5.3.16. On the morning of November 9, 1994—the day after the electoral landslide

ID: 3240987 • Letter: 5

Question

5.3.16. On the morning of November 9, 1994—the day

after the electoral landslide that had returned Republicans

to power in both branches of Congress—several key races

were still in doubt. The most prominent was the Washington

contest involving Democrat Tom Foley, the reigning

speaker of the house. An Associated Press story showed

how narrow the margin had become (120):

With 99 percent of precincts reporting, Foley trailed

Republican challenger George Nethercutt by just

2,174 votes, or 50.6 percent to 49.4 percent. About

14,000 absentee ballots remained uncounted, making

the race too close to call.

Let p = P(Absentee voter prefers Foley). How small

could p have been and still have given Foley a 20% chance

of overcoming Nethercutt’s lead and winning the election?

Solution:-

There are 14000 absentee ballots left.

The lead is 2174, so the opponent needs to lead by at least 2175 in these 14000 ballots.

He does so if he gets at least 8088 votes.

If P(x >= 8088) = 0.20, then the correspoding z score to this is,

z = 0.841621234

Now, as

z = (x - u) / s

and

u= n p
s = sqrt(n * p(1-p))

Then

0.841621234 = (8088 - 14000 * p) / sqrt(14000 * p(1-p))

0.841621234 = (8088 - 14000p) / (sqrt (14000p - 14000p2))

Solving for p here,

p = 0.574197 or 0.57

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