Final Answer:
Probability that signal exceeds 240 µV: ≈ 10.56% Probability signal > 240 µV given it's > 210 µV: ≈ 41.61%
Step-by-step explanation:
To determine the probability that the signal is larger than 240 micro volts given that it is greater than 210 micro volts, we can utilize
Given a Gaussian random variable N(200, 16²) representing the analog signal received, we're asked to find the probability that the signal exceeds 240 micro volts. Employing the properties of the Gaussian distribution, we determine the z-score as (240 - 200) / 16 = 2.5. Using a standard normal distribution table or a calculator with the cumulative distribution function (CDF) for the standard normal distribution, we find P(Z > 2.5) ≈ 0.1056 or 10.56%.
ii. For the second part, we're asked to compute the probability that the signal is larger than 240 micro volts, given that it is greater than 210 micro volts. First, we find P(X > 210) using the CDF of N(200, 16²) Gaussian distribution, yielding approximately 0.7734 or 77.34%. Then, to determine P(X > 240 | X > 210), we divide the probability that the signal exceeds both 240 µV and 210 µV by the probability that it exceeds 210 µV, resulting in approximately 0.4161 or 41.61%.
These calculations showcase the application of Gaussian probability distributions and conditional probability in analyzing the probabilities associated with an analog signal exceeding specific voltage thresholds in a Gaussian distribution scenario.