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TABLE 134 The Distribution, Fio, and Expected hventoryMQ) Functions for the Stan

ID: 375784 • Letter: T

Question

TABLE 134 The Distribution, Fio, and Expected hventoryMQ) Functions for the Standard Nor mal Distribution Function 0968 0455 0561 0686 0833 1004 1.4367 1.5293 1.6232 17 183 9332 0001 000 1 0002 0002 0003 0005 1587 9554 -0.9 1841 9641 18143 0001 2420 2743 3085 3446 9772 9821 9861 9893 2.0085 2.1065 2.2049 2.3037 -0.7 1429 -0.6 0002 0003 0004 0005 0008 0011 0007 2304 2668 3069 3509 3989 4509 5069 5668 6304 6978 7687 2.4027 2.5020 2.6015 .0013 9933 0026 0035 0047 0062 0082 0107 4207 4602 5000 5398 5793 6179 0 9965 2.7011 28008 9974 9981 9987 2.9005 0037 0049 0065 0085 3.0004 3.1003 3.2002 3.3001 6915 9993 7580 7881 8159 9997 9998 9998 .7 3.4001 0228 0287 0359 0446 0548 0668 0808 3.5001 1.0004 1.0833 1.1686 1.2561 1.3455 36000 3.7000 3.8000 3.9000 1.0000 4.0000 1.8 0183 0232 0293 0367 8643 8849 1.0000 Page 401

Explanation / Answer

a)

= 225

= 70

z = (x-)/ = (375-225)/70 = 2.1429

Lookup table 13.4 corresponding to z value as determine above, corresponding F(z) = 0.9839   (derived by interpolation)

Therefore, probability that it will sell more than 375 copies = 1-0.9839 = 0.0161

b)

50% of mean forecast, x = 225*50% = 112.5

z = (112.5-225)/70 = -1.6071

Using Excel, corresponding F(z) = 0.0540

Therefore, probability that this expose will be a dog = 0.0540

c)

Upper limit of 20% of mean forecast = 225*(1+20%) = 270

For upper limit, z-statistic = (270-225)/70 = 0.6429

Using Table 13.4, corresponding F(z) = 0.7398   (calculated from the table by using interpolation)

Lower limit of 20% of mean forecast = 225*(1-20%) = 180

For upper limit, z-statistic = (180-225)/70 = -0.6429 (negative)

Using Table 13.4, corresponding F(z) = 0.2602   (calculated from the table by using interpolation)

Probability that demand will be within 20% of mean forecast = 0.7398 - 0.2602 = 0.4797

d) Underage cost, Cu = 20-11 = 9

Overage cost, Co = 5 (shipping cost incurred in ruturning)

Critical ratio, F(z) = Cu/(Cu+Co) = 9/(9+5) = 0.6429

Using Table 13.4, Corresponding value of z = 0.3661

Optimal quantity = + z = 225 + 0.3661*70 = 251