Question

In an election, suppose that 35% of voters support creating a new fire district. If we poll 166 of these voters at random, the probability distribution for the proportion of the polled voters that support creating a new fire district can be modeled by the normal distribution pictured below. The marks on the horizontal axis show 0, +1, -1, +2, -2.. standard deviations from the mean on the sampling distrubution. The boxes correspond to the mean, +2 standard deviations from the mean, and -2 standard deviations from the mean. Fill in the boxes accurate to two decimal places 2? 2o
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media%2F27a%2F27a59f03-b319-4ef3-97e2-a1
0 0
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Answer #1

The sampling distribution of sample proportion is normal with mean p and variance p ( 1- p )/n

here p = 0.35 and n = sample size = 166

So that mean = = p = 0.35

variance =

Therefore -2* =   (after rounding up to two decimal places)

and    + 2* = (after rounding up to two decimal places)

Therefore first 0.28 , then 0.35 and then 0.42 are the answers.

2) We want to find P( p > 0.14 )

Let's write the given information

P = 0.12 , n = 100

mean = = P = 0.12

standard deviation =

P( p > 0.14 ) = 1 - P( p < 0.14) ----------(1)

Using excel:

P( p < 0.14) = "=NORMDIST(0.14,0.12,0.0325,1)" = 0.7309

Plug this value in equation 1, we get

P( p > 0.14 ) = 1 - 0.7308 = 0.2691

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