Hi can someone help me? The subject is pricing and revenue management.

Demand # Days Relative (Number of papers (Total days the demand Frequency Probability Probability Probability Sales=x Sales s x Sales > x sold) was sold) O 4 0.04 0.04 96 1 7 0.07 0.71 89 2 9 0.09 0.2 80 3 12 0.12 0.32 13 0.13 0.45 17 0.17 0.62 6 13 0.13 0.75 10 0.1 0.85 8 7 0.07 0.92 9 5 0.05 0.97 10 3 0.03 7 Sum 100 So given this demand, how many papers should Maria order? If she stocks only one newspaper, that paper will generate considerable profit because it will sell almost every day (all days except the four out of 100 days when demand was zero papers). But on many of these days (89% of the time) there is demand for more than one paper, so she is foregoing some potential revenue. Conversely, if she stocks 10 papers, the 10th paper sells infrequently (only 3 out of 100 days). Does the 10th paper cover its cost? Given that on all 100 days Maria purchases the 10th paper for $0.90 (100 days * $.90), so she spent $90 on the 10th paper. But this paper only generated revenue of $4.50 (3 days * $1.50). Thus the 10th paper does not cover its costs. You can generalize this idea with the formula that follows. Assume P is the probability of selling a paper (i.e., the probability that demand is greater than or equal to the number of papers Maria stocks). For a paper to be profitable (or at least break even), the probability that the paper sells times the selling price must equal or exceed the cost of the paper. P * sales price > costs . P *$1.50 = $.90 . P = $0.90/$1.50 = 0.60 As long as the probability that Maria can sell a newspaper is greater than or equal to 0.60, that newspaper will cover its fixed costs and actually make a profit. If Maria stocks one newspaper, she will sell that newspaper whenever the demand is one or greater, or 1 minus the demand for zero, or 96% of the time. If she stocks two newspapers, she sells two newspapers whenever demand is 2 or greater; i.e., not 1 or less, or 89% of the time, and so on.Instructions Scenario Maria. a young entrepreneur. sells newspapers daily at the corner of Campus Road and Hoy Road in Ithaca. NY. Every evening. she purchases newspapers for $0.90 and sells them the next day for $1.50. For example. on Sunday. she orders the papers that she will sell on Monday. Keep in mind that on Tuesday. those Monday newspapers are worthless; the inventory is perishable. Maria commits to a stock of papers prior to realizing demand {which is uncertain]. - If Maria stocks too many papers. some may go to waste. - If Maria does not stock enough papers. she may fail to realize potential revenue. Fortunately. Maria has collected data to help her decide how many papers she should buy. She created a chart showing the number of papers she sold over a 100-day period. The first column lists the demand. or the number of papers sold in a day. The second column lists the number of days that demand was sold. For example. on four different days the demand was zero; Maria did not sell any papers. There were seven days on which she sold one paper. and on nine days she sold two papers. and so on. Maria can now transform these frequencies of sales into relative frequencies or probabilities. To calculate the probability that Maria sells zero papers on any given day. divide the number of days she sold zero papers by the total number of observations: 4H0!) = 0.04 To calculate the probability that she sells one paper or less that is. that she sells1 paper or zero papers add the days zero or one paper was sold [4 + '1'] and divide by the total observations. or add the probability of selling zero papers to the probability of selling one papen [4 + TLJ'I00 = .04 + .0? = 0.11 The probability that she sells more than one paper is the demand for two to ten papers. or1 minus the demand for 01 papers. I - 0.11: 0.39