤ 12) = 1 – 0.1099398 = 0.8900602.
Part c
Mean = λ = 13/30 = 0.433.
The probability of getting more emergency calls than what the company can handle is
P (X > 3) = 1 – P (X < 3) = 1 – 0.000675 = 0.999325.
Part a
In the problem given, the random variable that has been described follows a Binomial Distribution. The distribution is given by
Part b
Here mean is given by (n*p) = 0.75 * 20 = 15. Thus, the probability of 20 customers to be satisfied in a sample of 25 customers i
Part c
The expected number of dissatisfied customers for n = 50 is 50 * 0.25 = 13.
Part d
The required probability is given by BINOM.DIST(100, 150, 0.75, TRUE) = 0.013618601
Part a
The problem given in this question follows a normal distribution with mean = µ and variance = .
Part b
P (X < 50000)
= P (Z < 2.5)
= 0.00621
The required proportion of tyres, which fail before the warranty expires, is 0.00621.
Part c
P (X > 58500)
= P (Z > 1.5)
= 1 – 0.93319
= 0.668.
= 6.68 %.
Therefore the claim that the tyre will last longer than 58,500 km has a chance of atleast 10% is wrong.
Part d
P (X< 54700)
= P (Z < 1.5)
= 0.06681
The probability that the average lifetime of these tyres is less than 54700 km is 0.06881.
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