P(z ≤ 0. 82) or p-value corresponding to z ≤ 0. 82 comes to be approximately 80%.
What is a z-score?
Z-score indicates how much a given value differs from the standard deviation.
[tex]z=\frac{x-\mu}{\sigma}[/tex]
Where [tex]\mu[/tex] = mean
[tex]\sigma[/tex]=standard deviation
P(z ≤ 0. 82) means the p-value corresponding to z≤ 0. 82.
From the standard normal table
P(z ≤ 0. 82) ≈80%.
Hence, P(z ≤ 0. 82) or p-value corresponding to z ≤ 0. 82 comes to be approximately 80%.
To get more about the z-score visit:
https://brainly.com/question/25638875
