Question

Find the probability that the random Gaussian distributed voltage, representing a signal plus noise, is greater than ±1.5 standard deviations displaced from the mean value for the case of a signal to noise power ratio of 2, and a signal power based on a variance of the signal=.5 watts

Answer 1

Answer:

$P(X>3.06)=P(\frac{X-\mu}{\sigma}>\frac{3.06-\mu}{\sigma})=P(Z>\frac{3.06-2}{0.707})=P(z>1.5)$

And we can find this probability using the complement rule and excel or the normal standard table:

$P(z>1.5)=1-P(z$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the heights of a population, and for this case we know the distribution for X is given by:

$X \sim N(2,\sqrt{0.5}= 0.707)$  

Where $\mu=2$ and $\sigma=0.707$

We are interested on this probability

$P(X>2+ 1.5*0.707)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X>3.06)=P(\frac{X-\mu}{\sigma}>\frac{3.06-\mu}{\sigma})=P(Z>\frac{3.06-2}{0.707})=P(z>1.5)$

And we can find this probability using the complement rule and excel or the normal standard table:

$P(z>1.5)=1-P(z$