Continuous Distribution Problems for Reliability in Engineering – Solutions.
    1. The lifetime of all Everglo bulbs is normally distributed with mean 400 hours and standard deviation 50 hours.
    a. What percentage of the bulbs will burn out in less than 360 hours?
    b. After how many hours will 80% of the bulbs burn out?

Let X = lifetimes. Given X ~ N(400,50)

    a. We want P(X < 360) =NORMDIST(360,400,50,1)=0.2119

        Alternatively by hand and tables we need to convert X to standard Z using Z = (X – )/:

        P(X < 360) = P[Z < (360 – 400)/50] = P[Z < -0.80] = 0.2119 obtained directly from looking up in Normal tables.

    b. We want k such that P(X < k ) = 0.80; use NORMINV(0.8,400,50) = 442.1 hours

        Alternatively first compute a z value to satisfy P(Z < z1 ) = 0.80. Looking in the body of normal tables for 0.8000
        we get the closest number to be 0.7995 yielding z1 = 0.74. Now form an equation connecting z1 and k then solve
        for k: z1 = 0.74 = (k – 400)/50. We have k – 400 = 0.74*50 = 37 giving k = 400+37 = 437 which is within rounding
        of 442.1

    2. High-Tech Inc., produces an electronic component, GX-7, that has an average life span of 4500 hours. The
       lifespan is normally distributed with a standard deviation of 500 hours. The company is considering a 3800 hour
       warranty on the GX-7. If this warranty policy is adopted, what proportion of GX-7 components should High-Tech
       expect to replace under warranty?

Let X = lifetimes. Given X ~ N(4500,500) and k = 3800

    a. We want P(X < 3800) =NORMDIST(3800,4500,500,1)=0.081. Thus High-Tech can expect to replace about 8% of
       the GX-7 components.

    3. The inside diameter of a manufactured cylinder is believed to follow a normal distribution centered on 5.00
       inches with a standard deviation of .03 inches. The diameter of the cylinder must be within 4.95 inches (the
       lower spec limit) and the 5.07 inches (the upper spec limit). If the diameter is outside of these limits then the
       cylinder is considered to be nonconforming. What proportion of the manufactured cylinder is nonconforming?

Let D = Diameter. Given D ~ N(5,0.03)

    a. We want P(4.95 < D < 5.07) =NORMDIST(5.07,5,0.03,1) - NORMDIST(4.95,5,0.03,1) =0.9424. This is the
       proportion conforming. Proportion NON-conforming = 1 – 0.9424 = .0576 ~ 5.8%

    4. The thickness of a manufactured sheet of metal is believed to follow a normal distribution with a mean of .30
       inches and a standard deviation of .02 inches. The lower spec. limit is 2.5 standard deviations below the mean.
       The upper spec. limit is 3.0 standard deviations above the mean. Nonconforming sheets of metal have
       thicknesses outside of theses tow spec limits. What proportion of the sheets of metal is nonconforming?

Let T = thickness of sheet. Given T ~ N(0.3,0.02)

    a. We want P(-2.5*sigma < T < 3.0*sigma) =NORMSDIST(3.0) - NORMSDIST(-2.5) =0.9924. This is the proportion
       conforming. Proportion NON-conforming = 1 – 0.9924 = .00756 ~ 0.76%

    5. The time intervals (in operation hours) between successive failures of air conditioning equipment in certain
       aircraft are believed to follow an exponential distribution. Assume that the mean time between failures is 300
       hours. What is the probability that air conditioning equipment in a particular aircraft will fail after operating 200
       Let T = time between failures. Given T~ Expon(1/300). Note that mean is reciprocal of the parameter lambda
       and vice-versa. We want P(T > 200) = 1 - P(T < 200) =1 - EXPONDIST(200,1/300,1) =1 - 0.4866 = 0. 5134

    6. Clearvision Company manufactures picture tubes for color TV sets and claims that the life spans of their tubes
       are exponentially distributed with a mean of 1800 hours. What percentage of the picture tubes will last no
       more than 1600 hours?

Let T = life span of TV sets. Given T~ Expon(1/1800). Note that mean is reciprocal of the parameter lambda and vice-
versa. We want P(T < 1600) =EXPONDIST(1600,1/1800,1) =0.5889.

    7.   United Motors claims that one of its cars, the Starbird 300, gets city driving mileages that are normally
         distributed with a mean of 20 mpg. and a standard deviation of 1 mpg. Let x denote the city driving mileage of
         a randomly selected Starbird 300.
         a. Assuming United Motors’ claim is correct, find P(x ≤ 27).
         b. If you purchase a Starbird 300 and your car gets 27 mpg in city driving, what do you think of United Motors’
             claim? Explain your answer.

    a) Given X ~ N(20,1) We want P(X < 27) = NORMDIST(27,20,1,1) = 1.000

    b) Based on the assumption of N(20,1) P(X >= 27) = 0.0000 which implies that it is impossible to get a mileage as
       extreme as 27 mpg if the assumption about the average is correct. Thus we have reason to doubt the claim by
       United Motors and conjecture that the true average mpg is perhaps closer to 27 than to 20.

   8.    A tire company has developed a new type of steel-belted radial tire. Extensive testing indicated the population
        of mileages obtained by all tires of this new type is normally distributed with a mean of 40,000 miles and a
        standard deviation of 4,000 miles. The company wishes to offer a guarantee providing a discount on a new set
        of tires if the original ties purchased do not exceed the mileage stated in the guarantee. What should be the
        guaranteed mileage if the tire company desires that no more than 2% of the tires will fail to meet the
        guaranteed mileage?

Let X = tire mileage. Given X ~ N(40000,4000)

   a. We want W such that P(X < W ) = 0.02; use NORMINV(0.02,40000,4000) = 31,785 miles

   9. The maintenance department in a factory claims that the number of breakdowns of a particular machine follow
      a Poisson distribution with a mean of two breakdowns every 500 hour. Let x denote the time (in hours)
      between successive breakdowns.
      a. Assuming the relationship between the given Poisson distribution and an exponential distribution, find the
          probability that the time between successive breakdowns is at most 5 hours.
      b. Making the same assumption, find the probability that the time between successive breakdowns is between
          100 and 300 hours.
      c. Suppose the machine breaks down five hours after its most recent breakdown. Based on your answer to
          part a, do you believe the maintenance department’s claim? Explain.

   a) Let X = time between breakdowns. Given X~ Expon(1/250). Note that 2 per 500 hours implies 1 breakdown
      every 250 hours and that the parameter lambda is its reciprocal. We want P(X < 5) =EXPONDIST(5,1/250,1)

   b) We want P(100 < X < 300) =EXPONDIST(300,1/250,1) – EXPONDIST(100,1/250,1) = 0.6988 – 0.3297 = 0.3691

   c) Since, if the maintenance department’s claims is true there is only a 2% chance that time between failures is at
      most 5 hours we would have reason to doubt the claim and conclude the time between failures is perhaps more
      a smaller number of hours (occurs more frequently).