Analysis of Scientific Data - Semester 2, 2020 Week 5 Tutorial Part A - Binomial Distribution a) A couple decides that they will have four children. Let X be the number of girls they will have. Assuming that the probability of a girl is 0.50, independent across births, what is the distribution of X? X ~ Binomial(4,0.5) b) Let X be the number of towns in which it will rain tomorrow among five neighbouring towns. Is X a Binomial random variable? No. There are only two options to choose from, whether it rains or not, but since they are neighbouring towns, it makes each case not independent so it would not be a random variable. c) Suppose 10% of people are left-handed and let X be the number of left-handed people in sample of 20 individuals. What is the probability of at least one left-handed person in the sample? 1-P(0) = 1-(0.9)^20 X ~ Binomial(20, 0.1) Want P(X>=1)=1-P(X=0)=1-0.9^20=0.878 d) Suppose a drug has a 20% chance of making a person drowsy. Out of a sample of 80 people who each take the drug, what is the probability that no more than 10 of them experience drowsiness? X ~ Binomial(80,0.2) Want P(X <= 10)=P(X=0)+P(X=1)+P(X=2)+ ... +P(X=10) sum(dbinom(x=0:10, size=80, prob=0.2)) 0.0564609 e) In Week 4, we looked at the expected value and standard deviation of X, the number of days in a week that a piece of equipment is working. The probabilities given in the table actually came from a Binomial distribution. What are the parameters of the Binomial distribution? E(x)=2.92=np for binom. Since n=5, p=2.02/5=0.584 So X ~ Binomial(5, 0.584) Reference Several of these questions are adapted from Section 6.6 of Mind on Statistics, a useful source of other practice questions. See the ECP for details.
Part B - Normal Distribution Based on the student survey data, suppose that pulse rates while completing the survey come from a Normal distribution with mean 71.7 bpm and standard deviation 11.7 bpm. a) What is the probability that a random student has a pulse rate of at least 90 bpm? P(X>=90)=P(Z>=(71.7-90)/11.7) (71.7-90)/11.7 = - 1.564 pnorm(-1.564)= 0.05890878 b) What is the probability that a random student has pulse rate between 60 and 80 bpm? P(X <= 80)=pnorm(0.7094017)=0.7609624 P(X>=60)=pnorm(-1)=0.1586553 So P(60 <= x <= 80)=P(Z <= 0.7094)-P(Z> =- 1)=0.6023071 c) What value would put a student in the bottom 10% of pulse rates? qnorm(-0.1) =- 1.281552 -1.281552*11.7+71.7=56.7 d) In a random sample of 5 students, what is the probability that at least 3 of them have a pulse rate over 90 bpm while completing the survey? sum(dbinom(x=3:5,size=5,prob =. 0589)) =0.001867087 You may find the following output from R useful in answering the questions on this sheet. > dbinom(x=4,size=4,prob =. 5) > pnorm(-2.114) [1] 0.0625 > sum(dbinom(x=1:4,size=4,prob =. 5)) [1] 0.9375 > 1-(1 -. 5)^4 [1] 0.9375 [1] 0.05890878 > sum(dbinom(x=3:5,size=5,prob =. 0589)) > pnorm(0.7094017) [1] 0.001867087 [1] 0.7609624 > dbinom(x=10,size=80,prob =. 2) > pnorm(-1) [1] 0.02774334 [1] 0.1586553 > sum(dbinom(x=1:10,size=80,prob =. 2)) [1] 0.05646089 >