I would like to ask about this question:
For an upcoming election, reports suggest that 52 percent of voters favour candidate A. If 70 million voters will be voting soon, approximate the probability that A wins, assuming the candidate who receives more than half of the votes wins.
I tried to use normal approximation but the z value I got is -334 which is not very useful.
p=0.52
q=1-p=0.48
mean = np = (0.52)(70,000,000)=36,400,000
variance = npq = (0.52)(70,000,000)(0.48) = 17,472,000
std dev = 4179.952153
So Probability to win = 1 - P(x<=35,000,000)
Using normal approximation
P(x<= 35,000,000)= P(z <= (35,000,000-(36,400,000))/4179.952153)= P(z<=-334.9)
??? In the standard normal distribution, z values typically ranges from -3 to 3?
I am thinking could I use n = 70 and >35 success trials instead, by expressing them in millions?
Thanks!
You can assume that X is the number (in millions), of voters who vote for Candidate A. This will give you P(X>35). Using continuity correction, this then becomes P(X>35.5). You should get P(Z> -0.22) after normal approximation. This is simply 0.5 + 0.08706(from the Z table) = 0.58706.