Just before a referendum on a school budget a local newspape

Just before a referendum on a school budget, a local newspaper 427 voters to predict whether the budget will pass. Suppose the budget has the support of 52% of the voters. What is the probability that the newspapers sample will lead it to predict defeat?

Solution

The standard deviation of the sampling distirbution of p^ is

s(p^) = sqrt(p(1-p)/n) = sqrt(0.52*(1-0.52)/427) = 0.024177331

Predicting defeat means less than 0.50 of voters.

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    0.5      
u = mean =    0.52      
          
s = standard deviation =    0.024177331      
          
Thus,          
          
z = (x - u) / s =    -0.827221168      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -0.827221168   ) =    0.204055858 [ANSWER]

Just before a referendum on a school budget, a local newspaper 427 voters to predict whether the budget will pass. Suppose the budget has the support of 52% of

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