Question

in a certain large metropolitan city there are two candidates running for mayor, ronald chump and halle minton. 48% of the voters are in favor of ms. minton. an exit poll (this is a random sample taken of people exiting polling place) of 3300 people was taken. x = the number of people who voted for ms. minton.
a) what is the mean of x? μx =
b) what is the standard deviation of x? σx =
(round to 2 decimal places)
c) using the normal approximation to binomial, what is the probability that more than 1660 people voted for ms. minton?
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Explanation:

Part A: Mean of \( X \)

Step 1: Identify distribution and parameters

\( X \) follows a binomial distribution \( X \sim \text{Binomial}(n, p) \), where \( n = 3300 \) (sample size) and \( p = 0.48 \) (probability of voting for Ms. Winton). The mean of a binomial distribution is \( \mu = np \).

Step 2: Calculate the mean

\( \mu_X = n \times p = 3300 \times 0.48 \)
\( \mu_X = 1584 \)

Step 1: Recall the formula for binomial standard deviation

The standard deviation of a binomial distribution is \( \sigma = \sqrt{np(1 - p)} \), where \( n = 3300 \), \( p = 0.48 \), so \( 1 - p = 0.52 \).

Step 2: Substitute values and calculate

\( \sigma_X = \sqrt{3300 \times 0.48 \times 0.52} \)
First, calculate \( 3300 \times 0.48 \times 0.52 = 3300 \times 0.2496 = 823.68 \)
Then, \( \sigma_X = \sqrt{823.68} \approx 28.70 \) (rounded to 2 decimal places)

Step 1: Check conditions for normal approximation

For binomial \( X \sim \text{Binomial}(n, p) \), \( np = 3300 \times 0.48 = 1584 \geq 5 \) and \( n(1 - p) = 3300 \times 0.52 = 1716 \geq 5 \), so approximation is valid. We approximate \( X \) with \( Y \sim \text{Normal}(\mu = 1584, \sigma = 28.70) \).

Step 2: Apply continuity correction

For \( P(X > 1660) \), with continuity correction, we find \( P(X > 1660) \approx P(Y > 1660.5) \) (since \( X \) is discrete, "more than 1660" means \( X \geq 1661 \), so continuity correction adds 0.5 to 1660).

Step 3: Calculate the z-score

The z-score is \( z = \frac{x - \mu}{\sigma} = \frac{1660.5 - 1584}{28.70} \)
\( z = \frac{76.5}{28.70} \approx 2.67 \)

Step 4: Find the probability using z-table

\( P(Y > 1660.5) = 1 - P(Y \leq 1660.5) = 1 - \Phi(2.67) \), where \( \Phi(z) \) is the cumulative distribution function of the standard normal. From z-table, \( \Phi(2.67) \approx 0.9962 \), so \( 1 - 0.9962 = 0.0038 \)

Answer:

\( \mu_X = \boxed{1584} \)

Part B: Standard Deviation of \( X \)