Home $>2023$ Fall2 MAT120 (TTh11:30) > Assessment 3.4 - Inverse Normal Distributions Score: $0 / 30 \quad 0 / 9$ answered Question 2 A manufacturer knows that their items have a normally distributed lifespan, with a mean of 5.6 years, and standard deviation of 1 years. The $4 \%$ of items with the shortest lifespan will last less than how many years? Give your answer to one decimal place. Question Help: $\square$ Message instructor $D$ Post to forum

Step 1 ：We are given that the lifespan of a certain item follows a normal distribution with a mean of 5.6 years and a standard deviation of 1 year. We are asked to find the lifespan that corresponds to the bottom 4% of items.

Step 2 ：This problem can be solved using the concept of inverse normal distribution or z-score. The z-score is a measure of how many standard deviations an element is from the mean.

Step 3 ：In this case, we are looking for the z-score that corresponds to a cumulative probability of 0.04. Using a standard normal distribution table or a z-score calculator, we find that the z-score corresponding to a cumulative probability of 0.04 is approximately -1.75.

Step 4 ：We can then use the formula for the z-score to find the corresponding lifespan: \(Z = \frac{X - \mu}{\sigma}\) where Z is the z-score, X is the value we are looking for, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.

Step 5 ：Rearranging the formula to solve for X gives us: \(X = Z\sigma + \mu\)

Step 6 ：Substituting the given values into the formula gives us: \(X = -1.75*1 + 5.6 = 3.85\)

Step 7 ：So, the 4% of items with the shortest lifespan will last less than approximately 3.9 years (rounded to one decimal place).

Step 8 ：Therefore, the lifespan that corresponds to the bottom 4% of items is \(\boxed{3.9}\) years.