For the standard normal distribution, if P (z≤a) = 0. 7116, then the value of P(z≥a) is 0.2884.
The standard normal distribution is a special type of normal distribution where the mean is 0, and the standard deviation is 1.
As it is given that the value of P(z≤a)=0.7116, therefore, for P(z≤a) we can write,
[tex]P(z\leq a) = 0.7116\\\\P(z > a) =1- 0.7116\\[/tex]
But as we know the fact that in continuous distribution, the probability of a single point is 0, therefore, we can write
[tex]P(Z \leq z) = P(Z < z) )[/tex]
Similarly, for P(z>a)=0.7116 we can write,
[tex]P(z > a) = 1-0.7116\\\\P(z \geq a) = 1-0.7116\\\\P(z \geq a) = P(z=a) + P(z > a) = 1- 0+0.7116\\\\P(z \geq a) = 0.2884[/tex]
Hence, For the standard normal distribution, if P (z≤a) = 0. 7116, then the value of P (z≥a) is 0.2884.
Learn more about Standard normal distribution:
https://brainly.com/question/25447725