Question: The combined math and verbal scores for females taking the SAT-I test are normally distributed with a mean of 998 and a standard deviation of 202 (based on date from the College Board). If a college includes a minimum score of 900 among its requirements, what percentage of females do not satisfy that requirement?  Edit

Answer: Ans:
Test statistic, z= (x-mean)/sd=(x-998)/202

if x= 900 , then z= (900-998)/202=-0.48

P(x<900)=P(z<-0.48)
=0.5-P(-0.48 =0.5-0.1844
=0.3156

percentage of females do not satisfy that requirement=0.3156*100%=31.56%(Ans.)  Edit