In a Q ​system, the demand rate for strawberry ice cream is normally​ distributed, with an average of 305 pints per week. The lead time is 5 weeks. The standard deviation of weekly demand is 14 pints. Refer to the standard normal table for​ z-values.
a. The standard deviation of demand during the 5​-week lead time is ______ pints. ​(Enter your response rounded to the nearest whole​ number.)
b. The average demand during the 6-week lead time is _____pints. ​(Enter your response as an​integer.)
c. The reorder point that results in a​cycle-service level of 96 percent is _____pints. ​(Enter your response rounded to the nearest whole​ number.)
Z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09
0.0 0.5000 0.5040 0.5080 0.5120 0.5160 0.5199 0.5239 0.5279 0.5319 0.5359
0.1 0.5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636 0.5675 0.5714 0.5754
0.2 0.5793 0.5832 0.5871 0.5910 0.5948 0.5987 0.6026 0.6064 0.6103 0.6141
0.3 0.6179 0.6217 0.6255 0.6293 0.6331 0.6368 0.6406 0.6443 0.6480 0.6517
0.4 0.6554 0.6591 0.6628 0.6664 0.6700 0.6736 0.6772 0.6808 0.6844 0.6879
0.5 0.6915 0.6950 0.6985 0.7019 0.7054 0.7088 0.7123 0.7157 0.7190 0.7224
0.6 0.7258 0.7291 0.7324 0.7357 0.7389 0.7422 0.7454 0.7486 0.7518 0.7549
0.7 0.7580 0.7612 0.7642 0.7673 0.7704 0.7734 0.7764 0.7794 0.7823 0.7852
0.8 0.7881 0.7910 0.7939 0.7967 0.7996 0.8023 0.8051 0.8079 0.8106 0.8133

Respuesta :

Answer: a. 31.304. b. 1525. c. 1589.17

Explanation:

Lead time = 5 weeks

Standard deviation of weekly demand = 14 pints

a. ✓L × Standard deviation Weekly

= ✓5 × 14

= 2.236 × 14

= 31.304

b. Average demand during the 5-week lead time will be:

= Leadtime × weekly demand

= 5 × 305

= 1525

c. Note that the Z value at 96% service level is 2.05

R=dL+z*sd*sqrt(L)= (305 × 5)+ (2.05 × 14 × ✓5)

= 1525 + 64.17

= 1589.17