According to the results of the 2004 presidential election, 59% of New Yorkers voted for Kerry. Imagine the set of all possible samples of size 100 from all New Yorkers who voted. For each sample of size 100, the value of the sample proportion p^^\^ is the percentage of the people in the sample that voted for Kerry. Use 3 decimal places. (a) What would the average of all these sample proportions be? .59 Correct: Your answer is correct. (b) What would the standard deviation of all these sample proportions be? 0.0492 Correct: Your answer is correct. (c) Would the distribution of all these sample proportions be Normal?

For each New Yorker, there are only two possible outcomes. Either they voted for Kerry, or they voted against Kerry. So we can solve this problem using concepts of the binomial probability distribution.

Binomial probability distribution:

Probability of x sucesses on n repeated trials, with p probability. If n is large enough, say larger than 20, the distribution of all these samples proportion is binomial.

Average of all the sample proportions: [tex]E(X) = p[/tex]

Standard deviation of all the sample proportions: [tex]s= \sqrt{\frac{p(1-p)}{n}}[/tex]

In this problem, we have:

Samples of size 100, so [tex]n = 100[/tex].

59% of New Yorkers voted for Kerry, so [tex]p = 0.59[/tex].

(a) What would the average of all these sample proportions be?

[tex]E(X) = p = 0.59[/tex]

(b) What would the standard deviation of all these sample proportions be?

[tex]s= \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.59*0.41}{100}} = 0.0492[/tex]

(c) Would the distribution of all these sample proportions be Normal?

Sample large enough, so yes.