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.
The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The area within 0.9 standard deviations of the mean is the p-value of Z = 0.9(0.8159) subtracted by the p-value of Z = -0.9(0.1841), hence:
0.8159 - 0.1841 = 0.6318 = 63.18%.
More can be learned about the normal distribution at https://brainly.com/question/4079902
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