Question

Suppose that 80 percent of the voters in a particular region support a candidate. Find the probability that a sample of 1600 voters would yield a sample proportion in favor of the candidate within 1 percentage points of the actual proportion.

Answer 1

Given,

p = 0.80 , n = 1600

We have to calculate probability that sample proportion is within 1% point of actual proportion.

That is P( 0.80 - 0.01 < < 0.80 + 0.01 ) = ?

P( 0.79 < < 0.81) = ?

Using central limit theorem,

P( < p ) = P( Z < - p / sqrt( p (1 - p) / n) )

So,

P( 0.79 < < 0.81) = P( < 0.81) - P( < 0.79)

= P( Z < 0.81 - 0.80 / sqrt( 0.8 * 0.2 / 1600) ) - P( Z < 0.79 - 0.80 / sqrt( 0.8 * 0.2 / 1600) )

= P (Z < 1) - P (Z < -1)

= 0.8413 - 0.1587

= 0.6827