Assume that adults have IQ scores that are normally distributed with a mean of $\mu=105$ and a standard deviation $\sigma=15$. Find the probability that a randomly selected adult has an IQ between 93 and 117.
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The probability that a randomly selected adult has an IQ between 93 and 117 is (Type an integer or decimal rounded to four decimal places as needed.)

Step 1 ：Given that adults have IQ scores that are normally distributed with a mean of \(\mu=105\) and a standard deviation \(\sigma=15\). We are asked to find the probability that a randomly selected adult has an IQ between 93 and 117.

Step 2 ：To solve this, we need to convert the IQ scores to z-scores, which are a measure of how many standard deviations an element is from the mean. The formula for calculating a z-score is \[z = \frac{x - \mu}{\sigma}\] where \(x\) is the value from the dataset, \(\mu\) is the mean of the dataset, and \(\sigma\) is the standard deviation of the dataset.

Step 3 ：Calculate the z-scores for 93 and 117. Let's denote these as \(z1\) and \(z2\) respectively.

Step 4 ：\(z1 = \frac{93 - 105}{15} = -0.8\) and \(z2 = \frac{117 - 105}{15} = 0.8\)

Step 5 ：We can use a z-table to find the probabilities associated with these z-scores. The probability that the IQ is between 93 and 117 is the probability that the z-score is between the z-scores of 93 and 117. This is calculated by finding the probability at the z-score of 117 and subtracting the probability at the z-score of 93.

Step 6 ：From the z-table, the probabilities associated with \(z1\) and \(z2\) are approximately 0.2119 and 0.7881 respectively.

Step 7 ：The probability that a randomly selected adult has an IQ between 93 and 117 is \(p = p2 - p1 = 0.7881 - 0.2119 = 0.5762\)

Step 8 ：Final Answer: The probability that a randomly selected adult has an IQ between 93 and 117 is approximately \(\boxed{0.5762}\)