Question

1.
In a city with 10,000 voters, there will be an election where the
voters choose either Michael or Natasha. Currently, 4,900 voters
support Michael and 5,100 voters support Natasha. Suppose that we
collect a simple random sample of 2,000 voters from this city.

What is the expected value for the percentage of voters in the
sample who will vote for Michael?

What is the standard error for the percentage of voters in the
sample who will vote for Michael? Sufficient work must be shown to
justify your answer.

Using the normal approximation, estimate the chance that more
than 50% of those in the sample will support Michael? Sufficient
work must be shown to justify your answer.

Answer #1

p1= 4900/10000= 0.49

p2= 5100/10000= 0.51

n= 2000

Expected value for the

Expected percentage of voters in the sample who will vote for
Michael

=np1= 2000*0.49= 980

b) standard error of mean= standard deviation/sqrt (n)

Variance= n*p*q= 2000*0.49*0.51= 499.8

Standard deviation= sqrt (499.8)= 22.36

Standard error of mean= 22.36/sqrt (2000) = 22.36/44.72= 0.5

P(X>490)=

Since μ=980 and σ=22.36 we have:

P ( X>490 )=P ( X−μ>490−980 )=P ( X−μσ>490−98022.36)

Since Z=(x−μ)/σ and (490−980)/22.36=−21.91 we have:

P ( X>490 )=P ( Z>−21.91 )

: Use the standard normal table to conclude that:

P (Z>−21.91)=1

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