Assignment Questions:
Question 1
- The following data represent the number of years patients survived after being diagnosed with terminal cancer:
0.4, 0.5, 0.6, 0.6, 0.6, 0.8, 0.8, 0.9, 0.9, 0.9,
1.2, 1.2, 1.3, 1.4, 2.1, 2.4, 2.5, 4.0, 4.5, 4.6
Stem
leaf
- Construct a stem-and-leaf display
0 4, 5, 6, 6, 6, 8, 8, 9, 9, 9
1 2, 2, 3, 4
2 1, 4, 5
4 0, 5, 6
- Supposedly you are inserting the above stem-and-leaf display in a report to be submitted to management, write a short comment on the diagram.
Answers: This chart shows that half of the patients are diagnosed as Advanced cancer, and survival
rate is less than one year.
Secondly, the diagnosed patients can survive for 4 years or longer.
- The following data shows the weight (in kg) of 13 crabs found in a restaurant on a particular evening:
3, 4, 1.2, 1.7, 2.4, 2.4, 1.1, 0.9, 0.8, 1.2, 1.6, 0.7, 1.2, 1.3
- Compute the mean and median
n
0.7, 0.8, 0.9, 1.1, 1.2, 1.2, 1.2, 1.3 1.6, 1.7, 2.4, 2.4, 3.4
n
2
1
i
I =1
Mean = X = X + X + + X / n
= 0.7 + 0.8 + 0.9 + 1.1 + 1.2 + 1.2 + 1.2 + 1.3 + 1.6 + 1.7 + 2.4 + 2.4 + 3.4 / 13
= 1.531
Median = n + 1 / 2 = 13 + 1 / 2 = 7 (average of1.2).
- Determine the shape of the distribution based on the sample data. Explain your conclusion.
Answer: The mean is more than the median, and it is a right-skewed distribution and the diagram
contains a long curve tail on the right side.
Question 2
- It is noted that 8% of Kaplan students are left handed. If 20 (TWENTY) students are randomly selected,
calculate the
- Probability that none of them are left-handed,
n – x
x
n = 20 x = 0 π = 0.08
20 – 0
0
P (x) = nCx (π) (1 – π)
P (x = 0) = 20C0 (0.08) (1 – 0.08)
= 0.1887
- Probability that at most 2 are left-handed,
20 – 2
2
20-1
1
20 – 0
0
P (x >2) = P (x = 0) + P (x = 1) + P (x = 2)
= 20C0 (0.08) (1 – 0.08) + 20C1 (0.08) (1 – 0.08) + 20C2 (0.08) (1 – 0.08)
= 0.7879
- Standard deviation for the number of left-handed students
μ = Mean = nπ
= 20 (0.08)
= 1.6
Variance = nπ(1 – π)
= 20 x 0.08 x (1 – 0.08)
= 1.
SD = √nπ(1 – π472)
= √1.472
= 1.213
- If 50 (FIFTY) classes of 20 (TWENTY) students are randomly selected, what is the probability that 10 (TEN)
classes have no left-handed students?
P (X = 0) = P (no left-handed students) = 0.1887
50 – 10
10
P (have left-handed students) = 1- 0.1887 = 0.8113
P (10 class have no left-handed students) = 50C10 (0.1887) (1 – 0.08)
= 0.137
Question 3
- Superior Construction Pte Ltd is a successful company dealing with many major projects in Singapore.
Recently, it has submitted its biddings for two major Government projects. Project A worth about $120 million and the company believes it has 40% chance of securing the project. Project B worth $1.8 billion and there is 30% chance the company can win the project. Both projects are independent of each other. What is the probability that the company:
- will secure Project A or B but not both
P (A) = 40% = 0.4 P (A’) = 60% = 0.6 (lose)
P (B) = 30% = 0.3 P (B’) = 70% = 0.7 (lose)
P (A or B but not both) = P (A) x P(B’) + P (A’) x P(B)
= 0.4 x 0.7 + 0.6 x 0.3
= 0.46
- will not secure Project A or will not secure Project B
P (not secure Project A or not secure Project B) = 1- P (Project A and Project B)
= 1 – (0.4 x 0.3)
= 1- 0.12
= 0.88
- Do you agree that “if two events are mutually exclusive then these two events will be independent”?
Why?
Answer: Disagree.
Project A and Project B can happen at the same time, and depend on each other, not mutually
Exclusive
- Provide one business-related example each, with explanation, for mutually exclusive and independent
events.
Answer: Example of events mutually exclusive: toss a coin . The result of flipping one of the “Head” or
“Tail” coin Because both results may not occur simultaneously.
Example of independent events: A person can throwing a coin and be an engineer. Because
both can be concurrent.
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