According to the 2011 Gallup daily tracking polls Mississipp

According to the 2011 Gallup daily tracking polls, Mississippi is the most conservative U.S. state, with 53.4% of its residents identifying themselves as conservative. What is the probability that at least 100 but fewer than 115 respondents of a random sample of 200 Mississippi residents identify as conservative?

0.1685

0.3370

0.7085

0.8770

Solution

Normal Approximation to Binomial Distribution
Mean ( np ) =200 * 0.534 = 106.8
Standard Deviation ( npq )= 200*0.534*0.466 = 7.0547
Normal Distribution = Z= X- u / sd                   
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 100) = (100-106.8)/7.0547
= -6.8/7.0547 = -0.9639
= P ( Z <-0.9639) From Standard Normal Table
= 0.16755
P(X < 115) = (115-106.8)/7.0547
= 8.2/7.0547 = 1.1623
= P ( Z <1.1623) From Standard Normal Table
= 0.87745
P(100 < X < 115) = 0.87745-0.16755 = 0.7085