Question

One theory concerning the Dow Jones Industrial Average is that it is likely to increase during U.S. presidential election years. From 1964 through 2000, the Dow Jones Industrial Average has increased in eight of the ten US. presidential election years. Assuming that this indicator is a random event with no predictive value, you would expect that the indicator would be correct 50% of the time. What is the probability of the Dow Jones Industrial Average increasing in eight or more of the ten U.S. presidential election years if the true probability of an increase in the Dow Jones Industrial Average is

a. 0.50?

b. 0.70?

c. 0.90?

d. (optional) Based on the results of (a) through (c), what do you think is the probability that the Dow Jones Industrial Average will increase in a US. presidential election year?

Answer 1

Here we assuming that the Dow Jones Industrial Average has increased is a random event with no predictive value, so we would expect that the indicator would be correct 50% of the time.

Let X = the number of times the Dow Jones Industrial Average has increased out of 10.

So that X follows binomial distribution with n = 10, and p is a) 0.5 , b) 0.70, c) 0.9

Here we want to find P( the Dow Jones Industrial Average increasing in eight or more of the ten U.S. presidential election years) for a, b, and c

P(X >= 8 ) = 1 - P( X <= 7) .....( 1 )

a) n = 10, p= 0.5

Let's use excel:

P( X <= 7) = "=BINOMDIST(7,10,0.5,1)" = 0.9453

Plug this value in equation ( 1 ), we get

P(X >= 8 ) = 1 - 0.9453 = 0.0547

b)

n = 10, p= 0.70

P( X <= 7) = "=BINOMDIST(7,10,0.70,1)" = 0.6172

Plug this value in equation ( 1 ), we get

P(X >= 8 ) = 1 - 0.6172= 0.3828

c)

n = 10, p= 0.9

Let's use excel:

P( X <= 7) = "=BINOMDIST(7,10,0.9,1)" = 0.0702

Plug this value in equation ( 1 ), we get

P(X >= 8 ) = 1 - 0.0702= 0.9298

d) Yes, the probability is increase. Because when we increase p from 0.5 then the more weight is concentrated to the extreme values of X.