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Sagot :
With a mean of 1000 and a standard deviation of 600, the probability that the demand is going to be withing 25 percent of its forecast is 0.3230.
a. Mean = 1000
sd = 600
p(1000x 1-25%) - p(1000x 1+25%)
using the z test
d-μ/σ
[tex]p(z < \frac{1250-1000}{600} )-p(z < \frac{-1000}{600} )\\\\(z < 0.4167)-(z < -0.4167)[/tex]
find values using excel sheet formula
NORMSDIST(0.4167) - NORMSDIST(-0.4167)
=m0.6615 - 0.3385
= 0.3230
b. The probability that the forecast would be more than 40 percent
1000x 1+40%
= p(D>1400)
= 1- NormDist(0.667)
= 0.225
c. Cu = 121-72 = 49
Co = 72.50 = 22
The critical ratio calculation
49/22 +49 = 0.6901
Normsinv(0.6901) = 0.496
1000+0.496x600
= 1297
The units that Flextrola has to order is 1297.
d. The expected sales of Flextrola
1200-1000/600
= 0.3333
loss function from z = 0.333 is 0.254
600x0.254 = 152. 4
1000-152.4 = 847.6
the expected sales are 847.6
e 1200- 847.6
= 352. 4
The units of inventory that can be sold is 352.4
Read more on inventory here: https://brainly.com/question/24868116
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