Question

Given a random variable X which is normally distributed with mean 15 and Standard deviation 10 find?

(a). P( X<20)

(b). P(X>12)

(c). P(12

Answer 1

(a) $P(X<20)=P(Z<\frac{20-15}{10})=P(Z<0.5)=0.6915.$

(b)

$P(X>12)=P(Z>\frac{12-15}{10})=P(Z>-0.3)=1-P(Z<-0.3)=\\=1-0.3821=0.6179.$

(c) $P(12<X<20)=P(X<20)-P(X<12)=0.6915-0.3821=0.3094.$