Question

in a recent election 58% of people supported reelecting the incumbent. suppose a poll is done of 1480 people.
if we used the normal as an approximation to the binomial, what would the mean and standard deviation be?
mean:
standard deviation:
give your answers to at least 3 decimal places.
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Explanation:

Step1: Identify the binomial parameters

For a binomial distribution, the number of trials \( n = 1480 \) and the probability of success \( p = 0.58 \) (since 58% of people supported the incumbent).

Step2: Calculate the mean of the binomial distribution

The formula for the mean (\(\mu\)) of a binomial distribution is \(\mu = np\).
Substitute \( n = 1480 \) and \( p = 0.58 \):

$$ \mu = 1480 \times 0.58 = 858.4 $$

Step3: Calculate the standard deviation of the binomial distribution

The formula for the standard deviation (\(\sigma\)) of a binomial distribution is \(\sigma = \sqrt{np(1 - p)}\).
First, calculate \( 1 - p = 1 - 0.58 = 0.42 \).
Then substitute \( n = 1480 \), \( p = 0.58 \), and \( 1 - p = 0.42 \):

$$ \sigma = \sqrt{1480 \times 0.58 \times 0.42} $$

First, calculate the product inside the square root: \( 1480 \times 0.58 \times 0.42 = 858.4 \times 0.42 = 360.528 \)
Then take the square root: \( \sigma = \sqrt{360.528} \approx 18.9876 \approx 18.988 \) (rounded to three decimal places)

Answer:

Mean: \( 858.4 \)
Standard deviation: \( 18.988 \)