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In a survey of 1125 people, 794 people said they voted in a recent presidential election. Voting records show that 68% of eligible voters actually did vote. Given that 68% of eligible voters actually did vote, (a) find the probability that among 1125 randomly selected voters, at least 794 actually did vote. (b) What do the results from part (a) suggest?

In A Survey Of 1125 People 794 People Said They Voted In A Recent Presidential Election Voting Records Show That 68 Of Eligible Voters Actually Did Vote Given T class=

Sagot :

Based on the question, let's list down the information.

Sample Size = 1125 people

Mean = 68% of 1125 = 1125 x 0.68 = 765 people

Sample Value x = at least 794 people

To be able to solve the probability, we need to get the z-value of the value x at 794 and get the area to the right of it because of the phrase "at least". However, before we can get the z-value, we need to solve for the standard deviation first. To solve, we have the formula below:

[tex]\sigma=\sqrt[]{np(1-p)}[/tex]

where n = sample size = 1125 people and p = actual proportion of the population who voted = 0.68. Let's plug in these values to the formula above.

[tex]\sigma=\sqrt[]{1125(0.68)(1-0.68})[/tex]

Then, solve.

[tex]\begin{gathered} \sigma=\sqrt[]{1125(0.68)(0.32)} \\ \sigma=\sqrt[]{244.8} \\ \sigma=15.646 \end{gathered}[/tex]

The standard deviation is 15.646.

Now, let's solve for the z-value. The formula is:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

Plugging in the mean, sample value, and standard deviation to the formula above, we have:

[tex]z=\frac{794-765}{15.646}[/tex]

Then, solve.

[tex]z=\frac{29}{15.646}=1.85[/tex]

The calculated z-value is 1.85. Looking at the normal distribution table, the area covered to the right of 1.85 is 0.0322. Hence, the probability that among 1125 randomly selected voters, at least 794 actually did vote is 0.0322 or 3.22%

To interpret this, we have some people are being less than honest because P (x ≥ 794) is less than 5%. (Option D)

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