The data in a data set are normally distributed with a mean of 180 and a standard deviation of 30. Estimate the percent of the data that are less than 120 or greater than 240.
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+6248 \(\text{we normalize the data by finding }\\ z = \dfrac{X - \mu}{\sigma}\\ z_1 = \dfrac{120-180}{30} = -2\\ z_2 = \dfrac{240-180}{30} = 2\\ \text{using the 68, 95, 99.7 rule we have that}\\ P[-2 \leq z \leq 2] = 0.9545 \\ \text{we want the probability outside this area so we have}\\ P([z<-2) \wedge (2 < 2)] = 1-0.9545 = 0.0455\)
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