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 \(\boxed{0.819}\).