Ex: The mean shelf life of a spice is 13.4 weeks and the standard deviation is 1.8 weeks.
(a) Would a shelf life of 20 weeks be unusual? Why?
(b) Find the z-score for a shelf life of 15.5 weeks.
(c) What percentage of shelf lives would be expected to be between 13.4 and 15.5 weeks?

Step 1 ：Given: mean (\(\mu\)) = 13.4 weeks, standard deviation (\(\sigma\)) = 1.8 weeks

Step 2 ：Calculate the z-score for a shelf life of 20 weeks: \(z = \frac{x - \mu}{\sigma}\)

Step 3 ：\(z = \frac{20 - 13.4}{1.8} = 3.67\)

Step 4 ：A shelf life of 20 weeks is unusual because its z-score (3.67) is outside the typical range of z-scores (-2 to 2)

Step 5 ：Calculate the z-score for a shelf life of 15.5 weeks: \(z = \frac{x - \mu}{\sigma}\)

Step 6 ：\(z = \frac{15.5 - 13.4}{1.8} = 1.17\)

Step 7 ：Using a standard normal distribution table or calculator, find the area between z-scores of 0 and 1.17, which is approximately 38.2%

Step 8 ：\(\boxed{\text{About 38.2\% of shelf lives would be expected to be between 13.4 and 15.5 weeks}}\)