Solution :
Using standard normal table,
P(-a
Z
a) = 0.4314
P(Z
a) - P(Z
-a) = 0.4314
2P(Z
a) - 1 = 0.4314
2P(Z
a) = 1 + 0.4314 = 1.4314
P(Z
a) = 1.4314 / 2 = 0.7157
P(Z
0.57) = 0.7157
z = 0.57
Assume that z scores are normally distributed with a mean of O and a standard deviation...
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stats
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