Question

Use Table 8.1, a computer, or a calculator to answer the following. Suppose a candidate for public office is favored by only 47% of the voters. If a sample survey randomly selects 2,500 voters, the percentage in the sample who favor the candidate can be thought of as a measurement from a normal curve with a mean of 47% and a standard deviation of 1%. Based on this information, how often (as a %) would such a survey show that 50% or more of the sample favored the candidate? (Round your answer to two decimal places.) 99.87 x % Need Help? Talk to a Tutor

Answer 1

We are given the distribution here as:

$X \sim N(\mu = 47,\sigma = 1)$

The probability that 50% of more of the sample favored the candidate is computed here as:
P(X >= 50)

Converting it to a standard normal variable, we have here:

P(Z >= (50 - 47) / 1)

= P(Z >= 3)

= 1 - P(Z < 3)

This is computed from the standard normal tables as:

= 1 - 0.9987

= 0.0013

Therefore 0.13% is the required probability here.