Question 15 • SVLC ALGEBRA 2B - STANDARD (15272)
Convert the tree height to a z-score: (37 − 25) ÷ 6 = 2. The table shows that about 0.9772, or 97.72%, of trees are at or below 2 standard deviations above the mean. Thus, the probability of a height at least 37 meters is 1 − 0.9772 = 0.0228, or 2.28%, which rounds to 2.3%.
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