Question

In a recent election, 68% of people supported reelecting the incumbent. Suppose a poll is done of 1380 people. If we use the normal as an approximation to the binomial, what would the mean and standard deviation be? Give your answer to at least 3 decimal places.

Answer 1

The normal approximation to the binomial distribution gives us a mean of 938.400 and a standard deviation of approximately 20.045 for a poll of 1380 people with 68% support for the incumbent.

Step-by-step explanation:

The question asks about using the normal approximation to the binomial distribution to determine the mean and standard deviation of a survey sample in an election. Given that 68% of people supported reelecting the incumbent and the poll size is 1380 people, the mean (μ) and standard deviation (σ) can be calculated using the formulas for a binomial distribution:

• Mean (μ) = n * p
• Standard deviation (σ) = sqrt(n * p * (1 - p))

Where 'n' is the sample size and 'p' is the probability of success (in this case, the support for reelecting the incumbent).

Therefore:

• The mean (μ) = 1380 * 0.68 = 938.4
• The standard deviation (σ) = sqrt(1380 * 0.68 * 0.32) = sqrt(1380 * 0.2176) which equals approximately 20.045.

Thus, the mean number of people who support reelecting the incumbent out of a poll of 1380 people is 938.400, and the standard deviation is approximately 20.045 to three decimal places.