MODULE 10 - CATEGORICAL DATA ANALYSIS - TOPIC 6

A survey is conducted to determine whether if student use of public transport is independent to campus. The result of the survey is tabulated for the following possible responses and their respective campuses, Parramatta and Campbelltown. Campus Parramatta Campbelltown Total Would use public transport 100 60 160 Would not use public transport 20 20 40 Total 120 80 200 If one of student is selected at random and we let R be the event that the student’s campus is Par ramatta and let T be the event that the student would use public transport find the following probabilities: a) P (T) b) P(T  R) c) P(T  R) d) P (R | T) e) P (T | R) a) 0.80 b) 0.50 c) 0.90 d) 0.625 e) 0.833 2.2 A survey was undertaken to determine what sport ‐ Football (Soccer) or Rugby League was pref erred by each gender to play. 100 people were surveyed with the following results Football Rugby League

Male 48 32 Female 12 8 a) If a person is selected at random, what is the probability that the person prefers playing R ugby League? Probability (Rugby League) = (Number of people preferring Rugby League) / (Total number of people surveyed) Number of people preferring Rugby League = 32 (males) + 8 (females) = 40 Total number of people surveyed = 100 Probability (Rugby League) = 40 / 100 = 0.4 So, the probability that a person prefers playing Rugby League is 0.4. b) If a male is selected at random, what is the probability that he prefers playing Football? Probability (Male prefers Football) = (Number of males preferring Football) / (Total number of males) Number of males preferring Football = 48 Total number of males = 48 (males) + 32 (males) = 80 Probability (Male prefers Football) = 48 / 80 = 0.6 So, the probability that a male prefers playing Football is 0.6. c) Are the events 'Preferred Code of football to play' and 'Gender of Respondent' statisticall y dependent or independent events?

To find the probability that a randomly selected personal loan is in default (D) given that it is a homeowner (H), you can use Bayes' theorem. Bayes' theorem states: P ( D ∣ H )= P ( H ∣ D ) ⋅ P ( D ) / P ( H ) You already have the following information:  P ( D )=0.10 (probability of being in default)  P ( DC )=0.90 (probability of not being in default)  ( H ∣ D )=0.20 (probability of being a homeowner given that it's in default)  P ( H ∣ DC )=0.70 (probability of being a homeowner given that it's not in default) First, calculate the overall probability of being a homeowner (P(H)) using the law of total probability: P ( H )= P ( H ∣ D ) ⋅ P ( D )+ P ( H ∣ DC ) ⋅ P ( DC ) P ( H )=0.20 ⋅ 0.10+0.70 ⋅ 0.90 Now, you can calculate P ( D ∣ H ) using Bayes' theorem: P ( D ∣ H )= P ( H ∣ D ) ⋅ P ( D ) / P ( H ) Plug in the values: P ( D ∣ H )=0.20 ⋅ 0.10 / 0.20 ⋅ 0.10+0.70 ⋅ 0.90 Now, calculate the numerator and denominator: Numerator: 0.20 ⋅ 0.10=0.020.20 ⋅ 0.10=0.02 Denominator: 0.20 ⋅ 0.10+0.70 ⋅ 0.90=0.02+0.63 =0.650.20 ⋅ 0.10+0.70 ⋅ 0.90=0.02+0.63 =0.65 Now, divide the numerator by the denominator:

P ( D ∣ H )= 0.02 / 0.65 =0.0308 So, the probability that a randomly selected personal loan is in default (D) given that it is a homeowner (H) is approximately 0.031, or about 3.1%.