S-ACC544 Assignment 1: Probability and Statistics in Decision Support

This assignment solution covers key concepts in probability and statistics within the context of decision support tools. It includes definitions of probability and statistical independence, along with calculations and interpretations related to probability distributions, variance, and standard deviation. The document provides detailed solutions to problems involving daily sales of sourdough bread and the probability of orange sales exceeding certain thresholds. Additionally, it addresses hypothesis testing, specifically examining the statistical significance of the distance from homes to the nearest fire station. Confidence intervals are calculated and analyzed for different sample sizes and confidence levels, offering insights into maintaining control over production time. The solutions include both calculations and interpretations, making it a comprehensive resource for students studying decision support tools and statistical analysis. Desklib offers more solved assignments and resources for students.

Running Head: DECISION SUPPORT TOOLS
Decision Support Tools
Name of the student:
Name of the university:
Course ID:

1DECISION SUPPORT TOOLS
Table of Contents
Q 1...................................................................................................................................................2
Q 1. a)...........................................................................................................................................2
Q 1. b)..........................................................................................................................................2
Q 1. c)...........................................................................................................................................3
Q 1. d)..........................................................................................................................................4
Q 2...................................................................................................................................................4
Q 2. 3............................................................................................................................................4
Q 3...................................................................................................................................................5
Q 3. a)...........................................................................................................................................5
Q 3. b)..........................................................................................................................................8
References:....................................................................................................................................10

2DECISION SUPPORT TOOLS
Q 1.
Q 1. a)
Probability:
Probability is the ratio of the probable number of outcomes for any event with respect to equally
likely total number of possible outcomes of that given event (Durrett, 2010).
Probability of an event = Probable number of outcomes of an event
Total number of outcomes
Measurement of probability:
A probability measure or a probability distribution of any random experiment is a ‘real-valued
function’ defined on the ‘collection of events’ that satisfies some axioms such as-
1. P (S) = 1, where, S = Total probable event,
2. P (A) ≥ 0, for any event A,
3. , where {Ai: i € I} is countable.
The probability of any event ranges between 0 and 1. Its value cannot be negative or greater than
1 in any way.
Let us assume that ‘S’ is a ‘non-empty’ and ‘countable’ set of event. ‘g’ is a non-negative real-
valued function defined on ‘S’.
Therefore, probability measure is defined by, P (A) = μ( A)
μ( S) , if 0 < μ(S) < ∞.
Q 1. b)
‘Statistical independence’:
Any two events A and B are said to be ‘statistically independent’ to each other if and only if the
conditional probability of A given B (P (A|B)) satisfies-

3DECISION SUPPORT TOOLS
Identification:
If two events (A and B) are independent to each other, then, we can write-
If n events (A1, A2… An) are independent to each other, then-
Q 1. c)
1. c. 1.
1. c. 2.
The average daily sales is found to be = 2.85.
1. c. 3.
The probability of selling daily 2 or more than 2 loaves = 80
100 = 0.8.
1. c. 4.
The probability of selling daily 2 or less than 2 loaves = 40
100 = 0.4.

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1. c. 5.
The variance of the distribution = 36.6475.
1. c. 6.
The standard deviation of the distribution = 6.05372.
Q 1. d)
The mean of sales of oranges = 4700.
Standard deviation of sales of oranges = 500.
a) The probability that sale of oranges would be greater than 5500 oranges =
P (X ≥ x) = P ( x−μ
σ ≥ 5500−4700
500 ¿=P( x−μ
σ ≥ 800
500 )=P(Z ≥ 1.6)=1−0.9452=0.0548 (Dibley
et al., 1987).
b) The probability that sales would be greater than 4500 oranges =
P (X ≥ x) = P ( x−μ
σ ≥ 4500−4700
500 ¿=P ( x−μ
σ ≥ −200
500 )=P (Z ≥−0.4)=1−0.3446=0.6554
c) The probability that sales would be lesser than 4900 oranges =
P (X ≤ x) = P ( x−μ
σ ≤ 4900−4700
500 ¿=P ( x−μ
σ ≤ 200
500 )=P(Z ≤0.4)=1−0.3446=0.6554
d) The probability that sales would be less than 4300 oranges =
P (X ≤ x) = P ( x−μ
σ ≤ 4300−4700
500 ¿=P ( x−μ
σ ≤ −400
500 )=P(Z ≤−0.8)=1−0.7881=0.2119
Q 2.

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Q 2. 3.
Answer 2. 3. 1.
The probability of being a randomly selected population female
= Total females population
Total population = 11582494
23232413 =0.498549
Answer 2. 3. 2.
The probability of being a randomly selected population aged between 25 and 54
= Total 25−54 population
Total population = 9629106
23232413 =0.414469
Answer 2. 3. 3.
The ‘joint probability’ that a person chosen at random from the ‘population’ is a male and age is
between 55 to 64 years old
= Total number of male population whose age groupis 55−64 years
Total population = 1363331
23231413 =0.058682
Answer 2. 3. 4.
The ‘conditional probability’ that a person chosen at random from the ‘population’ of belonging
in the age-group of 25 to 64 years given that the person is a ‘female’
= Total female population of age−group 25−64 years
Total female population
¿ Total female population of agegroup 25−54 years+Total female population of agegroup54−64 years
Total female population
= ( 4725976+1384036)
11582494 =0.263761
Q 3.
Q 3. a)

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3. a. 1.
The average time to produce any product under monitoring ( X ) = 30 hours.
The standard deviation of time to produce any product under monitoring (σ) = 10 hours.
The number of random samples (n) = 64.
The critical Z-statistic with 5% level of significance = 1.96.
The statistical formula to calculate 95% confidence intervals of mean is-
X ± Zα /2* σ
√ n (Gardner & Altman, 1986)
The upper control limit (UCL) is found to be = 32.45 hours.
The lower control limit (LCL) is found to be = 17.55 hours.
3. a. 2.

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While management prefers using lesser samples of only 16 observations, then the calculated 95%
‘confidence limits’ are-
UCL (Upper control limit) = 34.90 hours
LCL (lower control limit) = 25.10 hours.
3. a. 3.
‘Management’ is regarding three substitute methods for maintaining ‘higher control’ over time.
Sampling by using 16 observations and ‘confidence intervals’ 90%
The 90% confidence limits are = (25.88787, 34.11213).
Sampling using 64 observations and confidence intervals 95%

8DECISION SUPPORT TOOLS
The 95% confidence limits are = (27.55, 32.45).
Sampling using 36 observations and confidence intervals 95%
The 95% confidence limits are = (26.73, 33.27).
The process to maintain 95% ‘confidence limits’ with selected sample size of 64 observations
provide narrowest control limits that are (27.55 hours, 32.45 hours).

9DECISION SUPPORT TOOLS
Q 3. b)
3. b. 1.
Hypotheses:
Null hypothesis (H0): There exist no significant difference between population mean distance
(5.8 km.) and sample mean distance (5.5 km.) from home to nearest fire station.
Alternative hypothesis (HA): There exists significant distance between population mean distance
(5.8 km.) and sample mean distance (5.5 km.) from home to nearest fire station.
3. b. 2.
Critical value:
For critical p-value (level of significance) (α = 0.05), the critical Z-statistic is found to be =
1.644854.
3. b. 3.
Interpretation of the hypothesis test:
The critical p-value (0.05) is less than calculated p-value = 0.158655. Again, critical Z-statistic
(1.644854) is greater than calculated Z-statistic (1). Therefore, from both point of view, no
statistical significance was found in the hypothesis test. Failing to reject the ‘Null hypothesis’ at
5% ‘level of significance’, it can be interpreted that there exists no statistical significance
between average population mean and hypothetical mean (Chen, 2011). Therefore, there is
enough evidence to refute contention that the mean difference is not greater than claimed 5.5 km.
3. b. 4.

10DECISION SUPPORT TOOLS
Graphical presentation:
References:
Chen, Z. (2011). Is the weighted z‐test the best method for combining probabilities from
independent tests?. Journal of evolutionary biology, 24(4), 926-930.
Dibley, M. J., Staehling, N., Nieburg, P., & Trowbridge, F. L. (1987). Interpretation of Z-score
anthropometric indicators derived from the international growth reference. The American
journal of clinical nutrition, 46(5), 749-762.
Durrett, R. (2010). Probability: theory and examples. Cambridge university press.
Gardner, M. J., & Altman, D. G. (1986). Confidence intervals rather than P values: estimation
rather than hypothesis testing. Br Med J (Clin Res Ed), 292(6522), 746-750.