Explore BrainMass

# Probability- lifetime of light bulbs

Not what you're looking for? Search our solutions OR ask your own Custom question.

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

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.

## SOLUTION This solution is FREE courtesy of BrainMass!

See attached file
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%

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

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