The noise voltage in an electric circuit follows a Gaussian

The noise voltage in an electric circuit follows a Gaussian distribution with mean equal to zero and variance equal to 10^-8. Find the probability that the value of the noise exceeds 10^-4. Given that the value of the noise is positive, calculate the probability that the value of the noise exceeds 10^-4.

Solution

a)

Find the probability that the value of the noise exceeds 10^-4.

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    0.0001      
u = mean =    0      
          
s = standard deviation =    0.0001      
          
Thus,          
          
z = (x - u) / s =    1      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   1   ) =    0.158655254 [ANSWER]

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b)

Given that the value of the noise is positive, calculate the probability that the value exceeds 10^-4.

As the mean is 0, then half of the noise values is positive, P(x>0) = 0.5.

Thus,

P(x>0.0001|x>0) = P(x>0.0001)/P(x>0) = 0.158655254/0.5

= 0.317310508 [ANSWER]