The noise voltage in an electric circuit can be modeled as a gaussian random var

Question

The noise voltage in an electric circuit can be modeled as a gaussian random variable with a mean equal to o and variance 10-8 a. What is the probability that the noise exceeds 10-4? b, what is the probability that the noise is between-2 x 10-4 and 10-4? c. Given that the noise is positive what is the probability that it exceeds 10-4? d. The noise is passed through a half-wave rectifier (whose output is the input if the input is positive and 0 otherwise). Find the CDF of the rectified noise and then its PDF. e. What is the expected value of the rectified noise in part (d)? f. Now suppose the noise is passed through a full wave rectifier. What is the PDF of the rectified noise in this its expected value.

Explanation / Answer

Assuming the normal distribution , we are given that

mean = 0 and var = 10^-8 so sd = 0.0001

we know that the z score is given as
Z = (X-Mean)/SD

= (0.0001 - 0)/0.0001 = 1

hence from the z tables we check
P ( Z>1 )=1P ( Z<1 )=10.8413=0.1587

b)

between -0.0002 and 0.0001

Z 0.0001 is 0
and

Z (-0.0002) = -0.0002/0.0001 = -2

To find the probability of P (2<Z<1), we use the following formula:
P (2<Z<1 )=P ( Z<1 )P (Z<2 )
We see that P ( Z<1 )=0.8413.
We see that P ( Z<2 )=0.9772 so,
P ( Z<2)=1P ( Z<2 )=10.9772=0.0228
At the end we have:
P (2<Z<1 )=0.8185

c)

we know that the z score is given as
Z = (X-Mean)/SD

= (0.0001 - 0)/0.0001 = 1

hence from the z tables we check
P ( Z>1 )=1P ( Z<1 )=10.8413=0.1587

please note that the properties of half wave and full wave rectifies are beyond the scope of this subject , as they fall under electronics and communications subject

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