Question

Part 1 Binomial Distribution [Mark 20%/cancer type, 40% total mark]

Five year survival chance from any cancer depends on many factors like availability of treatment options, expertise of attending medical team and more. Five year survival rate is also an important measure and it is used by medical practitioners to report prognosis to patients and family. We will be analyzing five year survival rate of two types of cancer, very aggressive and very treatable cancer and to have comparative analysis of cancer in Norway.

1. Five year survival rate of breast cancer is 87.7%
2. Five year survival rate of esophageal cancer is 16.5%

(NOTE: due to limitation imposed by our available probability distribution table assume survival rate for breast cancer is 90% and for esophageal cancer is 20%)

To simplify our comparative analysis, we will assume 480 patients were admitted in January 2018. For each type of cancer:

1. Calculate the number of patients that are expected to survive at least 5 years, calculate variance and standard deviation expected for 5 year survival.
2. For a selected 15 patients in each category, calculate probability of at least 7 patients surviving past 5 years, compare your calculated value with estimated number from table in page 329. How accurate is our estimation? Do you prefer estimation or calculation methods?
3. Use the table on page 329 of your textbook to determine survival probability for 0 to 15 patients and complete table below.

Selected number of patient will survive 5 years	Probability of breast cancer patient will survive 5 years	Probability of esophageal cancer patient will survive 5 years
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15

4. Draw frequency distribution bar graph for each type of cancer
5. What is the minimum/maximum number of survivors expected for given cancer type at 5 years. [HINT: use mean and 3 standard deviations to report outcome].

Part 2 Normal distribution [Mark 30%]

Daily discharge from phosphate mine is normally distributed with a mean daily discharge of 38 mg/L and a standard deviation of 12 mg/L. What proportion of days will the daily discharge exceed 58 mg/L?

Part 3 Normal approximation of binomial Probability Distribution [Mark 30%]

Airlines and hotels often grant reservation in excess to their available capacity, to minimize loss and maximize profitability due to no shows. Suppose that the records of Air Georgian shows that on average, 10% of their prospective passengers will not show up at departure gates. If Air Georgian sells 215 tickets  and their plane has capacity for 200 passengers.

1. Use binomial probability distribution to calculate, the mean of passengers showing up at the gate out of the 215 reservations made
2. Calculate the standard deviation.
3. Calculate standard z score for 200 passengers showing up at the gates.
4. Using normal probability distribution, determine probability of at least 200 passengers will show up!
5. Determine probability of more than 200 passengers showing up at the gate?
6. Determine probability of all 215 passengers showing up at the gate!

Answer 1

Part 1 : Cannot be completed as it has reference to a table in the textbook page 329, which is not been provided.

Part 2 Normal distribution [Mark 30%]

Daily discharge from phosphate mine is normally distributed with a mean daily discharge of 38 mg/L and a standard deviation of 12 mg/L. What proportion of days will the daily discharge exceed 58 mg/L?

Part 3 Normal approximation of binomial Probability Distribution [Mark 30%]

Airlines and hotels often grant reservation in excess to their available capacity, to minimize loss and maximize profitability due to no shows. Suppose that the records of Air Georgian shows that on average, 10% of their prospective passengers will not show up at departure gates. If Air Georgian sells 215 tickets and their plane has capacity for 200 passengers.

Use binomial probability distribution to calculate, the mean of passengers showing up at the gate out of the 215 reservations made
Probability of passenger who do not show up = 0.1
Probabiity of passanger who show up = 0.9

mean number of passenger who show up = np = 215*0.9 = 193.5

Calculate the standard deviation.

sd = np(1-p) = 215*(0.9)*(0.1)= 19.35

Calculate standard z score for 200 passengers showing up at the gates.

Using normal probability distribution, determine the probability of at least 200 passengers will show up!

Determine the probability of more than 200 passengers showing up at the gate?

Determine the probability of all 215 passengers showing up at the gate!