Question

The output voltage of a power supply is normally distributed with mean 12 V and standard deviation 0.11 V. If the upper and lower specifications for voltage are 12.15 V and 11.85 V respectively, what is the probability that the power supply selected at random will confirm to the specifications on voltage?

Answer 1

Answer:

82.62%

Explanation:

The z score is a score used in statistics to determine by how many standard deviations the raw score is above or below the mean. The z score is given by:

$z=\frac{x-\mu}{\sigma} \\\\where\ x=raw\ score,\mu=mean\ and\ \sigma=standard\ deviation.\\\\Given \ that\ \mu=12V, \sigma=0.11V.\\\\For\ x<12.15V:\\\\z=\frac{12.15-12}{0.11} =1.36\\\\For\ x>11.85V:\\\\z=\frac{11.85-12}{0.11} =-1.36$

From the normal distribution table, P(11.85 < x < 12.15) = P(-1.36 < z < 1.36) = P(z < 1.36) - P(z < -1.36) = 0.9131-0.0869 = 0.8262 = 82.62%