According to its label, a soft-drink can contains 500 mL. Currently, the filling machine is set so that the volume per can is normally distributed, with a mean of 502 mL and a standard deviation of 1.5 mL. If too many cans contain less than 500 mL, the company will lose sales. If too many cans contain more than 504 mL, the company will incur excess costs. Does the company need to recalibrate the filling machine? Justify your response.
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We are given `mu=502,sigma=1.5` .
Since the population is normal, the probability that a can has less than 500ml is the same as the probability that a can has more than 504ml.
To find the probability that a can has less than 500ml, we can convert the measure to a z-score: `z=(x-mu)/sigma==>z=(500-502)/1.5=-1.bar(3)`
Using a standard notrmal table (or algebra utility) we find `P(z<-1.bar(3))~~.0912`
Since P(x>504)=P(x<500) we have the probability that a random can is less than 500ml or more than 504ml is about 18.24%.
Is this enough to recalibrate the machines? This is a judgement call. 18% of your product either costs too much or causes lost sales. The amount of cost and lost sales should be compared to the cost of resetting the machine to make the decision.
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