Statistical Methods Issues - Assignment Example

The transformation has reduced the value of mean and standard deviation. Both are the non-linear transformation that leads to reduced skewness of the distribution of data.Since, µ = 60 and σ =10, therefore any negative value of X will be more than 6 standard deviations from the mean and the probability below that will be zero. For example: Pr{X < -10}Using Excel function NORMDIST area below Z = 1 is 0.8413.Therefore, 0.8413 (or 84.13%) of bags will weigh less than 516 grams.What proportion of bags weighs between 476 and 516 grams?

Using Excel function NORMDIST area below Z = 1 and Z = -1.5 are 0.8413 and 0.0668.Area between Z equal to -1.5 to 1 = 0.8413 – 0.0668 = 0.7745Therefore, 0.7745 (or 77.45%) of bags will weigh between 476 and 516 grammes.If the manufacturer wants to keep the mean weight at 500 grams but wishes to adjust the production process so that 2.5% of bags weigh less than 495 grams, calculate the value the standard deviation should take.Using Excel function NORMSINV the value of Z for 0.025 (or 2.5%) is -1.96.The value the standard deviation should take will be approximately 2.55.Paperback: = 240, and = 12Hardbacks: = 300, and = 16a) The probability that one paperback and one hardback together weigh less than 570 grammes, andCombined weight = 570 grammesCombined mean, µ= + = 240 + 300 = 540Standard deviation, σ = = Using Excel function NORMDIST area below Z = 1.5 is 0.9332.

Hence, Pr{combined weight < 570 gm} = 0.9332There is 93.32 percent chance that one paperback and one hardback together weigh less than 570 grammes.b) The probability that a selection of 9 different hardbacks weighs between 2676 and 2772 grammes, andCombined mean of 9 hardbacks = = 2700 grammesStandard deviation, σ = = 48Using Excel function NORMDIST area below Z = -0.5 and Z = 1.5 are 0.3085 and 0.9332.Hence, Pr{2676 < 9 different hardbacks < 2772 gm} = 0.9332 – 0.3085 = 0.

6247There is a 62.47 percent chance that a selection of 9 different hardbacks weighs between 2676 and 2772 grams.c) 95% limits for the weight of a random selection of 4 different paperbacksThe bookshop also supplies other bookshops, which usually order multiple copies of the same book. Using Excel function NORMSINV, Z value for 95% limit is 1.645 Combined mean weight of 4 paperbacks = 4 = = 96095% limits for the weight of a random selection of 4 different paperbacks is 999.48 grams.d) One such order is for 4 copies of a particular paperback.

Find 95% limits on the weights of such orders, and explain why are these different from the 95% limits in (c).Using Excel function NORMSINV, Z value for 95% limit is 1.645 Average weight of 4 particular paperbacks = = 240Standard deviation, σ = = 249.87 grammes.95% limits for the average weight of 4 copies of a particular paperback is 249.87 grams.These are different from part c because here 95% limits on average weight for 4 particular order is asked whereas in part c, 95% limits for the weight of a random selection of 4 different paperbacks is asked.

µ = 3000, and σ = 300, cost of stand = £3000, other cost = £2700p (order) = 0.05, profit on each order = £50The probability that attending the fair generates enough orders to cover your costs, Total cost = cost of stand + other cost = £3000 + £2700 = £5700Number of order required to cover total cost = 5700/50 = 114Therefore, number of enquiry = 114/0.05 = 2280Using Excel function NORMDIST area below Z = -2.4 is 0.0082.Hence, Pr{Enquiries > 2280} = 1 – 0.0082 = 0.9918There is 99.

18 percent chance that attending the fair generates enough orders to cover my costs.The probability that attending the fair leads to a profit of more than £2150Profit + Total cost = £2150 + £5700 = £7850Number of order required to cover profit = 7850/50 = 157Therefore, number of enquiry = 157/0.05 = 3140Using Excel function NORMDIST area below Z = 0.4667 is 0.6796.Hence, Pr{Enquiries > 3140} = 1 – 0.6796 = 0.3204There is a 32.04 percent chance that attending the fairleads to a profit of more than £2150.