Question

1. Suppose only 45% of the American people who registered to vote actually intended to vote in the 2016 presidential election.

1. You take a random sample of 2 Americans who are registered to vote. Using the appropriate method, find the chance that at most 1 of them actually intends to vote in the 2016 presidential election. Identify the method used.
2. You take a random sample of 20 Americans who are registered to vote. Using the appropriate method, find the chance that at least 10 of them actually intend to vote in the 2016 presidential election. Identify the method used.
3. You take a random sample of 200 Americans who are registered to vote. Using the appropriate method, find the chance that at least 100 of them actually intent to vote in the 2016 presidential election. Identify the method used.

Answer 1

P( actually vote)=0.45= p

Let X be the number of american people who actually vote

a) n=2

X~ BInomial ( n=2, p=0.45)

p( X<=1)= P( X=0) + P( X=1)

= ²C₀ (0.45)⁰(1-0.45)² + ²C₁ (0.45) (1-0.45)

= 0.7975

Here we use Binomial distribution

b)

n=20

X~ BInomial ( n=20, p=0.45)

p( X>=10)= 0.4086 ( using binomial table)

Here we use Binomial distribution

c) n=200

Since the sample size is quite large we will use Normal aproximation

Mean, np= 200*0.45 =90

Variance , np(1-p) = 200*0.45*(1-0.45)= 49.5

X~ Normal( 90, 49.5)

P( X>=100)= P( X> 109.5) ( continuity correction)

= P( )

= P( Z >2.77)

= 1- P( Z < 2.77)

= 1- 0.9972

= 0.0028

Normal approximation of binomial distribution is used