Using the normal distribution, it is found that 63.18% of the area under the curve of the standard normal distribution is between z = − 0.9 z = - 0.9.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

$Z = \frac{X - \mu}{\sigma}$

The z-score measures how many standard deviations the measure is above or below the mean.

Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

The area within 0.9 standard deviations of the mean is the <u>p-value of Z = 0.9(0.8159) subtracted by the p-value of Z = -0.9(0.1841)</u>, hence:

0.8159 - 0.1841 = 0.6318 = 63.18%.

More can be learned about the normal distribution at brainly.com/question/4079902

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2 years ago

Expand and simplify (x-6)(x+2)

maw [93]

I hope this helps you