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SPECIALITY TOYS Case study Assignment 2 Introduction to business statistics

1. Use the sales forecaster’s prediction to describe a normal probability distribution that can be used to approximate the demand distribution. Sketch the distribution and show its mean and standard deviation.

Solution: Specialty’s senior sales forecaster predicted an expected demand of 20,000 units with a .95 probability that demand would be between 10,000 units and 30,000 units. Mean=20,000 (expected demand of 20000 ) Probability P=0.95 Value of z for probability 0.95 based on symmetry of normal distribution curve as shown below is 0.95/2= 0.475 is 1.96 Normal probability distribution function Z=x-µ/σ 1.96=30000-20000/ σ σ (deviation)=5102. Normal Distribution curve

speciality toy rough sheet.xlsx

2. Compute the probability of a stock-out for the order quantities suggested by members of the management team.

suggested quantity

15000 18000 24000 28000

Expected demand=20000 Mean=20000 Standard deviation=5102 unit

order 15000 18000 24000 28000

Z=orderp(order p(out of 20000/5102 quantity) stock>order) -0.98000784 0.1635 -0.392003136 0.3483 0.784006272 0.7823 1.568012544 0.9416

0.8365 0.6517 0.2177 0.0584

Based on above observation of probability of order out of stock we can deduce that As quantity of order increases chances of stock getting out of stock gets lower.

3. Compute the projected profit for the order quantities suggested by the management team under three scenarios: worst case in which sales _ 10,000 units, most likely case in which sales _ 20,000 units, and best case in which sales _ 30,000 units.

Solution:

Cost price=$16 Selling price=$24 Profit normal=24-16=8 Profit for surplus inventory=16-5=-11 i.e loss of $11

order worst case scene(10000) 15000 8*10000-11*5000=25000 18000 8*10000-11*8000=-8000 8*10000-11*14000=24000 74000 8*10000-11*18000=28000 118000

most probable(20000) 8*15000=120000 8*18000=144000 8*2000011*4000=116000 8*2000011*8000=72000

best case(30000) 8*15000=120000 8*18000=144000 8*24000=192000 8*28000=224000

4. One of Specialty’s managers felt that the profit potential was so great that the order quantity should have a 70% chance of meeting demand and only a 30% chance of any stock-outs. What quantity would be ordered under this policy, and what is the projected profit under the three sales scenarios?

Solution:

Z value for 70% chance when demand met is 0.7580. (oq-20000)/5102 = 0.7580. oq = 20000 + 5102 * 0.7580 = 20000 + 3867 oq= 23867 units to be ordered. Hence ordered quantity when 70 percent demand is met is 23867. order

worst case scene(10000) 8*10000-11*13867=23867 72537

most probable(20000) 8*2000011*3867=117463

best case(30000) 8*23867=190936

For worst case scene of 10000 there will be loss of 72537. For most probable scene of 20000 there will be profit of 117463. For best case scene of 30000 there will be profit of 190936.

Q.Provide your own recommendation for an order quantity and note the associated profit projections. Provide a rationale for your recommendation

solution: Z=orderp(order p(out of order 20000/5102 quantity) stock>order) 15000 -0.98000784 0.1635 0.8365 18000 -0.392003136 0.3483 0.6517 24000 0.784006272 0.7823 0.2177

28000

1.568012544

0.9416

0.0584

Based on above observation of probability of order out of stock we can deduce that As quantity of order increases chances of stock getting out of stock gets lower.

order worst case scene(10000) 15000 8*10000-11*5000=25000 18000 8*10000-11*8000=-8000 8*10000-11*14000=24000 74000 8*10000-11*18000=28000 118000

most probable(20000) 8*15000=120000 8*18000=144000 8*2000011*4000=116000 8*2000011*8000=72000

best case(30000) 8*15000=120000 8*18000=144000 8*24000=192000 8*28000=224000

Based on above analysis we can say that probability of stock getting out of stock is lower at 24000 . Also at 20000 profit is maximum and it decreases to 72000 with stock increasing to 28000. Hence stock should be nearer to 20000 and around 24000 for best profitability.