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1) A computer repair company uses a particular spare part at an average rate of 5 per week. It is assumed that the request for the spare part occurs randomly and independently.
i) The manager decides to replenish the stock of this particular spare part to a constant level k at the beginning of each week. Find the least value of k such that on average there will be insufficient stock no more than once in a 26-week period.(k=9)
2)An internet service provider serves 5000 subscribers and during off peak period, there is a probability of 0.04 for each subscriber that he requires a line for a correction.
Using a suitable approximation, find the least number of lines required to be set up by the internet service provider such that the probability that a connection will fail during an off peak period due to all the lines being occupied is less than 1%.(232)
Thx to those who can help. Sometimes i cannot understand this qns as they are so wordy
Edited by lavastar 09 Jun `08, 11:54PM
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1) This question I don't understand what they are asking. So instead I'm using the answer k=9 and trying to work with that... and still cannot get the right answer lol. Just showing my workings below.
Looks like a Poisson distribution to me.
Let the no. of requests for spare parts be X.
X~Po(5)
P(Insufficient stock)
= P(X>k)
= 1 - P(X<=k) ***1 - [poissoncdf (5,k)]***
= Ans
Let the no. of times there is insufficient stock in a 26-week period be Y.
Y~B(26,Ans)
P(Y<=1) > ? ***binomcdf (26,Ans,1)***
The last part, I don't know what the question means by "on average".
2) Have to use normal approximation from binomial.
X~B(5000,0.4)
=> X~N(200,192) approximately
P(Z>a) < 0.01
P(Z<=a) > 0.99
***invNorm(0.99,200,sqrt192)***
= 232.235
Edited by Something random 10 Jun `08, 2:11AM
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1) A computer repair company uses a particular spare part at an average rate of 5 per week. It is assumed that the request for the spare part occurs randomly and independently.
i) The manager decides to replenish the stock of this particular spare part to a constant level k at the beginning of each week. Find the least value of k such that on average there will be insufficient stock no more than once in a 26-week period.(k=9)Let X be the average rate at which the company uses a particular spare part in 1 week.
X~Poi(5)
This simply means the average rate at which the company is using.
Now, k is the number of spare parts at the beginning of each week.
To have insufficient stock, the probability is P(X>k) = 1-P(X<=k)
Using GC (I don't know about TI but my CASIO fx-9860G could do it), generate a list of 1 - P(X<=k)
Now generate another list of 26[1-P(x<=k)]
You'll see that k is 9, where 26[1-P(x<=K)] is below 1.
2)An internet service provider serves 5000 subscribers and during off peak period, there is a probability of 0.04 for each subscriber that he requires a line for a correction.
Using a suitable approximation, find the least number of lines required to be set up by the internet service provider such that the probability that a connection will fail during an off peak period due to all the lines being occupied is less than 1%.(232)Let X is the number of subscribers that requires a line for a correction out of 5000 during an off peak period.
X~Bin(5000,0.04)
Since np > 5, nq > 5, npq > 5, X can be approximated to a normal distribution.
X~N(np, npq)
X~N(200, 192)
P(X>a) < 0.01, where a is the number of lines set up.
Remember to apply continuity correction. P(X>a+1) = P(X>a+0.5)
Using GC, a+0.5 = 232.234822
a=231.734822
By looking at the normal distribution curve (important), and since number of lines is a whole number, the least number is 232.
Edited by secretliker 12 Jun `08, 12:57AM
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