Suppose that the diameters of oak trees are normally distributed with a mean of 4 feet and a standard deviation of 0.375 feet. Step 4 of 5: If we wanted to look at the top 26% of trees, what would their minimum diameter be? [Round to 2 decimals] can someone tell me how they got 4.24 feet
+6248 \(\text{First we need to convert "top 26%" to a z-score}\\ \text{We do this by using a table or software of the }\\ \text{CDF of the standard normal distribution}\\ \text{We find that 74% has a corresponding z-score of }\\ z=0.643345\)
\(\text{We then use the mean and deviation to convert this z-score into a value}\\ d= \mu + \sigma z = 4 + 0.643345\cdot 0.375 = 4.24125 \Rightarrow 4.24ft\)
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+36915 .644*.375+4 = 4.2415
.644 is from POSITIVE z-score table value 74% (100%-26%) .644 standard deviations (.375) above the mean of 4 ft
