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

Five year survival chance from any cancer depends on manyfactors like availability of treatment options, expertise ofattending medical team and more. Five year survival rate is also animportant measure and it is used by medical practitioners to reportprognosis to patients and family. We will be analyzing five yearsurvival rate of two types of cancer, very aggressive and verytreatable cancer and to have comparative analysis of cancer inNorway.

1. Five year survival rate of breast cancer is87.7%
2. Five year survival rate of esophageal cancer is16.5%

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

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

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

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

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

Part 2 Normal distribution [Mark30%]

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

Part 3 Normal approximation of binomial ProbabilityDistribution [Mark 30%]

Airlines and hotels often grant reservation in excess to theiravailable capacity, to minimize loss and maximize profitability dueto no shows. Suppose that the records of Air Georgian shows that onaverage, 10% of their prospective passengers will not show up atdeparture gates. If Air Georgian sells 215 tickets  andtheir plane has capacity for 200 passengers.

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