Question

The lifespan of a fluorescent lamp used in greenhouses is normally distributed (Gauss) with an average of 600 hours and a standard deviation of 40 hours.
a) Calculate the probability that a randomly chosen lamp has a lifespan greater than 695 hours,
b) What times of last life are within the 85th percentile?
c) How long will 30% of the lamps last for the longest?
d) How long will 30% of the lamps last less time?
Ans.
a) 0.0087,
b) 641.6 hrs maximum,
c) 620.8 hrs minimum,
d) 579.2 hrs maximum

Answer 1

Mean, = 600 hours

SD, = 40 hours

Let X denote the lifespan of a flourescent lamp

(a) The probability that a randomly chosen lamp has a lifespan greater than 695 hours = P(X > 695)

= P{Z > (695 - 600)/40}

= P(Z > 2.375)

= 0.0088

(b) Corresponding to 85th percentile, the z value is 1.0365

Thus the last life = 600 + 1.0365*40 = 641.46 hrs maximum

(c) Corresponding to 30% of the lamps with longest life, the z value = 0.5245

Thus the required life = 600 + 0.5245*40 = 620.98 hrs minimum

(d) Corresponding to 30% of the lamps with least time, the z value = -0.5245

Thus the required life = 600 + (-0.5245)*40 = 579.02 hrs maximum