Problem 3, It is known that among UCLA students 60% support candidate A for the

Question

Problem 3, It is known that among UCLA students 60% support candidate A for the student council, while only 40% support candidate B. In order not to waste the time of too many students, it was decided that instead of holding general elections, n students will be selected at random and the outcome of the elections will be based on the majority vote among them (the candidate receiving the most votes wins). Suppose that n is small compared to the entire student population, so that the votes of the n selected students are essentially i.i.d. 1. Let Sn be the nuber of students among the n who voted for candidate A. How is Sn distributed? 2. Write the event that candidate A wins the elections in terms of Sn. 3. If n = 500, find a lower bound on the probability that the outcome of the elections is just (candidate A wins)? Use Chebyshev's inequality elections 1s just (candidate A wins

Explanation / Answer

let p=proportion of students support candidate A=60%=0.6

the votes of the n selected students are essentially iid

1. Sn be the number of students out of n who voted for candidate A.

so the outcomes of Sn are dichotomous that is : either voted for A or not voted A

the outcomes are iid since the votes of the n selected students are essentially iid

and the probability of voting candidate A is fixed at p=0.6 in each trial

hence voting candidate A can be regarded as a success and voting candidate B can be regarded as a failure.

hence Sn is binomially distributed with parameters n and p=0.6

so Sn~Bin(n,0.6) [answer]

2. candidate A will win if he receives most votes that is if he receives more votes than B.

that is if he receives more than half of n votes.

hence the event that candidate A will win is denoted by Sn>n/2 [answer]

3. for n=500 we have Sn~Bin(500,0.6)

so u=mean of Sn=500*0.6=300

variance=sigma2=500*0.6*(1-0.6)=300*0.4=120

so here the event that A will win is Sn>500/2 that is Sn>250

now for chebyshev's one sided inequality we have

P[Sn<u-k]<=sigma2/(sigma2+k2) where k>0

or, -P[Sn<u-k]>-sigma2/(sigma2+k2)

or, 1-P[Sn<u-k]>1-sigma2/(sigma2+k2)

or, P[Sn>u-k]>1-sigma2/(sigma2+k2)

now we have u=300 and sigma2=120   

we choose k=50

so P[Sn>300-50]>1-120/(120+502)

or, P[Sn>250]>1-0.0458=0.9542

or, P[Candidate A will win]>0.9542

hence the lower bound of the probability is 0.9542 [answer]

4. now we have P[Sn>u-k]>1-sigma2/(sigma2+k2)

now Sn~Bin(n,0.6)

so u=0.6n sigma2=0.6*0.4*n=0.24n

we have P[Sn>n/2]>0.975

so u-k=n/2=0.5n or, 0.6n-k=0.5n so k=0.1n

and 1-sigma2/(sigma2+k2)=0.975

or, 1-0.24n/(0.24n+(0.1n)2)=0.975

or, 0.24n/(0.24n+0.01n2)=1-0.975=0.025

or, (0.24n+0.01n2)/0.24n=1/0.025=40

or, 1+(0.01/0.24)n=40

or, n=(40-1)*0.24/0.01=936

hence the lower bound for n is 936 [answer]

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.