A voltage source outputs a Gaussian random voltage with a me

A voltage source outputs a Gaussian random voltage with a mean mu=5 and a variance sigma?=16. What is the probability that the output voltage is greater than zero? What is the probability that the output voltage is greater than zero but less than or equal to the mean? What is the probability that the output voltage is greater than double the mean?

Solution

a)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    0      
u = mean =    5      
          
s = standard deviation =    4      
          
Thus,          
          
z = (x - u) / s =    -1.25      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   -1.25   ) =    0.894350226 [answer]

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b)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    0      
x2 = upper bound =    5      
u = mean =    5      
          
s = standard deviation =    4      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -1.25      
z2 = upper z score = (x2 - u) / s =    0      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.105649774      
P(z < z2) =    0.5      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.394350226   [answer]

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c)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    10      
u = mean =    5      
          
s = standard deviation =    4      
          
Thus,          
          
z = (x - u) / s =    1.25      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   1.25   ) =    0.105649774 [answer]