Let X be a random variable denoting the lifetime of bulbs
X~N(210,562)
"z=\\frac {x-\\mu}{\\sigma}"
i) P(X>300)
"=\\frac {300-210}{56}=1.607"
=1.607
P(z>1.607) =0.054 from the standard normal tables
=0.054
ii) P(z<100)
"=\\frac{100-210}{56}=-1.964"
P(Z<-1.964)=0.025 from the z tables
=0.025
iii) P(150<X<250)
"z_1=\\frac{150-210}{56}=-1.071"
P(z<-1.071)=0.142 from the standard normal tables
"z_2=\\frac{250-210}{56}=0.714"
P(z<0.714)=0.762 from the z tables
=0.762-0.142
=0.620
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