PLEASE SHOW ALL WORK :) The output voltage of a power supply is normally distrib

Question

PLEASE SHOW ALL WORK :)

The output voltage of a power supply is normally distributed with mean 5 V and standard deviation 0.02 V. if the lower the upper specifications for voltage are 4.95 V and 5.05 V, respectively,

A. What is the probability that a power supply selected at random will conform to the specifications on voltage?

B. Suppose we wanted to improve the process. can shifting the mean reduce the number of nonconforming units produced? How much would the process variability need to be reduced in order to have all but one out of 1000 units conform the specification?

Explanation / Answer

A. P(X < A) = P(Z < (A - mean) /standard deviation)

P(power supply confirms to the specification) = P(4.95 < X < 5.05)

= P(X < 5.05) - P(X < 4.95)

= P(Z < (5.05-5)/0.02) - P(Z < (4.95-5)/0.02)

= P(Z < 2.5) - P(Z < - 2.5)

= 0.9938 - 0.0062

= 0.9906

B) Let us ussume the variability is Y

For all but one of 1000 units to confirm the specification,

P(4.5 < X < 5.5) = 0.999

P(X < 5.5) = 0.9995 and P(X < 4.5) = 0.0005

So, P(Z < (5.5 - 5)/Y) = 0.9995

From standard normal distribution table,

0.05/Y = 3.29

Y = 0.015

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.