GEM2900 - Understanding Uncertainty andStatistical Thinking

David Chew and David Nott

Department of Statistics and Applied ProbabilityNational University of Singapore

Exercise sheet 4, Exercise 1

There are 500 students taking GEM2900 this semester.Someone taking GEM2900 is trying to estimate the proportionof 1st year students (freshmen) in the class. A random sampleof 60 students was drawn and 12 students were freshmen.

(a) Based on this sample, what is your best estimate of theproportion of freshmen taking GEM2900?

(b) Compute and explain a measure of the reliability of thisestimate.

Exercise sheet 4, Exercise 1

(a) The proportion of freshment taking GEM2900 in thesample is a reasonable estimate, namely p = 12/60 = 0.2.

(b)

Standard error =

p (1 p

n=

0.2 0.8

60 0.0516

This standard error is an estimate of the standard deviationof the estimator p (considered as a random variable thatvaries between samples).

Exercise sheet 4, Exercise 2

Assume the midterm exam for GEM2900 includes 100 multiplechoice questions with five possible responses to each question.Suppose a student did not study and decided to guess theanswer for all questions in a completely random fashion.

(a) What is the expected number of questions that the studentguesses correctly?

(b) What is the standard deviation of the number of questionsthat the student guesses correctly?

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Exercise sheet 4, Exercise 2

(a) The number of correct guesses X is a binomial randomvariable Bin(n,p) where n = 100 and p = 1/5. For abinomial random variable you know that E(X ) = np. So theexpected number of guesses correct isnp = 100 1/5 = 20.

(b) Again considering the number of correct guesses X as abinomial random variable with parameters n and p we havethat the variance of X is Var(X ) = np(1 p). Hence thestandard deviation is the square root of this,

sd(X ) =np(1 p) =

100 1

5 4

5= 4

Exercise sheet 4, Exercise 3

A friend of yours claims to have the ability to read your mind.He asks you to flip a fair coin 5 times without showing him theresult, recording the sequence of coin flips on a piece of paper(again without showing him the sequence). Then he asks you toconcentrate on the sequence on the piece of paper and he willbe able to tell you what the sequence is by reading your mind.(a) If we assume that your friend has no mind reading ability

and that he is just guessing, what is the distribution of thenumber of correct guesses out of 5?

(b) What is the probability that all 5 guesses will be correct?

Exercise sheet 4, Exercise 3

1. Assuming your friend has no ability to read minds, hischance of guessing any flip correct is 0.5, independentlyfor each of the flips. Hence the number of guesses correctX is binomial, X Binomial(5,0.5).

2. The probability of 5 guesses correct isP(X = 5) =

(12

)5= 132 .

Exercise sheet 4, Exercise 4

After playing the game in exercise 3 above with you your friendfailed to correctly predict all 5 coin tosses. He continues toinsist that he can read your mind, claiming that he wasntwarmed up" yet. He suggests that you play the game with him10 more times (that is, you generate 10 more sequences of 5coin tosses where for each sequence he must guess the resultof all 5 tosses correctly).

What is the chance that he will guess all 5 tosses correctly onat least one of the 10 occasions?

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Exercise sheet 4, Exercise 4

Let Ai be the event of failing to win on game i , i = 1, ...,10. LetA be the event of failing to win on all 10 occurrences. From thelast exercise we know that P(Ai) = 31/32. Then

P(A) = P(10i=1Ai)

=10i=1

P(Ai)

=

(3132

)10and

P(A) = 1 P(A) = 1(

3132

)10 0.27.

Exercise sheet 4, Exercise 5

A lecturer administers an exam to his class of 500 GEM2900students. He drops all the exams on the floor so that theanswer booklets and the cover sheets identifying the studentsare separated, so that he has no way of knowing which studentsubmitted which answer booklet. He decides to pair up theanswer books and cover sheets randomly.

What is the (approximate) probability that at least one studentgets assigned their correct mark (that is, what is the chancethat at least one cover sheet and answer booklet are correctlymatched)?

Exercise sheet 4, Exercise 5

We can think of the event of a correct match as a rare event,and apply the law of rare events. The chance of a match forone particular answer book is 1/500 if the cover sheets andanswers are paired randomly. With 500 answer books, theexpected number of matches is 500 1/500 = 1.

The number of matches should approximately follow a Poissondistribution with mean 1 if the law of rare events holds. Hencethe probability of no match is the probability of zero for aPoisson random variable with mean 1, which is exp(1).Hence the probability of one or more correct match is1 exp(1) 0.632.