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

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ashishdwivedilVT

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

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