Question

Use the empirical rule to find a reasonable estimate for P(z ≤ 0. 82). 10% 30% 50% 80%.

Answer 1

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.

$z=\frac{x-\mu}{\sigma}$

Where $\mu$ = mean

$\sigma$=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:

brainly.com/question/25638875