The mean clotting time of blood is 7.45 seconds with a standard deviation of 3.6 seconds. What is the probability that an individual's clotting time will be less than 7 seconds or greater than 9 seconds?

Assume a normal distribution.

The probability is $\square$.
(Round to the nearest thousandth.)

Step 1 ：We are given that the mean clotting time of blood is \(7.45\) seconds with a standard deviation of \(3.6\) seconds. We are asked to find the probability that an individual's clotting time will be less than \(7\) seconds or greater than \(9\) seconds.

Step 2 ：We can use the z-score to find the probability. The z-score is a measure of how many standard deviations an element is from the mean. The formula for the z-score is: \(z = \frac{X - \mu}{\sigma}\) where \(X\) is the value we are interested in, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.

Step 3 ：We need to calculate two z-scores: one for \(X = 7\) and one for \(X = 9\).

Step 4 ：For \(X = 7\), the z-score is \(z1 = \frac{7 - 7.45}{3.6} = -0.125\).

Step 5 ：For \(X = 9\), the z-score is \(z2 = \frac{9 - 7.45}{3.6} = 0.431\).

Step 6 ：We can use a z-table to find the probabilities associated with these z-scores. The probability associated with \(z1\) is \(0.450\) and the probability associated with \(z2\) is \(0.333\).

Step 7 ：The final probability is the sum of the probabilities of the clotting time being less than \(7\) seconds and the clotting time being greater than \(9\) seconds. This is calculated by adding the probabilities associated with the z-scores for these two values.

Step 8 ：So, the final probability is \(0.450 + 0.333 = 0.783\).

Step 9 ：Final Answer: The probability that an individual's clotting time will be less than \(7\) seconds or greater than \(9\) seconds is approximately \(\boxed{0.784}\).