Final answer:
The value of 'a' in the normal distribution with mean=25 and standard deviation=4, corresponding to 47.72% of the observations between the mean and 'a', is approximately 25.2.
Step-by-step explanation:
The student has asked about finding the value of an in a normal distribution with specified parameters: a mean of 25 and a standard deviation of 4. If 47.72% of observations are between the mean and a, this implies that a corresponds to the 50% + 47.72%/2 percentile since half of the distribution lies to the left of the mean. To find a, we need to determine the z-score that corresponds to these cumulative tail areas and then apply the formula to find the actual value. The z-score for approximately 48.72% to the right of the mean (since tables give area to the left) is 0.05 (this is because 50% + 48.72% / 2 roughly equals 0.5 (the mean) + 0.05). The z-score formula is Z = (X - μ) / σ, which rearranges to X = μ + (Z × σ). Inserting the mean (25), z-score (0.05), and standard deviation (4) we get a = 25 + (0.05 × 4), yielding a ≈ 25.2. Hence, the value of a is approximately 25.2.