Question

in a recent election 51% 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.
question help: written example message instructor

Explanation:

Step1: Identify binomial parameters

Assume the proportion \( p \) of people supporting is 0.5 (since it's a common default if not given, or maybe from context, but here we need to note that maybe the problem had a typo, but let's assume \( n = 1480 \) and \( p = 0.5 \) for a binomial distribution \( X \sim B(n,p) \). The mean of a binomial is \( \mu = np \), standard deviation is \( \sigma=\sqrt{np(1 - p)} \).

Step2: Calculate the mean

Given \( n = 1480 \), \( p = 0.5 \) (assuming, since the problem might have a typo but let's proceed). Then mean \( \mu = 1480\times0.5 = 740 \).

Step3: Calculate the standard deviation

\( \sigma=\sqrt{1480\times0.5\times(1 - 0.5)}=\sqrt{1480\times0.5\times0.5}=\sqrt{1480\times0.25}=\sqrt{370}\approx19.235 \)

Answer:

Mean: \( 740.000 \)
Standard deviation: \( 19.235 \)

(Note: If the proportion \( p \) was different, say if there was a value like 0.55 or others, we would use that. But since the problem's text is a bit unclear, assuming \( p = 0.5 \) as a common case for such problems when proportion is not fully visible. If the actual \( p \) was given, we adjust. For example, if \( p = 0.5 \), the above holds.)