SOLUTION: National polls from October 2016 showed that preferences for the 2016 presidential election differed along gender lines. Poll results showed that a greater percentage of likely fem

Algebra ->  Probability-and-statistics -> SOLUTION: National polls from October 2016 showed that preferences for the 2016 presidential election differed along gender lines. Poll results showed that a greater percentage of likely fem      Log On


   

Question 1190694: National polls from October 2016 showed that preferences for the 2016 presidential election differed along gender lines. Poll results showed that a greater percentage of likely female voters supported Clinton/Kaine, while a greater percentage of likely male voters supported Trump/Pence. A summary of poll results is shown in the table below.
Men Women
Clinton/Kaine 0.44 0.49
Trump/Pence 0.47 0.43
(c) If 260 women across the US are randomly selected, how many would you expect to support Clinton/Kaine?
(d) If 240 men across the US are randomly selected, how many would you expect to support Trump/Pence?
Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

Given table
MenWomen
Clinton/Kaine0.440.49
Trump/Pence0.470.43

Though I think this is a slightly better table because it has each column adding to 1 (representing 100%)
MenWomen
Clinton/Kaine0.440.49
Trump/Pence0.470.43
Other0.090.08
Each column is separate.

-----------------------------------

Part c

There are 260 women and the chart shows 49% of them supporting Clinton/Kaine.

49% of 260 = 0.49*260 = 127.4
This rounds to 127

Answer: 127 women

-----------------------------------

Part d

There are 240 men and the chart says 47% of them support Trump/Pence

47% of 240 = 0.47*240 = 112.8
This rounds to 113

Answer: 113 men