bus2 This Question: 1 pt This Quiz: 5 pts possible A oenter that studies voting

Question

bus2 This Question: 1 pt This Quiz: 5 pts possible A oenter that studies voting in a country indioated that ofee voting-agevonens castbalotsina Presidential election talso indicated inhe went 48.1% of voting-age voters votedfor acemoaat a a How many of the employees would you expect to have voted for a Demoorat as a representate? 0201 mype integer or a decimal) b AI of the employees indicated that they voted inthe presidentalelecton Detemminethe probabaty o this assuming they folowed the natonal trend (Round to four decimal places as needed) (Round to four decimal places as needed) d. Based on your calculations in parts bland c do the employees refect the national trend? Support your answer wth statistical calaulations and measoning O A. No, because the probabilties found in parts band care not that small, which means thattheassumption thatthe employees folowod the national frondis nat valid B. No, because the probabilaes found in parts band care very small, which means that the assumpton that he employees filowed he natonil land is not vald O c. Yes, because the probablites found in pans band care very smal, which means that assumpton thatthe employees folowedthe national trend oould be vaid O D. Yes, because the probabilites found in parts band c are not thatsmal, which mears eat ee assumpten patthe employees followef the nationa tond could be vaid. MacBook Pro

Explanation / Answer

Normal Distribution
Proportion ( P ) =0.45
Standard Deviation ( sd )= Sqrt (P*Q /n) = Sqrt(0.45*0.55/600)
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 0.41) = (0.41-0.45)/0.0203
= -0.04/0.0203 = -1.9704
= P ( Z <-1.9704) From Standard Normal Table
= 0.02439
P(X < 0.47) = (0.47-0.45)/0.0203
= 0.02/0.0203 = 0.9852
= P ( Z <0.9852) From Standard Normal Table
= 0.83774
P(0.41 < X < 0.47) = 0.83774-0.02439 = 0.8133                  
b.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 0.41) = (0.41-0.45)/0.0352
= -0.04/0.0352 = -1.1364
= P ( Z <-1.1364) From Standard Normal Table
= 0.1279
P(X < 0.47) = (0.47-0.45)/0.0352
= 0.02/0.0352 = 0.5682
= P ( Z <0.5682) From Standard Normal Table
= 0.71504
P(0.41 < X < 0.47) = 0.71504-0.1279 = 0.5871                  

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