has five (5) processing plants, all receiving raw sugar in bulk. The amount of s

Question

has five (5) processing plants, all receiving raw sugar in bulk. The amount of sugar, X, that one plant can process in one day can be modeled as having an expo- nential distribution, with a mean of 4 (measurements in tons), for each of the 5 plants. (a) Find the probability that a plant process more than 3 tons on a given day. o.+ 72 (b) If the plants operate independently, find the probability that exactly 3 of the 5 plants process more than 3 tons on a given day. o.293 (c) Consider now a particular plant. Find the amount of raw sugar, m, that should be stocked for that particular plant so that P(X > m) = 0.05W.9812

Explanation / Answer

As given in the question, X = quantity (in ton) of raw sugar processed by a single plant in a day.

Also given is that X ~ Exp(4).

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 < ………………………………(1)

CDF (cumulative distribution function), F(t) = P(X t) = 1- e-t/ ……………….…(2)

From (2), P(X > t) = e-t/ ……………….……………………………………………(3)

If X ~ B(n, p). i.e., X has Binomial Distribution with parameters n and p, where n = number of trials and p = probability of one success, then probability mass function (pmf) of X is given by

p(x) = P(X = x) = (nCx)(px)(1 - p)n – x, x = 0, 1, 2, ……. , n ………………………..(4)

Part (a)

We want the probability that a single plant processes more than 3 tons in a day.

This probability = P(X > 3)

= e-3/4

= 0.4724 ANSWER

Part (b)

Given that quantity processed by the 5 plants are independent of each other, if Y = number of plants (out of 5) processing more than 3 tons in a day, then Y ~ B(5, p), where p = probability a single plant processes more than 3 tons in a day which, as found in Part (a) is 0.4724

Thus, the required probability = P(Y = 3)

= 5C3(0.4724)3(0.5276)2 [vide (4) of Back-up Theory]   

= 0.2935 [using Excel Function on Binomial Probability] ANSWER

Part (c)

We want the value of m such that P(X > m) = 0.05.

Vide (3) of Back-up Theory, the above implies, e-m/4 = 0.05.

Taking natural logarithm, - m/4 = ln(0.05) = - 2.99573

=> m = 2.99573 x 4 = 11.98 ANSWER

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