According to the National Center for Education Statistics 63

According to the National Center for Education Statistics, 63% of Texas students are eligible to receive free or reduced-price lunches. Suppose you randomly choose 295 Texas students.

Find the probability that no more than 66% of them are eligible to receive free or reduced-price lunches

Please show yoru work so i can learn this!!!

Solution

Normal Distribution
Proportion ( P ) =0.63
Standard Deviation ( sd )= Sqrt (P*Q /n) = Sqrt(0.63*0.37/295)
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
P(X > 0.66) = (0.66-0.63)/0.0281
= 0.03/0.0281 = 1.0676
= P ( Z >1.068) From Standard Normal Table
= 0.1428                  

P(X<=0.66) = 1 - P(X > 0.66) = 1 - 0.1428 = 0.8572