At a local restaurant, the amount of time that customers have to wait for their food is normally distributed with a mean of 48 minutes and a standard deviation of 2 minutes. What percentage of customers have to wait between 46 minutes and 52 minutes, to the nearest tenth?

Step 1 ：The problem is asking for the percentage of customers who wait between 46 and 52 minutes for their food. This is a problem of normal distribution. The mean (\(\mu\)) is 48 minutes and the standard deviation (\(\sigma\)) is 2 minutes. We need to find the probability (P) that the waiting time (X) is between 46 and 52 minutes.

Step 2 ：This can be calculated using the Z-score formula: \(Z = \frac{(X - \mu)}{\sigma}\). The Z-score gives us the number of standard deviations a particular score is from the mean.

Step 3 ：We can then use the Z-score to find the probability. However, since we are looking for the probability between two values, we need to find the Z-scores for both 46 and 52 minutes, and then find the area under the curve between these two Z-scores.

Step 4 ：Using the given mean of 48 and standard deviation of 2, the Z-scores for 46 and 52 minutes are -1.0 and 2.0 respectively.

Step 5 ：The area under the curve between these two Z-scores corresponds to the probability we are looking for. This is calculated to be approximately 0.8185946141203637.

Step 6 ：To find the percentage, we multiply this probability by 100, giving us 81.9%.

Step 7 ：Final Answer: The percentage of customers who have to wait between 46 and 52 minutes for their food is \(\boxed{81.9\%}\).