Results of a survey revealed that the monthly utility bills of a 3-bedroom house in Beckville had a mean of $\$ 115$ and a standard deviation of $\$ 11$. The distribution was considered bell-shảped.
Based on this information, answer the following.
Approximately what percent of the 3-bedroom homes had a monthly utility bill over $\$ 126$ ?
\%$
Approximately $95 \%$ of the 3-bedroom homes had a monthly utility bill between $\$$
and \$

Step 1 ：The problem is asking for the percentage of 3-bedroom homes that had a monthly utility bill over $126. This is a problem of finding the area under the curve of a normal distribution.

Step 2 ：The first step is to standardize the value of $126 by subtracting the mean and dividing by the standard deviation. This will give us the z-score, which we can then use to find the corresponding percentile in a standard normal distribution table.

Step 3 ：However, since we want the percentage of homes with a bill over $126, we need to subtract the percentile corresponding to the z-score from 1 (or 100%) to get the percentage of homes with a bill over $126.

Step 4 ：Given that the mean is $115 and the standard deviation is $11, we can calculate the z-score for a value of $126 as follows: \( z = \frac{value - mean}{std\_dev} = \frac{126 - 115}{11} = 1.0 \)

Step 5 ：The percentile corresponding to a z-score of 1.0 is approximately 0.8413. Therefore, the percentage of homes with a bill over $126 is \( 1 - 0.8413 = 0.1587 \) or approximately 15.87%.

Step 6 ：Final Answer: The percentage of 3-bedroom homes that had a monthly utility bill over $126 is approximately \(\boxed{15.87 \%}\).