Question

Gadget Inc. needs to improve production levels to meet the current customer demand of 5,300 units/week without the need for overtime. One supervisor proposed to give a bonus payment to the employees each week the production reaches the top 5% of current levels. Current production levels are normally distributed with a mean of 5,000 units/week and a standard deviation of 200 units/week. a) What percentage of weeks is currently Gadget Inc. being able to at least meet customer demand without overtime? b) How many units need to be produced in a given week for the employees to get the bonus payment? c) Management wants to close the facility during Easter weekend and use inventory to satisfy the demand. However, inventory is low and based on current production levels it is also required to have an average production of 5,100 units/week during the previous 5 weeks in order not to run out of product and lose sales. What is the probability of still having to work through Easter weekend? d) Trucho & Co., a competitor, produces the same units with similar equipment but untrained employees. Their production levels are normally distributed with a mean of 4,850 units/week and a standard deviation of 400 units/week. A customer has said the supplier with the higher average production in the last 10 weeks will land a new contract. What is the probability that Trucho & Co. will get the contract?

Answer 1

Mean=5000, SD=200

$a)P(X>5300)=P(Z>\frac{5300-5000}{200}=1.5)=0.0668=6.68\%$

$b)P(X>a)=0.05\Rightarrow Z=1.645=\frac{a-5000}{200}\Rightarrow a\geq5329$

c)Assuming easter weekend is next week and the company has to ramp up all the remaining production this week.

This week company will have to produce 5300+5*100=5800

we need what is the P(X<5800)

$P(X<5800)=P(Z<\frac{5800-5000}{200}=4)=0.999968\approx 100\%$

d)Assuming X is the sales ofGadget Inc and Y is the sales of Trucho, We need $P(\bar{Y}>\bar{X})$ or $P(\bar{Y}-\bar{X}>0)$

$P(\bar{Y}-\bar{X}>0)=P(Z>\frac{0-(4850-5000)}{\sqrt{\frac{200^2}{10}+\frac{400^2}{10}}})=P(Z>1.0606)=0.1444)$