Question

Suppose that 55 percent of the voters in a particular region support a candidate. find the probability that a sample of 1,000 voters would yield a sample proportion in favor of the candidate within 2 percentage points of the actual proportion.

Answer 1

Here the original proportion is $p=0.55$. The proportion approximately follows normal distribution. Mean is $\mu=p=0.55$.

Standard deviation is $\sigma=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.55(1-0.55)}{1000}}=0.0157$

The required probability is

$P(0.55-0.02<X<0.55+0.02)=P(0.53<X<0.57)$

$P(0.53<X<0.57)=P(\frac{0.53-0.55}{0.0157}<Z<\frac{0.57-0.55}{0.0157})$

$P(0.53<X<0.57)=P(-1.274<Z<1.274)$

Using standard normal tables,

$P(0.53<X<0.57)=0.7973$