MAS183 Ass. 2 - 29.5 out of 42. PDF

Preview

Summary

29.5 out of 42....

Description

Assignment 2 Question 1.

Test Results

a) Contingency Table

Sars-Covid-2 Diseased Healthy Total Positive 0.0096 0.0792 0.0888 Negative 0.0004 0.9108 0.9112 Total 0.01 0.99 1

b) What proportion of people that get a positive test result would actually have a SARS-COV-2 infection? P(Diseased/Positive) = 0.0096/0.0888 = 0.1081 10.81% of people with a positive test result would actually have SARSCOV-2 infection c) Consider those who get a negative test result. If we chose one of these people at random, what is the chance that the selected person doesn't have SARS-COV-2? P(Healthy/Negative) = 0.9108/0.9112 = 0.999 There is a 99.9% chance that the selected person from the negative test result group will not have Sars-Cov-2 d) In this population, what does the test do better: show who has SARSCOV-2, or show who doesn’t have it? Justify your answer. The test is better at showing who does not have Sars-Covid-2, as 99.9% of the negative diagnoses are correct compared to only 10.81% of the positive diagnosis (seen in questions b & c).

Question 2 a) Mean  = xP(Y = y) = 1(0.22) + 2(0.63) + 3(0.15) = 1.93 b) Standard deviation 2 = x2 P(X = x) -2 = 1.71 + 1.84 + 3.08 = 5.2225/3 2= √  2.21 = 1.49 Question 3 Does X follow a binomial distribution? No, it is not binomial distribution in this case as it does not fulfill the five conditions. While there is a definite number of trials (100) and each trial has only two outcomes; “succeed” (affected by Disease Q) or “fail” (unaffected by Disease Q), and the variable (X, or the number of diseased trees) is the number of successes out of the n trials it is not possible for the trials to succeed or fail independently of one another as the disease is soil-borne. This means that if one square is affected by the disease, the surrounding squares are also likely to be affected. Question 4 a) What is the probability distribution of X? X~Bin (25,0.1) b) i) The mean and standard deviation of X. =np = 25x0.1 = 2.5 =√np(1-p) =√25x0.1(1-0.1) = 1.5 ii) P( X ≥ 4) P(X ≥ 4) = 1- P(X ≤ 3) = 1-0.7636 = 0.2364 iii) P( 3 ≤ X < 9) P(3≤X...