Statistics for Analytical Decisions: Probability and Data Analysis

This statistics assignment provides a detailed analysis of probability distributions, both discrete and continuous, illustrated with examples such as coin tosses and bread sales. It covers calculating probabilities, average daily sales, and conditional probabilities. The assignment further explores population demographics, calculating probabilities based on Australian population data, and demonstrates hypothesis testing and confidence intervals using sample data. The solutions include calculations for standard error, control limits, and test statistics, offering a comprehensive understanding of statistical concepts and their application in analytical decision-making. The assignment includes calculations and explanations for various statistical concepts such as mean, standard deviation, confidence intervals, and hypothesis testing, providing a comprehensive understanding of statistical analysis.

Running Head: STATISTICS FOR ANALYTICAL DECISIONS
Statistics for Analytical Decisions
Name of the Student
Name of the University
Author Note

1STATISTICS FOR ANALYTICAL DECISIONS
Table of Contents
Answer 1..........................................................................................................................................2
Answer 2..........................................................................................................................................7
Answer 3..........................................................................................................................................8
References......................................................................................................................................10

2STATISTICS FOR ANALYTICAL DECISIONS
Answer 1
1. The probabilities that are possessed by the values of a random variable gives the
probability distribution of that particular random variable. A probability distribution can
be represented in the form of a function, a table or a graph1.
Probabilities possessed by discrete random variables is a discrete probability
distribution. In a discrete random variable, the probabilities are assigned to each values
and are greater than 0. The sum of all the allotted probabilities is equal to 12. For
example, in the tossing of two coins, the probability distribution of obtaining heads is
given by the following table:
Number of
Heads
0 1 2 Total
Probability 1/4 2/4 = 1/2 1/4
(1/4) + (1/2) +
(1/4) = 1
The probabilities possessed by a continuous random variable is a continuous
probability distribution. In a continuous random variable, there are no particular points to
which probabilities can be assigned. Thus, the probabilities at each point is zero.
However, the area under the range of a curve is denoted by the probability of that range
of values. The total area under the probability curve equals to 13.
For example, the waiting time in a bus stop for the arrival of a bus is a continuous
random variable and the probability of waiting is the probability distribution of the
waiting time.
1 Aidara, Nafy. "Introduction to probability and statistics." (2018).
2 Von Mises, R. (2014). Mathematical theory of probability and statistics. Academic Press.
3 Allen, A. O. (2014). Probability, statistics, and queueing theory. Academic Press.

3STATISTICS FOR ANALYTICAL DECISIONS
2. The following table gives a record of the sales over the last 100 days of the top selling
bread loaf of a baker.
NUMBER SOLD NUMBER OF DAYS
0 5
1 15
2 20
3 25
4 20
5 15
Total 100
To calculate the necessary probabilities given the following questions, the
following calculations are necessary to be conducted:
NUMBER SOLD NUMBER OF DAYS PROBABILITIES
0 5 (5 / 100) = 0.05
1 15 (15 / 100) = 0.15
2 20 (20 / 100) = 0.2
3 25 (25 / 100) = 0.25
4 20 (20 / 100) = 0.2
5 15 (15 / 100) = 0.15
Total 100 (0.05 + 0.15 + 0.2 + 0.25 + 0.2 + 0.15) = 1
1. The required probability that 3 or 4 loaves can be sold in any one day is given by:
P (Selling 3 loaves in one day) * P (Selling 4 loaves in one day) = 0.25 * 0.2 =
0.05.
2. To calculate the average daily sales of the period, the product of the probabilities
with the respective number of bread loaves sold has to be calculated. The sum of
this product gives the average daily sales. Thus,
The average daily sales = (0 * 0.05) + (1 * 0.15) + (2 * 0.2) + (3 * 0.25) + (4 *
0.2) + (5 * 0.15) = 2.85.
3. The required probability that 2 or more than 2 loaves of bread can be sold on any
one day is given by:

4STATISTICS FOR ANALYTICAL DECISIONS
P (Selling 2 loaves in one day) + P (Selling 3 loaves in one day) + P (Selling 4
loaves in one day) + P (Selling 5 loaves in one day) = 0.2 + 0.25 + 0.2 + 0.15 =
0.8
4. The required probability that 4 or less loaves of bread can be sold on any one day
is given by:
P (Selling 0 loaves in one day) + P (Selling 1 loaf in one day) + P (Selling 2
loaves in one day) + P (Selling 3 loaves in one day) + P (Selling 4 loaves in one
day) = 1 – P (Selling 5 loaves in one day) = 1 – 0.15 = 0.85
3. If a coin is tossed, there are two possible outcomes, Head and Tail. Let the event
of obtaining a head be denoted by H and the event of obtaining a tail be denoted by T.
Now, if a coin is tossed twice, then the number of possible outcomes will be 22 = 4. The
four possible outcomes that can be obtained after tossing a coin twice are {HH, HT, TH,
TT}. Here, HH denotes both the outcomes are heads, HT denotes that the first outcome is
head and the second is tail, TH denotes that the first outcome is a tail and the first is a
head and TT denotes both the outcomes are tails.
1. A head on the first toss can be obtained in 2 out of the four cases. Therefore, the
required probability of obtaining a head on the first toss is (2 / 4) = 0.5.
2. The number of cases in which tail is obtained on the second toss given that a head
has been obtained on the first toss is 1. Thus, the required probability is (1 / 4) =
0.25.
3. The number of cases in which two tails can be obtained is 1 out of 4. Thus, the
required probability of obtaining two tails is (1 / 4) = 0.25

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4. The number of cases in which tail is obtained on the first toss and a head is
obtained on the second toss is 1. Thus, the required probability is (1 / 4) = 0.25.
5. The number of cases in which tail is obtained on the second and a head is
obtained on the first toss is 1 out of 4. The number of cases in which tail is
obtained on the first toss and a head is obtained on the second toss is 1 out of 4.
Therefore, the total number of cases favourable to the event is 2 out of 4. Thus,
the required probability is (2 / 4) = 0.5.
6. The number of cases in which at least one head can be obtained within the first
two tosses is 3 out of 4. Thus, the required probability is (3 / 4) = 0.75.
4. It is given that the average number of sales of apples is 5000 and the standard
deviation of the sales has been obtained as 600. Let the number of apples sold be denoted
by X4.
1. The probability that the number of apples sold will be greater than 5600
apples is given by P (X > 5600). Now,
2. The probability that the number of apples sold will be less than 5240
apples is given by P (X < 5240). Now,
4 Bharti, S., & Bharti, B. (2014). Fundamentals of Statistics. Textbook of Hospital Administration, 345.

6STATISTICS FOR ANALYTICAL DECISIONS
3. The probability that the number of apples sold will be less than 4400
apples is given by P (X < 4400). Now,

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Answer 2
1. On searching the ABS website, the latest figures on the age and sex of Australian
population that has been obtained is given in the following table.
Table 2.1: Male and Female population of Australia in 20175
 The probability of selecting a male person at random = (12,204,419 / 24,598,933) =
0.496.
 The probability of selecting a person aged between 55 and 64 years at random =
(2,839,984 / 24,598,933) = 0.115.
 Probability of selecting a female between 15 and 24 years of age at random =
(1,566,973 / 24,598,933) = 0.064.
 The probability that a person selected at random will be aged 55 years or more =
((2,839,985 + 3,794,800) / 24,598,933)) = 0.270.
5 '3101.0 - Australian Demographic Statistics, Jun 2017' (Abs.gov.au, 2018)
<http://www.abs.gov.au/AUSSTATS/abs@.nsf/DetailsPage/3101.0Jun%202017?OpenDocument> accessed 14
March 2018

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Answer 3
(a) According to the Question,
Mean = 20 hours and Standard deviation = 5 hours.
A random sample of 64 observations were collected.
Confidence level = 95%
Therefore, Standard Error =
Standard Devistion
Square root of the sample ¿ ¿ 5
√ 64 =1.25 ¿
1) UCL = (Mean + 2 * Standard Error) = 20 + 2 * 1.25 = 22.5
LCL = (Mean – 2 * Standard Error) = 20 – 2 * 1.25 = 17.5
2) If the sample size is reduced to 16, then the control limits are given by (Mean ±
A3 * Standard Deviation), where A3 is the constant for finding the control limits
evaluated from the control limit tables. For a sample of size 16, A3 is given by
0.739. Therefore,
UCL = (Mean + A3 * Standard Deviation) = 20 + 0.739 * 5 = 23.7
LCL = (Mean – A3 * Standard Deviation) = 20 – 0.739 * 5 = 16.3
(b) H0: μ ≥ 9 against
H1: μ < 9
Here, μ = 9, n = 50
X = 10.22, σ = 5
df = 50 – 1 = 49
Tabulated value of t for a lower tail test with 49 df = -1.677.
Test Statistic6:
6 Gupta, S. C. (2016). Fundamentals of Statistics. Himalaya Publishing House.

9STATISTICS FOR ANALYTICAL DECISIONS
The value of the test statistic is greater than the tabulated value of t. Thus, H0 is accepted.
Average distance from the store within which the customers stay is more than 9 km.

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References
'3101.0 - Australian Demographic Statistics, Jun 2017' (Abs.gov.au, 2018)
<http://www.abs.gov.au/AUSSTATS/abs@.nsf/DetailsPage/3101.0Jun%202017?
OpenDocument> accessed 14 March 2018
Aidara, Nafy. "Introduction to probability and statistics." (2018).
Von Mises, R. (2014). Mathematical theory of probability and statistics. Academic Press.
Allen, A. O. (2014). Probability, statistics, and queueing theory. Academic Press.
Bharti, S., & Bharti, B. (2014). Fundamentals of Statistics. Textbook of Hospital Administration,
345.
Gupta, S. C. (2016). Fundamentals of Statistics. Himalaya Publishing House.