To gauge their fear of going to a dentist, a large group of adults completed the Modified Dental Anxiety Scale questionnaire. Scores ( $X$ ) on the scale ranges from zero (no anxiety) to 25 (extreme anxiety). Assume that the distribution of $X$ scores is normal with mean $\mu=15$ and standard deviation $\sigma=4.0$. Simplify the value of $\mathrm{L}=15-4.0$ and $\mathrm{U}=15+(2)(4.0)$ and then answer the following question:
What is the probability that a randomly selected adult scores between $L$ and $U$ ?

Step 1 ：Given that the distribution of scores is normal with mean \(\mu=15\) and standard deviation \(\sigma=4.0\), we are asked to find the probability that a randomly selected adult scores between \(L\) and \(U\) on the Modified Dental Anxiety Scale questionnaire.

Step 2 ：First, we calculate the values of \(L\) and \(U\). According to the question, \(L=15-4.0=11.0\) and \(U=15+(2)(4.0)=23.0\).

Step 3 ：Next, we convert these values to z-scores. The z-score is a measure of how many standard deviations an element is from the mean. It is calculated as \(z = \frac{x - \mu}{\sigma}\), where \(x\) is the element, \(\mu\) is the mean and \(\sigma\) is the standard deviation. For \(L\), the z-score \(z_L\) is \(-1.0\) and for \(U\), the z-score \(z_U\) is \(2.0\).

Step 4 ：Finally, we use the z-scores to find the probability that a randomly selected adult scores between \(L\) and \(U\). This can be done by finding the area under the normal distribution curve between these two z-scores. This area represents the probability we are looking for, which is approximately \(0.819\).

Step 5 ：Final Answer: The probability that a randomly selected adult scores between \(L\) and \(U\) on the Modified Dental Anxiety Scale questionnaire is \(\boxed{0.819}\).