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.
