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A forester measured the trees in a large woods up for sale. He concluded that the diameter of the trees was normally distributed with mean 12.4 inches and standard deviation 4.7 inches. Which of the following equations can be used to find the diameter x which is at the 90th percentile? (A) 0.90= (B) 1.28 = x-12.4 (4.7)² x 12.4 (4.7)² x-12.4 (C) 0.90= 4.7 x-12.4 (D) 1.28= 4.7 (E) 1.645= x-12.4 4.7

A forester measured the trees in a large woods up for sale. He concluded that the diameter of the trees was normally distributed with mean 12.4 inches and standard deviation 4.7 inches. Which of the following equations can be used to find the diameter x which is at the 90th percentile? (A) 0.90= (B) 1.28 = x-12.4 (4.7)² x 12.4 (4.7)² x-12.4 (C) 0.90= 4.7 x-12.4 (D) 1.28= 4.7 (E) 1.645= x-12.4 4.7

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter10: Statistics
Section: Chapter Questions
Problem 22SGR
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A forester measured the trees in a large woods up for sale. He concluded that the diameter of the trees was normally distributed with mean 12.4 inches and standard deviation 4.7 inches. Which of the following equations can be used to find the diameter x which is at the 90th percentile? (A) 0.90= (B) 1.28 = x-12.4 (4.7)² x 12.4 (4.7)² x-12.4 (C) 0.90= 4.7 x-12.4 (D) 1.28= 4.7 (E) 1.645= x-12.4 4.7
Transcribed Image Text:A forester measured the trees in a large woods up for sale. He concluded that the diameter of the trees was normally distributed with mean 12.4 inches and standard deviation 4.7 inches. Which of the following equations can be used to find the diameter x which is at the 90th percentile? (A) 0.90= (B) 1.28 = x-12.4 (4.7)² x 12.4 (4.7)² x-12.4 (C) 0.90= 4.7 x-12.4 (D) 1.28= 4.7 (E) 1.645= x-12.4 4.7
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