Solution
For this case we have the following distribution:
[tex]X\approx N(\mu=82,\sigma=0.5)[/tex]and we want to find this probability:
[tex]p(X>82)=1-p(X<82)[/tex]We can use the z score formula and we got:
[tex]z=\frac{X-\mu}{\sigma}=\frac{82-82}{0.5}=0[/tex]Using the normal distribution table we got:
[tex]P(Z>0)=1-0.5=0.5[/tex]Then the final answer is 0.5
