Probability and Statistics Problems

This assignment delves into various probability and statistics problems. It explores the Poisson distribution to analyze emergency calls received by a company. The Binomial distribution is applied to customer satisfaction rates in a sample. Additionally, it examines the normal distribution in tire lifespan analysis, calculating probabilities of failure and warranty claims. The assignment includes step-by-step solutions for each problem, demonstrating the application of statistical concepts.

Running head: BUSINESS STATISTICS
Business Statistics
Name of the Student
Name of the University
Author Note

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Table of Contents
Answer 1:.........................................................................................................................................2
Answer 2:.........................................................................................................................................9
Answer 3:.......................................................................................................................................11
Answer 4:.......................................................................................................................................13
Answer 5:.......................................................................................................................................13
Answer 6:.......................................................................................................................................14
Answer 7:.......................................................................................................................................15
References......................................................................................................................................16

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Answer 1:
a) The following are the variables from the data set which fall in the respective categories:
i) Qualitative and Normal variable: Address, Type, Agent and Address
ii) Qualitative and Ordinal variable: no variable
iii) Quantitative and Ratio variable: Price
iv) Quantitative and Ordinal variable: Bedrooms
b) The overall outcomes are shown in the following table:
Result Code Overall Outcome
PI, NB, VB Did not sell
SP, PN Sold prior to the auction
S,SN Sold at auction
SA,SS Sold after auction
W Withdrawn from sale

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c) The four properties where the number of bedrooms is missing are as follows:
Suburbs Address of the properties
Kirribilli 49/20, Carabella Street
North Sydney 307/54, High Street
Manly 1/19 – 23, Pittwater Road
Darlington 9/299, Abercrombie Street
The two properties of the above four, which are errors, are as follows:
Suburbs Address Price Result
North Sydney 307/54, High Street NA PN
Darlington 9/299, Abercrombie St. NA PN
The two above properties are errors due to the reasons, discussed as follows:
i. Both the properties were sold before auction was conducted, which implies at the time of the
auction, the properties were no longer available.
ii) The prices of both the properties were not revealed in the data set.

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d) The pivot table for gaining insight into the auction market of Sydney is as follows:
Count of
Result
Column
Labels
Labels (Row) PI PN S SN SP VB W Total
h 36 8 159 28 72 5 5 313
1 1 1
2 2 18 3 6 1 30
3 15 4 60 12 29 120
4 6 4 54 7 27 2 2 102
5 7 22 4 9 1 2 45
6 3 4 1 1 1 10
7 3 1 4
8 1 1
studio 2 2
(blank) 2 2
t type 1 3 15 3 1 23
2 1 3 1 5
3 1 2 10 2 1 16
4 1 1
5 1 1
u 6 5 68 3 46 1 129
1 9 12 21
2 4 3 45 1 30 1 84
3 1 12 2 4 19
4 1 2 3
(blank) 2 2
Total 43 16 244 31 121 6 6 467
e) The calculations for the clearance rate are as follows:
i) Based on the results of the above pivot table, it is seen that a total of 467 properties were listed
for the auction, originally, on that particular day.
ii) 396 properties (244+31+22+121) were sold during, before and after the auction.

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iii) The percentage of the properties that were sold as a whole = 396/467, which is equal to
84.8%.
f) The following discussion shows the comparison, conducted by John and Julie, regarding the
clearance rates:
i) The total number of four bedroom houses, that were listed for the auction, for that particular
day originally are as follows:
Houses – 102
Townhouse – 1
Unit/duplex – 4
Therefore, 106 of four bedroom properties were listed in total on that day.
ii) Of the above properties a total of 91 were sold during the auction, before the auction and after
the auction was held.
iii) The percentage of the four bedroom properties which were sold = 85.8% (91/106).
iv) The clearance rate of the four bedroom properties, being 85.8%, is higher than that of the
clearance rate of the properties, as a whole, listed, which is 84.8%.

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g) i) The two-way pivot table for all the listed properties is as follows:
Result Count
Label
(Column)
Labels of Row PI PN S SN SP VB W Total
h 36 8 159 28 72 5 5 313
studio 2 2
t 1 3 15 3 1 23
u 6 5 68 3 46 1 129
Total 43 16 244 31 121 6 6 467
ii) Based on the above two-way pivot table, the 100% component bar chart is as follows:
h
studio
t
u
0% 10% 20% 30% 40% 50% 60% 70% 80% 90% 100%
Outcomes of sales for all properties
PI
PN
S
SN
SP
VB
W
Percentage of Sales
Type of property
iii) Based on the100% component bar chart, it is observed that township properties and
unit/duplex type of properties have almost identical proportions of the properties that passed
during that particular week.

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h) i) The descriprive statistics is as follows:
ii) The Median price for Selling was 1566250 dollars and the Standard Deviation for the same
was 963279 dollars.
iii) The price of the cheapest of the houses that was sold in that particular week was $428500.
As can be seen from the table, the cheapest property is in San Remo. The property is a flat of
three bedrooms.
iv) The sample variance of the above data is as follows:
Sample Variance
Actual Number Value 927906035169.86
Scientific Notation 9.27906E+11

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i) The frequency distribution of the selling prices of the houses that are sold, as mentioned, can
be shown as follows:
Selling prices
Frequenc
y
700000 6
1500000 100
2300000 78
3100000 29
3900000 8
4700000 7
5500000 1
6300000 0
7100000 0
7900000 0
8700000 0
9500000 1
Above 0
Based on the above frequency distribution table, the Histogram of the selling prices is as follows:
700000
1500000
2300000
3100000
3900000
4700000
5500000
6300000
7100000
7900000
8700000
9500000
Above
0
20
40
60
80
100
120
Histogram
Selling prices
Frequency

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j) Instead of using mean price, during the quotation of the prices of the houses in Sydney, the
median prices are always considered. This is done because, mean takes all the samples in the
data and is extremely affected by the extreme values which are present in the series. However,
the median pricing is more appropriate as it shows the 50% point of the data and is not affected
by the extreme values unlike mean prices. Therefore, median pricing is more appropriate than ea
pricing.
In the above data, it can be seen that the mean price is 1790575 whereas that of the median price
is 1566250, which implies that fifty percent of the total observations, are above this value and the
rest are below. This observation is supported by the outcome as is shown in the above histogram.
Answer 2:
a) Let X be the event of the number of cars that needs repairing.
i) P(X=1) = 0.17
P(X=2) = 0.08
P(X>2) = 0.06
Therefore, P(X=0) = 1 – 0.17- 0.08-0.06 = 0.69
This implies that the probability that the car needs no repair is 0.69.
ii) P(X≤1) = 0.69+0.17 = 0.86

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This means that the probability that the car chosen will not require more than one repair is 0.86.
iii) P(X≥1) = 1 – 0.69 = 0.31
The result shows that the probability that the car will need some repairs in 0.31.
b)
No. cars X 6 7 8 9 10
Probability 0.15 0.25 0.3 0.23 0.07
Given the above table, the mean of cars that are repaired on a random day is as follows:
Mean = 6*0.15 + 8*0.30 + 9*0.23 + 10*0.07 = 7.82
Therefore, approximately 8 cars are repaired on any day.
The Standard Deviation can be seen from the following table:
The Standard Deviation = [Pi(Xi-μ)^2]1/2 = [84.89]1/2 = 9.21
The SD is 10 approximately.

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c) The sample is biased due to the following reasons:
A. The survey is gender biased as it primarily took into account females, thereby biasing the data
set.
B. The survey specifically was conducted for only a single day, ruling out the variations that can
occur in the numbers on any other day. It would have been better if the survey took into account
more than one (at-least four) days within a specific period as that would have made the results
more robust and reliable (Ross, 2014).
d) i) Paul should sample at least (300/900)*90 male members = 30 male members between the
range of 31 years to 50 years.
ii) He should survey (200/900)*90 female members = 20 female members.
Answer 3:
a) In the concerned problem, there are presence of two types of incidents, namely, “Day shift
work” and “Workers not turning up”. Let T denotes that the worker is not coming to work and D
shows the day shift work. This implies D’ shows the work in night shift and T’ shows that the
worker is coming for work.
The probability tree is as follows:

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P(T/D)
P(D) P(T’/D)
Workers P(T/D’)
P(D’)
P(T’/D’)
From the probability tree drawn above, the following results can be derived:
P(D) = 70%, which implies, P(D’) = 30%
P(T/D) = 2%, therefore, P(T’/D) = 98% and as P(T/D’) is 4%, so, P(T’/D’) = 96%.
This implies that the percentage of workers (Day shift) who are absent on a random day =
(70*2)% = 1.4% and the same for the Night shift ones is (30*4)% = 1.2%.
Therefore, the total percentage of the worers who are absent = 2.6%.
b) It is known that, P(D).P(T) = P(D∩T). This implies that absenteeism is not dependant on the
working shifts.

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Answer 4:
a) Z is assumed to be a standard normal variable:
i) So, P(Z>0.4) = 1 – P(Z<0.4)
= 1- 0.6554 = 0.3446
ii) P(-135Z  1.25) = 0.8944 – 0.0885 = 0.8059
b) i) Probability of him making 30 sales is equal to BINOM.DIST(30, 300, 0.12, FALSE) =
0.042100743.
ii) Probability of >30 sales is equal to 1-BINOM.DIST(31, 300, 0.12, FALSE) = 0.949997651.
c) i) Probability of occurrence of <2 collisions within a period of six months is
POISSON.DIST(2, 1.8, TRUE) = 0.730621086.
ii) The same for one collision in a period of two months can be given by:
[POISSON.DIST(1, 1.8, FALSE) ] /3 = 0.099179333.
Answer 5:
a) The type of distribution followed by the random variable in the concerned problem is Poisson
distribution. It is defined as follows:

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Pλ ( X=x ) = λx
x ! e− λ Ghasemi & Zahediasl, 2012)
Here, λ signifies the mean. Here, the value of mean is 13/30 = 0.433.
b) The probability of the company receiving a minimum of 13 calls of emergency within a
particular month is:
P(X>=13) = 1 – 0.109939814
So, the probability is 0.890060186.
c) Here, as λ = 0.433, so, the probability of the company getting more emergency calls than it
will be able to handle is P(X>3) = 1 – 0.000675043 which is equal to 0.999324957.
Answer 6:
a) The distribution followed by the random variable in the concerned problem is Binomial
distribution, which is: C(n, x) * p^x * q^(n-x) (Ghasemi & Zahediasl, 2012).
b) Here, the Mean = pn = 0.75*20 = 15.
This implies that in a samlple of 25 random customer, 20 of them will be satisfied =
C(25,20)*(0.75)20*(0.25)5 = 0.164537588.
c) The expected dissatisfied customer number (n=50) is 50*0.25 = 13.
d) The probability that <100 customers out of 150 will be satisfied is 0.013618601.

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Answer 7:
a) The distribution in the concerned problem is normal distribution. Here, μ signifies the mean
and σ shows the SD.
b) Here, the proportion of tyres that will fail before the warranty period is:
P(X<50000) = P[Z<(50000-55000)/ 2000] = P(Z<-2.5) = 0.621%.
c) The claims of the company can be evaluated with the help of the following calculations:
P(X>58500) = P[Z>(58500-55000)/2000] = 6.68%.
Based on the above result it can be concluded that the claim of the company is not right.
d) The probability that the average life-time of the tyres< 54,700 km is as follows:
P(X<54700) = P[Z<(54700-55000)/ {2000/(100)1/2}] = P(Z<-1.5)
= 0.066807201
= 6.68% (Approximately).

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References
Ghasemi, A., & Zahediasl, S. (2012). Normality tests for statistical analysis: a guide for non-
statisticians. International journal of endocrinology and metabolism, 10(2), 486.
Ross, S. M. (2014). Introduction to probability models. Academic press.