Step 1 :The question is asking for the percentage of people who are 51 years old or older in a normally distributed data set with a mean of 37 and a standard deviation of 7.
Step 2 :To solve this, we need to calculate the z-score for the age of 51. The z-score is a measure of how many standard deviations an element is from the mean.
Step 3 :The formula for the z-score is: \(Z = \frac{X - \mu}{\sigma}\) where: \(X\) is the value we are interested in (in this case, 51), \(\mu\) is the mean (in this case, 37), and \(\sigma\) is the standard deviation (in this case, 7).
Step 4 :After calculating the z-score, we can use a z-table or a statistical calculator to find the percentage of people who are 51 years old or older.
Step 5 :mean = 37, std_dev = 7, value = 51, z_score = 2.0, p_value = 0.02275013194817921
Step 6 :The p_value represents the probability of a value being greater than 51 in our normal distribution. This value is approximately 0.02275, which is closest to 2.5% among the options given in the question.
Step 7 :Final Answer: The closest percentage of people who are 51 years old or older in the given data set is \(\boxed{2.5\%}\).