Determine the probability that the bulb will burn in between 778 and 834 hours.

If the manufacturer of electrical firm has the life bulbs that have the life before the burn-out. This means that they are normally distributed with the mean and equal to 800 hours with the standard deviation of 40 hours. Determine the probability that the bulb will burn in between 778 and 834 hours.

Solution:

The light bubs distribution is given at the figure above. The z-values has corresponds to the x1 which is equal to 778 and x2 is equal to 834:

Z1  = 778 - 800  = -0.55

               40

And

P(778< x < 834) = P(-0.55 < z < 0.85)

                              = P (z < 8.85) – P(z<-0.55)

                              = 0.8023 – 0.2912

                               = 0.5111

Credit:ivythesis.typepad.com

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