Essays on Probability Distributions Assignment

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The paper "Probability Distributions" is a perfect example of a business assignment. In the event that the 60% figure was used in the projection of the expected future performance of the business. Then there will obviously be a variation in the probability distribution. In assessing the previous probability distribution, it is clear that the probability of the distribution followed a uniform normal distribution with a high concentration of the contracts for the special disk transcription demand lying between 2 to 5. Similarly, with a 60% increase in the demand, the percentages will change with relative changes being a reflection of the previous probabilities, the only difference being a decrease with a factor of 1.6 as shown below Demand For Special.

Disk Transcription Probability 0 0.08 1 0.16 2 0.32 3 0.64 4 0.32 5 0.032 6 0.032 7 or more 0.016 3 From this table, the expected number of customers wishing for the special facilities would be determined by finding the average of the total customers that have already requested for this facility. Out of 10, it is obtainable that the number of customers seeking these services increases consistently at the 3rd number. Consequently, multiplying this by a factor of 1.6 we obtain 1.6 x3 = 4.8/10 Therefore on average, the number of customers wishing for these facilities is 5. 4. The annual expected income following table 2 will be as follows: Demand For Special.

Disk Transcription Estimated number of transcriptions sold 0 24 1 48 2 96 3 128 4 96 5 10 6 10 7 or more 5 The total number of transcriptions sold stands at a total of 417. From these, we only take 60%, the projected analysis that was suggested from the total figure. This gives a total of 250 expected transcriptions. Having in mind that the transcriptions are being sold at an average cost of £ 100, we can safely determine that the annual expected income from the facilities will be £ 100 x 250 -£ 25,000 6 A) i) From the analysis of the data given in the question, it is clear that any of the rods produced will be of different lengths.

As such, to effectively determine the number of rods that meet the requirements specified as the average, we will require to have an approximate figure of what we deem to be the range of an average metal rod. From the question, an average metal rod will be of the lengths between 14.8 and15.3cm.

This indicates that within this range, then the metal rods will be good for sale. From these measurements, then we can deduce that there is a range of 15.3 – 14.8 = 0.5 This range indicates the possible number of nails that would fall under the category of normal given a normal distribution. Further, considering that each value stands for a value for the nails, the range also intimates that in every number of unit of metal rods, there will be 5 normal nails that are good for sale.

With this information, it is then possible to calculate the value of the proportion of nails sold as either crap or that require to be shortened. This will involve stretching the values for the range to determine the total number of nails that fall out of this range. From the figure given for the standard deviation, we can determine the average proportion of the rods that are sold as crap to be (0.2/14.8) x 5 = 6.76% this means that out of the nails that will be produced, there will be 6% of the nails that will be sold like scrap. Consequently, we follow the same calculations in determining the number of nails that will require to be shortened.

These will be calculated from the value of the standard deviation and the highest value that the nails attain. (0.2/15.3) x 5 = 6.53% This, therefore, means that of the nails that will be manufactured, 6.5% of the nails will be longer than the anticipated average which falls under the range of 14.8 to 15.3iii) If the firm wants to have a figure of less than 5% of the metal rods to be scrapped, then we need to recalculate the percentages of the number of the rods to be scrapped and work backward.

This will give us an equation that will take the form 0.05 = 5 x (0.2/Y). Here, we shall seek to find the value of Y. In making Y the subject of the formula, we have 0.01 = (0.2/Y)Y = 0.2/0.01Y = 20cmTherefore, in order to achieve this set percentage for the metal rods, then the rods are required to at least have a mean that averages around 20 cm.

B. i.) In determining the probability that all the candidates passed to the second interview, we shall have to consider the probability of all successive events. In the calculation of probabilities if all the events turned out as expected, then the probabilities become small as the chances increase. In this case, the total number of events is 6. This represents the number of candidates that took the interview. If the probability of each candidate passing the exam is 2/3, then the probability of all the candidates passing the interview will be calculated as: (2/3)^6 this value is 2/3 raised to power 6 = 0.0878ii. ) In calculating the probability that exactly five candidates pass in ti the second interview, we shall embark on the use of combinations to effectively calculate this probability.

This shall allow for the probability of five passing the interview while on flops the same second interview. The use of combinations is majorly used where the issue of the order in which the events occurred is not a priority. As in this case, the order in which the interviewees conduct the interview is not of importance.

What is really necessary is the probability. Therefore, 6C5/ (5! 1!)The division gives 6/120 = 1/20iii. ) The last question wants us to determine the probability of less than four passing to the second interview. This will involve calculating the probability of four passing and two failings in the second interview. Therefore, the calculations will look pretty similar with the only difference being the frequency of occurrence: 6C4/ (4! 2!)The division of this gives us 15/48 = 5/16.

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