Probability and Statistics Assignment

This assignment delves into concepts of probability and statistics. It presents problems involving the binomial distribution, requiring the calculation of probabilities for specific events. The assignment also explores normal distributions, calculating probabilities based on given values and utilizing Z-scores. Students are tasked with solving various scenarios, demonstrating their understanding of these statistical concepts.

DATA ANALYSIS
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TABLE OF CONTENTS
QUESTION 1.......................................................................................................................................3
1. Stating descriptive statistics of gross profit ................................................................................3
3. Drafting frequency tables for different locations and states where stores are situated................9
4. Presenting a frequency table of online stores by state ..............................................................11
5. Stating descriptive analysis of gross profit on the basis of gender ...........................................12
QUESTION 2.....................................................................................................................................13
a. Assessing the proportion of people in survey ...........................................................................13
b. Finding the probability of randomly selected person from the survey will be..........................13
c. Presenting that events income and political affiliation are independent ...................................14
QUESTION 3.....................................................................................................................................14
a. Stating the distribution X follows .............................................................................................14
b. Calculating mean and standard deviation of X..........................................................................14
c. Presenting the probability that exactly 4 of WWW surfers will remember advertisement........14
d. Probability that 6 or more learners will remember the advertisement.......................................15
e. Probability that 3 learners will remember the banner advertisement.........................................16
QUESTION 4.....................................................................................................................................18
a. Stating the proportion which is greater than $35000.................................................................18
b. Assessing proportion which is less than $20000.......................................................................18
c. Finding proportion which is greater than $20000 and less than $35000...................................18
d. Determining probability when average value of the probability is greater than $35000..........18
e. Calculating the dollar value that it is expected by only 10%.....................................................18
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QUESTION 1
1. Stating descriptive statistics of gross profit
Store No. GrossProfit $m
1 0.712
2 0.091
3 1.075
4 1.372
5 2.148
6 2.019
7 0.662
8 0.700
9 0.937
10 0.065
11 2.144
12 0.248
13 1.607
14 1.624
15 1.995
16 0.588
17 1.288
18 1.908
19 1.000
20 0.121
21 0.159
22 2.284
23 0.799
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24 0.911
25 0.813
26 0.976
27 1.612
28 1.380
29 0.498
30 0.084
31 1.036
32 0.960
33 1.180
34 0.974
35 1.315
36 0.974
37 0.167
38 0.937
39 1.456
40 0.018
41 0.840
42 1.000
43 1.159
44 0.104
45 0.936
46 1.968
47 2.536
48 0.417
49 1.299
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50 1.155
51 1.999
52 2.872
53 0.734
54 1.310
55 0.879
56 1.496
57 0.655
58 1.644
59 0.819
60 1.623
61 1.084
62 1.461
63 0.532
64 1.336
65 1.018
66 1.135
67 1.280
68 0.612
69 0.739
70 1.142
71 1.476
72 0.546
73 1.295
74 1.512
75 0.103
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76 0.185
77 0.636
78 0.172
79 1.024
80 1.545
81 0.291
82 0.092
83 0.480
84 0.983
85 1.881
86 2.626
87 0.568
88 0.879
89 1.083
90 0.828
91 1.560
92 1.428
93 1.404
94 1.072
95 0.183
96 1.412
97 0.612
98 0.496
99 1.480
100 0.403
101 0.856
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102 1.836
103 0.408
104 0.124
105 0.085
106 0.852
107 1.927
108 1.018
109 0.864
110 0.626
111 1.384
112 0.590
113 0.072
114 1.283
115 0.075
116 0.899
117 1.248
118 0.231
119 1.512
120 0.831
121 0.123
122 0.131
123 1.539
124 0.637
125 0.275
126 0.711
127 1.200
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128 1.227
129 1.963
130 0.496
131 0.424
132 1.152
133 1.481
134 2.285
135 0.292
136 0.888
137 2.324
138 0.196
139 0.180
140 1.416
141 0.115
142 0.995
143 2.352
144 1.259
145 1.464
146 0.504
147 0.447
148 2.620
149 1.168
150 0.032
Descriptive statistics
Mean 1.01
Standard error 0.053
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Median 0.975
Mode 0.937
Standard deviation 0.646
Sample variance 0.417
Kurtosis -0.158
Skewness 0.493
Range 2.85
Minimum 0.018
Maximum 2.872
Sum 151.493
Count 150
Confidence level 0.104
From the above descriptive analysis, it has been analyzedanalysed that average gross profit
which is earned by KBS supermarket from its 150 stores is $1.01 million. Along with this, out of
100%, half ofhalf of gross profitability aspect of the firm belongs from to $.975 million. Hence, by
considering such aspect, it can be said that average and middle level or median profit is very near to
each other. Besides this, KBS has generated gross profit of $ .937 million from most of the stores.
This aspect shows that large number of the customers have same attitude and perception level
towards the products or services offered by supermarket. Further, standard error and sample
variance of the profit is 0.053 and 0.417. Above descriptive analysis also presents that gross profit
firm will be deviated from the amount of $.646 million in the near future. From 150, gross profit
margin of most of the stores are is high which shows that they have managed their direct
expenditure more effectually. Besides this, sales of the firm is continuously increasing which in
turn results into in higher profitability.
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2.
A. Average gross profit: Mean or average profit which is attained by KBS supermarket from 150
stores is $1.01 million.
B. Minimum gross profit: By doing descriptive analysis from the given set of data of 150 stores, it
has been assessed that minimum gross profit earned by the store is $.018 million. Minimum value
presents the lowest profit margin which is earned by the store of KBS. Further, it presents that the
level to which gross profit margin of the firm will increase in the near future is $.018 million. By
considering such aspect, owner of the supermarket can frame competent strategic and policy
framework.
C. Maximum gross profit: Highest sales which is generated by one of the KBS stores is $2.872
million. Hence, by taking into account such aspect, it can be said that maximum level to which
gross profit margin of the firm will incline by is $2.872 million in the near future.
D. On the basis of the output of descriptive statistics, KBS needs to close the stores which have
minimum gross profit such as $.018 million.
E. Store stands out as unusual in terms of gross profit margin: Profit of store number 40 is very
unusual out of 150 store of KBS.
3. Drafting frequency tables for different locations and states where stores are situated
Locations
Location Gross Profit (in $m)
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Country 0-0.5 11
0.5-1 12
1-1.5 8
1.5-2 4
2-2.5 3
Country Result 38
Mall 0-0.5 14
0.5-1 13
1-1.5 21
1.5-2 8
2-2.5 3
2.5-3 2
Mall Result 61
Strip 0-0.5 12
0.5-1 16
1-1.5 14
1.5-2 6
2-2.5 1
2.5-3 2
Strip Result 51
Total Result 150
Frequency table on the basis of location:
Data
Location Count - Location Count - Location
Country 38 25.33%
Mall 61 40.67%
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Strip 51 34.00%
Total Result 150 100.00%
Frequency table on the basis of states:
State Frequency
ACT 6
NSW 40
NT 4
Qld 27
SA 22
Tas 5
Vic 30
WA 16
Total Result 150
12
38
61
51
150 Country
Mall
Strip
Total Result

4. Presenting a frequency table of online stores by state
Frequency table of online stores according to states are enumerated as below:
State Online Channel Frequency
ACT No 3
Yes 3
ACT Result 6
NSW No 11
Yes 29
NSW Result 40
NT No 2
Yes 2
NT Result 4
Qld No 9
Yes 18
Qld Result 27
SA No 3
13
NSW
NT
Qld
SA
Tas
Vic
WA
Total Result
0
40
80
120
160
Total
6

Yes 19
SA Result 22
Tas No 2
Yes 3
Tas Result 5
Vic No 11
Yes 19
Vic Result 30
WA No 4
Yes 12
WA Result 16
Total Result 150
5. Stating descriptive analysis of gross profit on the basis of gender
Descriptive analysis of the gross profits of KBS stores operated by males and females are
enumerated as below:
Aspects of descriptive
statistics
Male Female
Mean 1.058 0.7565
Standard error 0.056 0.136
Median 1.009 0.558
Mode 0.937 xxx
Standard deviation 0.633 0.667
Sample variance 0.401 0.444
Kurtosis -0.126 1.163
Skewness 0.418 1.176
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Range 2.854 2.561
Minimum 0.018 0.065
Maximum 2.872 2.626
Sum 133.34 18.156
Count 126 24
Confidence level (95%) 0.112 0.282
The above mentioned descriptive analysis presents that there is significant difference
takes place between the gross profitability margin of the stores which arethat is operated by males
and females. Mean and median profit of KBS sore store run by male candidates are $1.058 million
and $1.009 million. On the other hand, average and middle level profit of the female stores are
$.757 million and $0.56 million. By considering such aspect, it can be said that gross profit margin
of the male stores are is higher than female ones. However, out of 150, 126 stores are run by the
males, whereas, only 24 stores are operated by females. In this regard, descriptive analysis clearly
shows that females have performed their activities more effectively and efficiently. Range of the
gross profit margin of the male and female operated stores are $2.854 million and $2.561 million.
Both these figures are very near to each other. Thus, it can be said that females have generated high
level of gross profit margin from 24from 24 stores.
QUESTION 2
A. ASSESSINGA. ASSESSING THE PROPORTION OF PEOPLE IN
SURVEY
Formula = Number of favourable outcomes / total number of outcomesI. Labour supporters: Probability (p) = 134 / 300II. High-Income earners: p =140 / 300
III. Labour supporters and low-income earners: p = 90 / 300
b. Finding the probability of randomly selected person from the survey will be
I. Labour supporter that has a high-income: p = 44 / 140
II. A low-income earner or a liberal supporter but not both
p(A) + p(B) -P (A n B)
= 160 / 300 + 126 / 300 -34 /300
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c. Presentingc
Presenting that event’s income and political affiliation are independent
An event is said to be independent when occurrence of one eventcase is not dependent on
another. On the basis of cited case situation, it can be said that income level is not dependent on
politicalon political affiliation. Thus, no impact will be placed by political affiliation on the event
income.
QUESTION 3
a. Statinga. Stating the distribution X follows
On the basis of the cited case situation, distribution X follows binomial distribution in which
there is are two probabilities. Moreover, binomial distribution may result into in two outcomes, that
is, such as success and failure. In this, probability of success in is 3, whereas probabilityand that of
failure is 7.
b. Calculating mean and standard deviation of X
Probability of success (p): .30
Probability of failure (q): .70
Mean: np
N: sample size
p: 3 / 10 = .30
Mean = 10*.30
= 3
Standard deviation:
= 3(1-p)
= 3 (1 - .30)
= 2.1
c. Presenting the probability that exactly 4 of WWW surfers will remember advertisement
P (x = 4) = 10c4 (0.3)4 (0.7)6
= 10! / 4! 6! (0.3)4 (0.7)6
= 10 * 9 * 8 * 7* 6! / 4! 6! (0.3)4 (0.7)6
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By removing 6! fromFrom numerator and denominator, we will get the following equation:
= 10 * 9 * 8 * 7 / 4* 3* 2 * 1 (0.3)4 (0.7)6
= 5040 / 24 (0.3)4 (0.7)6
= 210 * .0081 * .1176
= 0.20
Hence, .20 is the probability in relation the aspect that 4 surfers will remember
advertisement that would be placed on WWW.
d. Probability that 6 or more learners will remember the advertisement
Probability of 6 learners
P (x = 6) = 10c6 (0.3)6 (0.7)4
= 10! / 6! 4! (0.3)6 (0.7)4
= 10 * 9 * 8 * 7* 6! / 4! 6! (0.3)6 (0.7)4
By removing 6! fromFrom numerator and denominator, we will get the following equation:
= 10 * 9 * 8 * 7 / 4* 3* 2 * 1 (0.3)6 (0.7)4
= 5040 / 24 (0.3)6 (0.7)4
= 210 * .0007 * .2401
= 0.035
Probability of 7 learners
P (x = 7) = 10c7 (0.3)7 (0.7)3
= 10! / 7! 3! (0.3)7 (0.7)3
= 10 * 9 * 8 * 7! / 3! 7! (0.3)7 (0.7)3
By removing 7! fromFrom numerator and denominator, we will get the following equation:
= 10 * 9 * 8 / 3* 2 * 1 (0.3)7 (0.7)3
= 720 / 6 (0.3)7 (0.7)3
= 120 * .0002 * .343
= 120 * 0.007 * 0.1029
= 0.0082
Probability of 8 learners
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P (x = 8) = 10c8 (0.3)8 (0.7)2
= 10! / 8! 2! (0.3)8 (0.7)2
= 10 * 9 * 8! / 8! 2! (0.3)8 (0.7)2
By removing 8! fromFrom numerator and denominator, we will get the following equation:\
= 10 * 9 / 2 * 1 (0.3)8 (0.7)2
= 90 / 2 (0.3)8 (0.7)2
= 45 * .0001* .49
= 0.0022
Probability of 9 learners
P (x = 9) = 10c9 (0.3)9 (0.7)1
= 10! / 9! 1! (0.3)9 (0.7)1
= 10 * 9! / 9! 2! (0.3)9 (0.7)1
By removing 9! fromFrom numerator and denominator, we will the get following equation:\
= 10 / 1 (0.3)9 (0.7)1
= 10* (0.3)9 (0.7)1
= 10 * .0000* 0.7
= 0
Probability of 10 learners
Approximately 0
Total probability = (0.035 +0.0082 +0.0022 + 0 + 0)
= 0.0454
e. Probability that 3 learners will remember the banner advertisement
Probability of 3 learners
P (x = 3) = 10c3 (0.3)8 (0.7)2
= 10! / 3! 7! (0.3)3 (0.7)7
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