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788 OPERATIONS RESEARCH V4 IL (45-15) 30 30 p= = 0.462. IP + IL (80-45) + (45-15) 35 + 30 65 This value of probability means that the vegetable seller must be 0.462 sure of selling at least an additional unit before it would pay him to stock that unit. In Fig. 9.1 is shown the 0.462 probability on the normal distribution curve of the past demand. The required 0.462 probability is the shaded area. The vegetable seller should stock additional boxes until he reaches point A; if he stocks more units, the probability will fall below 0.462. Z=1 u = 30 =9 538 of Area 462 of Areal 30 Fig. 9.1 Since the shaded area under the normal curve in Fig. 9.1 is 0.462 of the total area, the open area must be 1.000 -0.462 = 0.538 of the area under the curve. From table C-2 in the appendix, the normal deviatez=0.1 for value of 0.538. This means that point Ais0.1 standard deviation to the right of the mean. Since the standard deviation of the distribution of past demand is 9 boxes, point A can be located as follows: Point A = mean + standard deviation = 30 + 0.1 X 9 = 30.9 - 31 boxes. Thus the fruit seller should stock 31 boxes. 9.8 ADDITIONAL EXAMPLES : EXAMPLE 9.8-1 Under an employment promotion programme, it is proposed to allow sale of newspapers on the buses during off-peak hours. The vendor can purchase the newspapers at a special concessional rate of 25 paise per copy against the selling price of 40 paise. Any unsold copies are, however, a dead loss. The vendor has estimated the following probability distribution for the number of copies demanded: Number of copies 15 16 17 18 19 20 Probability 0.04 0.19 0.33 0.26 0.11 0.07 (a) How many copies should he order so that his expected profit will be maximum ? (b) Compute EPPI (e) The vendor is thinking of spending on a small market survey to obtain additional information regarding the demand levels. How much should he be willing to spend on such a survey ? [P.T.U.M.Tech. April , 2012; M.D.U. Rohtak B.E. (Mech.) Dec., 2006]