Let X be a random variable representing the lifetime of bulbs
X~N(210,562)
z=σx−μ
i) P(X>300)
=56300−210=1.607
P(z>1.607) =0.054 from the standard normal tables
=0.054
ii) P(z<100)
=56100−210=−1.964
P(Z<-1.964)=0.025 from the z tables
=0.025
iii) P(150<X<250)
z1=56150−210=−1.071
P(z<-1.071)=0.142
z2=56250−210=0.714
P(z<0.714)=0.762
=0.762-0.142
=0.620
Comments