Step 1 :The problem is asking for the probability that a single randomly selected value is greater than 38.4. This is a problem of finding the probability of a single value from a normal distribution.
Step 2 :The formula for the z-score is: \[Z = \frac{X - \mu}{\sigma}\] where X is the value we are interested in, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.
Step 3 :We can calculate the z-score for 38.4 using the given mean (\(\mu = 38\)) and standard deviation (\(\sigma = 2.8\)). This gives us a z-score of approximately 0.14285714285714235.
Step 4 :We then use a z-table or a statistical function to find the probability that a randomly selected value is greater than this z-score. This gives us a probability of approximately 0.443201503183532.
Step 5 :Final Answer: \[P(X>38.4) \approx \boxed{0.4432}\]