Question

# A music website claims the mean length of their songs is 3.75 minutes. Suppose the length of songs from this website follows a Normal distribution with standard deviation 0.5 minutes. If 7 songs from this website are randomly selected, then there is about a 90% probability that the sample mean will fall in which interval

Answer

(3.4377, 4.0623)

- Identify the given parameters: Mean (mu) = 3.75, Standard deviation (sigma) = 0.5, Sample size (n) = 7, Confidence level = 0.9
- Find the standard error of the mean (SEM): SEM = sigma / sqrt(n) = 0.5 / sqrt(7) = 0.1897.
- Find the z-score corresponding to the desired confidence level: From Z-table, the z-score for a 90% confidence level is 1.645.
- Find the margin of error (ME): ME = z-score * SEM = 1.645 * 0.1897 = 0.3123.
- Calculate the interval: The interval is (mu - ME, mu + ME) = (3.75 - 0.3123, 3.75 + 0.3123) = (3.4377, 4.0623).