a) A candidate has 51% of the vote with a standard deviation of 5%. How many peo
ID: 3132555 • Letter: A
Question
a) A candidate has 51% of the vote with a standard deviation of 5%. How many people were polled, and what’s the candidate’s probability of winning the election?
For the standard deviation, I used the formula ø = sqrt [np(1-p)], and got that n = 101 voters, but I don't know if that's right. Additionally, I don't know how to find the probability of winning.
b) In a new poll, the candidate has 52% of the vote, but the standard deviation is now 2%. What’s the new sample size, and what’s the candidate’s new probability of winning?
c) For the poll in part b, how many people would need to be polled for the standard deviation to be 1%?
Explanation / Answer
a)
As
n = p(1-p)/SE^2
Then
n = 0.51*(1-0.51)/0.05^2 = 99.96 = 100 [ANSWER]
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We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 0.5
u = mean = p = 0.51
s = standard deviation = sqrt(p(1-p)/n) = 0.049989999
Thus,
z = (x - u) / s = -0.200040012
Thus, using a table/technology, the right tailed area of this is
P(z > -0.200040012 ) = 0.579275356 [ANSWER]
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B)
As
n = p(1-p)/SE^2
Then
n = 0.52*(1-0.52)/0.02^2 = 624 [ANSWER]
**************************
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 0.5
u = mean = p = 0.52
s = standard deviation = sqrt(p(1-p)/n) = 0.02
Thus,
z = (x - u) / s = -1
Thus, using a table/technology, the right tailed area of this is
P(z > -1 ) = 0.841344746 [ANSWER]
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c)
As
n = p(1-p)/SE^2
Then
n = 0.52*(1-0.52)/0.01^2 = 2496 [ANSWER]
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