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Exponential Problem

This analysis focuses on the average life of a particular car battery, which follows an exponential distribution. It explores various probabilities and percentiles related to battery longevity.

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Exponential Problem

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  1. Exponential Problem The average life of a particular car battery is 40 months. The life of a car battery, measured in months, is known to follow an exponential distribution. • X = the life of a particular car battery, measured in months. •  = mean = 40 m = 1/  = 1/40 = 0.025 • X ~ Exp( 1/40 ) or Exp(0.025) • Graph of the probability distribution

  2. Exponential Problem • On average, how long would you expect one car battery to last?  = 40 months • On average, how long would you expect 3 car batteries to last, if they are used one after another? 3 car batteries last (3)(40) = 120 months

  3. Exponential Problem • Find the probability that one of the car batteries lasts more than 36 months. P(X > 36) = 1 – e-36*1/40 = 0.5934

  4. Exponential Problem • Find the probability that one of the car batteries lasts less than 36 months. P(X < 36)

  5. Exponential Problem • Find the probability that one of the car batteries lasts between 36 and 40 months. • Subtract the area to the right of 40 FROM the area to the right of 36. P(36 < X < 40) = e-36/40 – e-40/40 = 0.0387

  6. Exponential Problem • Find the 90th percentile. (90% of all batteries last less than this number of months.) • Let k = the 90th%ile • Area to the LEFT = 0.90 Percentile Formula: k = LN(1 – Area LEFT) = -m LN(1-0.9)/(-1/40) = 92.10 months

  7. Exponential Problem • 70% of the batteries last at least how long? • Let k = the 30th percentile (0.30 = Area to the LEFT) • Use the percentile formula Percentile Formula: k = LN(1 – Area LEFT) = -m LN(1-0.30)/(-1/40) = 14.27 months

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