Suppose that the time required to complete a 1040R tax form is normal distributed with a mean of 110 minutes and a standard deviation of 10 minutes. What proportion of 1040R tax forms will be completed in less than 91 minutes? Round your answer to at least four decimal places.

Step 1 ：We are given a normal distribution with a mean of 110 minutes and a standard deviation of 10 minutes. We are asked to find the proportion of tax forms that will be completed in less than 91 minutes.

Step 2 ：To solve this problem, we can use the Z-score formula which is \((X - μ) / σ\), where X is the value we are interested in, μ is the mean, and σ is the standard deviation. The Z-score tells us how many standard deviations away from the mean our value is.

Step 3 ：Substituting the given values into the Z-score formula, we get \((91 - 110) / 10 = -1.9\).

Step 4 ：Once we have the Z-score, we can use a Z-table to find the proportion of values that are less than our value. The proportion corresponding to a Z-score of -1.9 is approximately 0.0287.

Step 5 ：This means that about 2.87% of the tax forms will be completed in less than 91 minutes.

Step 6 ：Final Answer: The proportion of 1040R tax forms that will be completed in less than 91 minutes is approximately \(\boxed{0.0287}\) or \(\boxed{2.87\%}\).