The output voltage of a power supply is normally distributed withmean 5 V and st

Question

The output voltage of a power supply is normally distributed withmean 5 V and standard deviation .02 V. If the lower and upperspecifications for voltage are 4.95 and 5.05 V respectively, whatis the probability that a power supply selected at random willconform to the specifications on the voltage? (The answer to thisis .9876)
Reconsider the power supply manufacturing process in the previousquestion. Suppose we wanted to improve the process. Can shiftingthe mean reduce the number of nonconforming unit's produced? Howmuch would the process variability need to be reduced in order tohave all but one out of 1000 units conform to thespecifications?

I have the first part all I need is the second part of thequestion, thanks alot!

Explanation / Answer

Shifting the mean obviously cannot help. The mean outputvoltage falls exactly in between 4.95 and 5.05V, so shifting themean would result in a lower proportion of voltages falling withinthe specification range. We need to tighten the variabilityin order to reduce the number of nonconforming units. In effect you are doing what you did in part 1 in reverse, startingwith: (a) the probability a power supply selected at random will conformto the specifications on the voltage = 1/1000 = 0.001 (b) Now you need to know the z-score associated with a 2-tailedprobability of 0.001 The sum of the area below z = -3.29 and above z = 3.29 is equal to0.001 (c) Now work the z-score equation backwards solving for thestandard deviation z = (y - ) / = (5.05 - 5) / = 3.29 (5.05 - 5) / = 3.29 (5.05 - 5) / 3.29 = = 0.015 Thus the standard deviation must be decreased from 0.02V to 0.015Vto have all but one out of 1000 units conform to the specifications

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