Let's solve this problem step-by-step:
First, let's calculate the z-score that corresponds to the 3% shortest lifespans (i.e., the 2nd lowest z-score after the mean z-score of 0). To do this, we will use the inverse normal distribution table.
The inverse normal distribution table gives us the z-score that corresponds to a particular percentile. In our case, we want the second lowest z-score after the mean z-score, which means we want to find the largest z-score with a probability less than 0.03.
To find the corresponding z-score using the inverse normal distribution table, we first need to set the confidence level (here, it is 97%), and then find the corresponding z-score.
According to the inverse normal distribution table, the z-score that corresponds to a probability of 0.03 is approximately -1.645. (For reference, the mean z-score of 0 corresponds to a percentile of 0.5, i.e., the median lifespan.)
Now, let's solve for the lifespan itself. The z-score of -1.645 corresponds to a percentile of 0.03 (i.e., the second shortest 3% of items). So, the lifespan of the 3% of items with the shortest lifespan is given by the mean - 1.645 * the standard deviation, which is approximately 8.2 years.
Therefore, the 3% of items with the shortest lifespan will last less than 8.2 years.