Question

An electronic prognosis was created to detect faults to decrease the downtime and the number of unplanned repairs. The signal assumed to be normally distributed with a mean 1.5V and a variance of 0.000625V. What is the signal value that exceeds 90% probability?

Answer 1

Solution :

Given that,  

mean = = 1.5V

variance = 0.000625V

standard deviation = = 0.025V

Using standard normal table,

P(Z > z) = 90%

1 - P(Z < z) = 0.9

P(Z < z) = 1 - 0.9 = 0.1

P(Z < -1.28) = 0.1

z = -1.28

Using z-score formula,

x = z * +

x = -1.28 * 0.025 + 1.5

x = 1.468 = 1.47

signal value = 1.47