The data in a data set are normally distributed with a mean of 180 and a standard deviation of 30. Estimate the percent of the data that are less than 120 or greater than 240.
we normalize the data by finding z=X−μσz1=120−18030=−2z2=240−18030=2using the 68, 95, 99.7 rule we have thatP[−2≤z≤2]=0.9545we want the probability outside this area so we haveP([z<−2)∧(2<2)]=1−0.9545=0.0455