The Democrat and Chronicle reported that 25 of the flights a

The Democrat and Chronicle reported that 25% of the flights arriving at the San Diego airport during the first five months of 2001 were late (Democrat and Chronicle, July 23, 2001). Assume the population proportion is p =.25. Calculate sigma(p) with a sample size of 800 flights (to 4 decimals). What is the probability that the sample proportion will be within +/-.03 of the population proportion if a sample of size 800 is selected (to 4 decimals)? What is the probability that the sample proportion will be within +/-.03 of the population proportion if a sample of size 400 is selected (to 4 decimals)?

Solution

a)
Proportion ( P ) =0.25
Standard Deviation ( sd )= Sqrt (P*Q /n) = Sqrt(0.25*0.75/800) = 0.01531
b)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 0.22) = (0.22-0.25)/0.01531
= -0.03/0.01531 = -1.9595
= P ( Z <-1.9595) From Standard Normal Table
= 0.02503
P(X < 0.28) = (0.28-0.25)/0.01531
= 0.03/0.01531 = 1.9595
= P ( Z <1.9595) From Standard Normal Table
= 0.97497
P(0.22 < X < 0.28) = 0.97497-0.02503 = 0.9499                  
              
c)
Proportion ( P ) =0.25
Standard Deviation ( sd )= Sqrt (P*Q /n) = Sqrt(0.25*0.75/400) = 0.02165
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 0.22) = (0.22-0.25)/0.02165
= -0.03/0.02165 = -1.3857
= P ( Z <-1.3857) From Standard Normal Table
= 0.08292
P(X < 0.28) = (0.28-0.25)/0.02165
= 0.03/0.02165 = 1.3857
= P ( Z <1.3857) From Standard Normal Table
= 0.91708
P(0.22 < X < 0.28) = 0.91708-0.08292 = 0.8342