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    A certain factory has three machines A,B and C that produces one type of light bulbs.Past experience has shown that the lifetime of a light bulb by machine A can be modeled as an exponential random variable with an average of 20 days, whereas the lifetime of a bulb produced by a machine B can modeled as a normal random variable with mean of 15 days and variance of 5.76(days^2).As for the light bulbs produced by machine C, they usually last anywhere between 13 to 17 days.It is further known that 30% of the factory production of this type of light bulbs comes from machine A and 50% of production comes from machine B.Suppose all the light bulbs after production are packaged identically and stored in the same storage room.
    a)A bulb is selected at random from the storage room and tested.Find the probability that the bulb was produced by machine C given that it was still operating after 16 days.
    b) A random sample of size 10 bulbs is selected from the storage room, what is the probability that exactly 4 of the bulbs will have a lifetime less than 16 days?
    c)on average ,how many bulbs must be selected in order to get 7 bulbs with a lifetime between two and three weeks?
    d)Find the expacted lifetime of a lifetime of a light bulb produced by this factory.Hint:use the Law Of Total Expectation.

    © BrainMass Inc. brainmass.com December 24, 2021, 5:06 pm ad1c9bdddf
    https://brainmass.com/math/probability/probability-lifetime-light-bulbs-28063

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    A certain factory has three machines A,B and C that produces one type of light bulbs.Past experience has shown that the lifetime of a light bulb by machine A can be modeled as an exponential random variable with an average of 20 days, whereas the lifetime of a bulb produced by a machine B can modeled as a normal random variable with mean of 15 days and variance of 5.76(days^2).As for the light bulbs produced by machine C, they usually last anywhere between 13 to 17 days.It is further known that 30% of the factory production of this type of light bulbs comes from machine A and 50% of production comes from machine B.Suppose all the light bulbs after production are packaged identically and stored in the same storage room.
    a)A bulb is selected at random from the storage room and tested.Find the probability that the bulb was produced by machine C given that it was still operating after 16 days.
    b) A random sample of size 10 bulbs is selected from the storage room, what is the probability that exactly 4 of the bulbs will have a lifetime less than 16 days?
    c)on average ,how many bulbs must be selected in order to get 7 bulbs with a lifetime between two and three weeks?
    d)Find the expacted lifetime of a lifetime of a light bulb produced by this factory.Hint:use the Law Of Total Expectation

    a)A bulb is selected at random from the storage room and tested.Find the probability that the bulb was produced by machine C given that it was still operating after 16 days.
    Machine A:
    machine A can be modeled as an exponential random variable with an average of 20 days

    The exponential random variable is represented as :
    P(x)= lambda x e -lambda / x!
    where
    P(x)= probability of x
    lambda = average

    To calculate the probability that the bulb is operating after 16 days we need to calculate the probabilities of failure in 0,1,2,3---16 days
    Here lambda = 20

    x=
    0 P(0)= 2.06115E-09
    1 P(1)= 4.12231E-08
    2 P(2)= 4.12231E-07
    3 P(3)= 2.7482E-06
    4 P(4)= 1.3741E-05
    5 P(5)= 5.49641E-05
    6 P(6)= 0.000183214
    7 P(7)= 0.000523468
    8 P(8)= 0.001308669
    9 P(9)= 0.002908153
    10 P(10)= 0.005816307
    11 P(11)= 0.010575103
    12 P(12)= 0.017625171
    13 P(13)= 0.027115648
    14 P(14)= 0.03873664
    15 P(15)= 0.051648854
    16 P(16)= 0.064561067
    Total= 0.22107

    Therefore the probability of bulb lasting more than 16 days=
    1- {P(0)+ P(1)+P(2) + ----+ P(16)}= 0.7789 =1- 0.22107

    Machine B

    lifetime of a bulb produced by a machine B can modeled as a normal random variable with mean of 15 days and variance of 5.76(days^2).

    Mean=M = 15.00 days
    Variance =s^2= 5.76 days^2
    Standard deviation =s= 2.40 =square root of 5.76

    We need to find the probability that the bulb is operating after 16 days
    x= 16.00
    z=(x-M )/s 0.4167 =(16-15)/2.4
    Cumulative Probability corresponding to z= 0.4167 is= 0.6616
    Or Probability corresponding to x< 16.00 is Prob(Z)= 0.6616
    Therefore probability corresponding to x> 16.00 is 1-Prob(Z)= 0.3384 =1-0.6616

    Machine C
    for the light bulbs produced by machine C, they usually last anywhere between 13 to 17 days.

    If we assign equal probabilities to days 13,14,15,16 and 17 ( 5days) the probability for the bulb to last mre than 16 days

    The probability that the bulb is operating after 16 days ie in its 17th day= 1/5= 0.2

    Machine % of factory production probability of lasting more than 16 days ( calculated above) % of factory production x Probability of a bulb lasting more than 16 days
    A 30% 0.7789 0.2337 =30%*0.7789
    B 50% 0.3384 0.1692 =50%*0.3384
    C 20% 0.2 0.04 =20%*0.2
    0.4429

    Therefore total probability of bulb lasting more than 16 days= 0.4429

    Probability of bulb lasting more than 16 days and being produced by machine C= 0.04

    Therefore probability that bulb was produced by machine C given that it was still operating after 16 days=
    = 0.0903 =0.04/0.4429
    or 9.03%

    Answer: 9.03%

    b) A random sample of size 10 bulbs is selected from the storage room, what is the probability that exactly 4 of the bulbs will have a lifetime less than 16 days?

    Probability that bulb has lifetime more than 16 days= 0.4429
    Therefore probability that bulb has lifetime less than 16 days= 0.5571 =(1-0.4429)

    This problem can be solved using Binommial distribution
    P(r)= ncr p r * q n-r

    Here

    n= 10
    p= 0.5571
    q=1-p= 0.4429
    r= 4
    P(r)= ncr p r * q n-r = 0.1527 or 15.27%

    Answer: probability that exactly 4 of the bulbs will have a lifetime less than 16 days= 15.27%

    c)on average ,how many bulbs must be selected in order to get 7 bulbs with a lifetime between two and three weeks?

    Machine A:

    To calculate the probability that the bulb with a lifetime between 2 to 3 weeks we need to calculate the probabilities of failure in 15,16,17,18,19,20,21 days
    Here lambda = 20

    x=
    15 P(15)= 0.051648854
    16 P(16)= 0.064561067
    17 P(17)= 0.075954196
    18 P(18)= 0.084393552
    19 P(19)= 0.088835317
    20 P(20)= 0.088835317
    21 P(21)= 0.084605064
    Total= 0.53883

    Therefore the probability of bulb with a lifetime between 2 to 3 weeks= 0.53883

    Machine B

    lifetime of a bulb produced by a machine B can modeled as a normal random variable with mean of 15 days and variance of 5.76(days^2).

    Mean=M = 15.00 days
    Variance =s^2= 5.76 days^2
    Standard deviation =s= 2.40 =square root of 5.76

    We need to find the probability of bulb with a lifetime between 2 to 3 weeks or x= 14 to x= 21
    x= 14.00
    z=(x-M )/s -0.4167 =(14-15)/2.4
    Cumulative Probability corresponding to z= -0.4167 is= 0.3384
    Or Probability corresponding to x< 14.00 is Prob(Z)= 0.3384

    x= 21.00
    z=(x-M )/s 2.5 =(21-15)/2.4
    Cumulative Probability corresponding to z= 2.5 is= 0.9938
    Or Probability corresponding to x< 21.00 is Prob(Z)= 0.9938

    Therefore probability of a bulb lasting between 2 to 3 weeks= 0.6554

    Machine C
    for the light bulbs produced by machine C, they usually last anywhere between 13 to 17 days.

    If we assign equal probabilities to days 13,14,15,16 and 17 ( 5days) the probability for the bulb to last 15,16,17 dyas

    The probability that the bulb is operating between 2 to 3 weeks = 3/5= 0.6

    Machine % of factory production probability of lasting more than 16 days ( calculated above) % of factory production x Probability of a bulb lasting more than 16 days
    A 30% 0.53883 0.1616 =30%*0.53883
    B 50% 0.6554 0.3277 =50%*0.6554
    C 20% 0.6 0.12 =20%*0.6
    0.6093

    Therefore total probability of bulb lasting between 2 to 3 weeks= 0.6093 or 60.93%

    Expected no = Probability x Total no of bulbs

    Expected no = 7
    Probability= 0.6093

    Therefore total no of bulbs to be selected=Expected no / Probability = 11.49

    Which when rounded up= 12

    Answer: 12 bulbs

    d)Find the expacted lifetime of a lifetime of a light bulb produced by this factory.Hint:use the Law Of Total Expectation
    Mean life of a bulb produced by
    Machine A 20 (Given)
    Machine B 15 (Given)
    Machine C 15 Life is between 13 to 17 days

    Machine % of factory production mean life= % of factory production x Mean life=
    A 30% 20 6 =30%*20
    B 50% 15 7.5 =50%*15
    C 20% 15 3 =20%*15
    16.5

    expacted lifetime of a lifetime of a light bulb produced by this factory= 16.5

    Answer: 16.5 days

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 5:06 pm ad1c9bdddf>
    https://brainmass.com/math/probability/probability-lifetime-light-bulbs-28063

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