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Time Remaining: 01:07.04 Submit Quiz This Question: 1 pt 8 of 10 (0 complete) This Quiz: 10 pts possible Question Help Exit polling is a popular technique used to determine the outcome of an election prior to results being taled Suppose a referondum to increase funding for education sample if the population proportion of voters in the town in favor of the referendum is 0 527 Based on your result, comment on the dangers of using exit poling to call How likely are the results of your sample if the population proportion of voters in the town in favor of the referendurm is 0 527 is on the ballot in a large town (voting population over 100.000) An exit poll of 500 voters finds that 245 voted for the referendum. How likely are the results of your elections The probability that less than 245 people voted for the referendurm is Round to four decimal places as needed) Comment on the dangers of using exit polling to call elections. Choose the correct answer below A. The result is not unusual because the probability that p is equal to or more extreme than the sample proportion is greater than 5%. Thus it is not unusual for B. The result is not unusual because the probability that p is equal to or more extreme than the sample portion is less than 5% Thus it is unusual for a c. he result is unusual because the probability that p is equal to or more extreme than the sample roportion is geater than 5% us, t is not unusual for a a wrong call to be made in an election if exit polling alone is considered. wrong call to be made in an election if exit polling alone is considered wrong call to be made in an election if exit polling alone is considered Click to select your answer(s) Attempts Sample Tests and Quizzes

Explanation / Answer

a.
The sampling distribution is approximately normal

b.
Mean ( u ) =90
Standard Deviation ( sd )=(sd/Sqrt(n) = 18/ Sqrt ( 36 ) = 3
Number ( n ) = 36
Normal Distribution = Z= X- u / (sd/Sqrt(n) ~ N(0,1)  

P(X > 93.75) = (93.75-90)/18/ Sqrt ( 36 )
= 3.75/3= 1.25
= P ( Z >1.25) From Standard Normal Table
= 0.1056                  
              

c.
P(X > 83.85) = (83.85-90)/18/ Sqrt ( 36 )
= -6.15/3= -2.05
= P ( Z >-2.05) From Standard Normal Table
= 0.9798                  
P(X < = 83.85) = (1 - P(X > 83.85)
= 1 - 0.9798 = 0.0202                  
                  

d.
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 85.65) = (85.65-90)/18/ Sqrt ( 36 )
= -4.35/3
= -1.45
= P ( Z <-1.45) From Standard Normal Table
= 0.07353
P(X < 97.35) = (97.35-90)/18/ Sqrt ( 36 )
= 7.35/3 = 2.45
= P ( Z <2.45) From Standard Normal Table
= 0.99286
P(85.65 < X < 97.35) = 0.99286-0.07353 = 0.9193                  
              

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