Using the normal distribution, it is found that there is a 0.0139 = 1.39% probability that his systolic blood pressure will be greater than 160 mm.
In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
In this problem:
The probability that his systolic blood pressure will be greater than 160 mm is 1 subtracted by the p-value of Z when X = 160, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{160 - 138}{10}[/tex]
[tex]Z = 2.2[/tex]
[tex]Z = 2.2[/tex] has a p-value of 0.9861.
1 - 0.9861 = 0.0139
0.0139 = 1.39% probability that his systolic blood pressure will be greater than 160 mm.
You can learn more about the normal distribution at https://brainly.com/question/24663213