Answer only 4 please! A certain light bulb manufacturer makes two types of light
ID: 3366723 • Letter: A
Question
Answer only 4 please!
A certain light bulb manufacturer makes two types of light bulbs, a low-cost short life (S-type) bulb and a more expensive long life (L-type) bulb. The two types of bulbs look and feel the same so the manufacturer must be sure to carefully mark the boxes of bulbs. A box of bulbs is found on the floor of the plant that (you guessed it) has not been labelled. In order to determine what types of bulbs are in the box, a clever intern suggests that they take out a bulb from the box and run it until it burns out. After observing how long the bulb remains lit until it burns out, they should be able to make a good guess as to what type of bulbs are in the box. It is known that the length of time, X (in hours), that the bulb lasts can be modeled as an exponential random variable. In particular, for the two types of bulbs, the conditional PDFs are given by
where X is the random variable that measures the lifetime of the bulb in hours.The company makes 75% S-type bulbs and 25% L-type bulbs. If we observe that the bulb lasts for 350 hours, what is the probability that the bulb was S-type?
from here down is question 4------------------------------------------------
Continuing on from the previous light bulb problem, suppose that there is a vacation week coming up so that we will not be able to monitor the light bulb while the test is running. Instead, we will turn the light bulb on when we leave for vacation and then check on it again when we return several days later. At that point it will either still be lit or it will have burned out. The amount of time transpired between the beginning and the end of the test will be 230 h and we will not be watching the bulb at any point in time in between. As a result, there are only two possible observations in this experiment, (the bulb burnt out ? {X<230} ) or (the bulb is still lit ? {X>230}).
Given it is observed that the bulb burnt out over the vacation period, what is the probability that the bulb was an S-type bulb? As in the previous problem, assume that 75% of the bulbs are S-type, and 25% are L-type, and that the conditional distributions are exponentials with parameters of 100 for S-type and 1000 for L-type.
A certain light bulb manufacturer makes two types of light bulbs, a low-cost short life (S-type) bulb and a more expensive long life (L-type) bulb. The two types of bulbs look and feel the same so the manufacturer must be sure to carefully mark the boxes of bulbs. A box of bulbs is found on the floor of the plant that (you guessed it) has not been labelled. In order to determine what types of bulbs are in the box, a clever intern suggests that they take out a bulb from the box and run it until it burns out. After observing how long the bulb remains lit until it burns out, they should be able to make a good guess as to what type of bulbs are in the box. It is known that the length of time, X (in hours), that the bulb lasts can be modeled as an exponential random variable. In particular, for the two types of bulbs, the conditional PDFs are given by fxls(x)= TO exp(-100):1(x) and fxlL(x)=1000exp(-1000)u(x) type bulbs and 25% L-type bulbs. If we observe that the bulb lasts for 350 hours, what is the probability where Xis the random variable that measures the lifetime of the bubin hours. The company makes 75% that the bulb was S-type? QUESTION 4 Continuing on from the previous ght bulb problem suppose that there is a vacation week coming up so that we wil not be able to monitor the light bulb while the test is ?nning Instead we w I turn the light bulb on when we leave for vacation and then check on it again when we return several days later. At that point it will either still be lit or it will have burned out The amount of time transpired between the beginning and e end of the test will be 230 h and ill not be watching the bulb at any point in time in between. As a result, there are only two possible observations in this expenment, (the bulb bumt out 230)o (the bulb is still litX>230]). Given it is observed that the bulb burnt out over the vacation period what is the probability that the bulb was a type bulb? As in the previous problem, assume that 75% of the bulbs are S type, and 25% are L type, and that the conditional distributions are exponentials with parameters of 100 for S-type and 1000 for L-type. QUESTION 5 Referring to the previous problem, let the duration of the light bulb test be detemined instead by 310 hours. Suppose now that it is observed that the bulb is still lit at the end of the vacation, what is the probability that the bulb was an S type bulb? As in the p evious problems, assume that 75% o the bulbs are S-type and25% are L trpe, and that he condit ona distributors are e ponen als th parameters of 100 type and 1000 for L-type.Explanation / Answer
Solution Q4 ONLY
Let X and Y represent the life time of S-type, and L-type bulbs respectively.
We are given: X~ Exp(100) …………………………………………………..(1) and
Y ~ Exp(1000) …………………………………………………..(2)
Back-up Theory
If X ~ Exponential with parameter ? (average inter-event time), the pdf (probability density function) of X is given by f(x) = (1/?)e-x/?, 0 ? x < ? …………………………………(3)
CDF (cumulative distribution function), F(t) = P(X ? t) = 1- e-t/? ………………….…(4)
From (2), P(X > t) = e-t/? ……………….………………………………………………(5)
If A and B are two events such that probability of B is influenced by occurrence or otherwise of A, then
Conditional Probability of B given A, denoted by P(B/A) = P(B ? A)/P(A)..….(6)
P(B) = {P(B/A) x P(A)} + {P(B/AC) x P(AC)}………………………………….(7)
P(A/B) = P(B/A) x {P(A)/P(B)}……………………………..………………….(8)
Now, to work out the problem,
Let A be the event that the bulb is S-type. Then, trivially, AC is the event that the bulb is
L-type. Also, let B be the event that the bulb is burnt out before 230 hours. Then,
vide (1) and (4): P(B/A) = 1 - e-230/100 = 1 - e-2.30 = 1 – 0.8997 = 0.1003…………..(9)
vide (2) and (4): P(B/AC) = 1 - e-230/1000 = 1 - e-0.23 = 1 – 0.2055 = 0.7945………..(10)
Also given, P(A) = 0.75 ……………………………………………………(11) and hence
P(AC) = 0.25 …………………………………………………… (12).
So, vide (7), P(B) = (0.1003 x 0.75) + (0.7945 x 0.25) = 0.27385 …………………..(13)
Now, that the bulb burnt out over the vacation period, the probability that the bulb was an S-type bulb = P(A/B)
= P(B/A) x {P(A)/P(B)} [vide (4)]
= (0.1003 x 0.75)/0.27385 [vide (9), (11) and (13)]
= 0.2750 ANSWER