ESS210B
Prof. Jin-Yi Yu
Example 2
An electrical firm manufactures light bulbs that have a length of life that is normally distributed with mean equal to 800 hours and a standard deviation of 40 hours. Find the possibility that a bulb burns between 778 and 834 hours.
The Z values corresponding to x1 = 778 and x2 = 834 are
                    z1 = (778 – 800) / 40 = -0.55
                    z2 = (834 – 800) / 40 =0.85
Therefore,  P(778 < X < 834)
                = P(-0.55 < Z < 0.85)
                = P(Z <0.85) – P(Z<-0.55)
                = 0.8849 – 0.2912
                = 0.5111