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The output voltage of a power supply is normally distributed with mean 5 V and standard deviation 0.02 V. If the lower and upper specifications for voltage are 4.95 V and 5.05 V, respectively, what is the probability that a power supply selected at random will conform to the specifications on voltage

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Answer:

0.9876

Step-by-step explanation:

Given a normal distribution :

Mean(m) of distribution = 5V

Standard deviation (sd) = 0.02 V

Lower voltage specification = 4.95

Upper voltage specification = 5.05

Using the Zscore formula :

(x - m) / sd

x = 4.95

(4.95 - 5) / 0.02

-0.05 / 0.02 = -2.5

P(Z ≤ -2.5) = 0.0062

x = 5.05

(5.05 - 5) / 0.02

0.05 / 0.02 = 2.5

P(Z ≤ 2.5) = 0.9938

P(Z ≤ 2.5) - P(Z ≤ -2.5)

0.9938 - 0.0062 = 0.9876

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