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? (

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A: Given, 11, 13, 15, 17 and 20

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A:

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A:

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Q: B. -14 to 36 O C. -19 to 31 O D. 6 to 16

A: Given : Mean = u = 11 Variance = σ^2 = 25 Standard deviation = σ = √25 = 5

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A: Solution

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A: Solution: Given information: μ= 11  Population mean σ2= 25  Population variance σ= σ2σ=25σ= 5…

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A: The gives an average value of the data set and the standard deviation shows the dispersion of the…

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