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Question 6 In an election, 53% of voters are expected to vote for candidate A. C

ID: 3057364 • Letter: Q

Question

Question 6

In an election, 53% of voters are expected to vote for candidate A. Consider a sample of 75 of those voters and answer the below questions.

Question 6 options:

A. _____

What is the standard deviation of the proportion of voters voting for candidate A from the sample?

B. _____

Consider the proportion of voters voting for candidate A from the sample of n=75 voters.    What is the probability that the proportion is GREATER than 0.60?

C. _____

Consider the proportion of voters voting for candidate A from the sample of n=75 voters. What is the probability that the proportion is GREATER than 0.53?

D. _____

What is the mean of the distribution of the proportion of voters voting for candidate A from the sample?

Note: Below choices may NOT all be used as matches.

0.456

0.281

0.0576

0.11

0.50

0.530

No answer is correct.

A. _____

What is the standard deviation of the proportion of voters voting for candidate A from the sample?

B. _____

Consider the proportion of voters voting for candidate A from the sample of n=75 voters.    What is the probability that the proportion is GREATER than 0.60?

C. _____

Consider the proportion of voters voting for candidate A from the sample of n=75 voters. What is the probability that the proportion is GREATER than 0.53?

D. _____

What is the mean of the distribution of the proportion of voters voting for candidate A from the sample?

1.

Note: Below choices may NOT all be used as matches.

2.

0.456

3.

0.281

4.

0.0576

5.

0.11

6.

0.50

7.

0.530

8.

No answer is correct.

Explanation / Answer

Given n = 75 and p = 0.53
Using Binomial distribution
A) SD = sqrt(npq) = sqrt(75*0.53*0.47) = 4.3223

B) P(P > 0.60) = P(Z > (0.60 - 0.53)/4.3223)) = P(Z > 0.016195) = 0.49354

C) Consider the proportion of voters voting for candidate A from the sample of n=75 voters. What is the probability that the proportion is GREATER than 0.53?

P(P > 0.53) = P(Z > (0.53 - 0.53)/4.3223)) = P(Z > 0) = 0.50

D) mean = np = 75*0.53 = 39.75

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