Using the normal distribution, the probabilities are given as follows:
a) 0.6826 = 68.26%.
b) 0.9759 = 97.59%.
c) 0.0004 = 0.04%.
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]
Item a:
The probability is the p-value of Z = 1(0.8413) subtracted by the p-value of Z = -1(0.1587), hence:
0.8413 - 0.1587 = 0.6826
Item b:
The probability is the p-value of Z = 2(0.9772) subtracted by the p-value of Z = -3(0.0013), hence:
0.9772 - 0.0013 = 0.9759.
Item c:
This probability is P(|Z| > 3.5), which is 2 multiplied by the p-value of Z = -3.5, which is of 0.0002.
Hence:
2 x 0.0002 = 0.0004 = 0.04%.
More can be learned about the normal distribution at https://brainly.com/question/4079902
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