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The height of a certain type of palm tree is normally distributed with mean 45 feet and standard deviation 10 feet. A random sample of fifteen of these palm trees are observed. It is of interest to determine the probability that a single palm tree has a height of more than 50 feet. Identify the correct formula for solving. P (X > 50) (50-45) 10 VI5 = P Z > P (X > 50) (45-50) 10 V15 = P(Z > P(X > 50) = P (Z > (50-45) 10 О Р(X > 50) — P(z> (45-50) 10 P (X > 50) (50-45) 100 V15 = P ( Z > О Р(X > 50) (50-45) 100 = P O P(X > 50) = P ( z > (50-45) 10 None of the above /Unable to solve

The height of a certain type of palm tree is normally distributed with mean 45 feet and standard deviation 10 feet. A random sample of fifteen of these palm trees are observed. It is of interest to determine the probability that a single palm tree has a height of more than 50 feet. Identify the correct formula for solving. P (X > 50) (50-45) 10 VI5 = P Z > P (X > 50) (45-50) 10 V15 = P(Z > P(X > 50) = P (Z > (50-45) 10 О Р(X > 50) — P(z> (45-50) 10 P (X > 50) (50-45) 100 V15 = P ( Z > О Р(X > 50) (50-45) 100 = P O P(X > 50) = P ( z > (50-45) 10 None of the above /Unable to solve

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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The height of a certain type of palm tree is normally distributed with mean 45 feet and standard deviation 10 feet. A random sample of fifteen of these palm trees are observed. It is of interest to determine the probability that a single palm tree has a height of more than 50 feet. Identify the correct formula for solving. P (X > 50) = P ( z (50–45) 10 V15 P (X > 50) = P Z > (45–50) 10 V15 Р (X > 50) — Р (Z> (50–45) 10 V15 O P(X > 50) = P ( Z (45–50) 10 (50–45) P (X > 50) = P (z Z > 100 V15 O P(X > 50) = P ( Z (50–45) 100 O P(X > 50) = P ( Z (50–45) 10 O None of the above / Unable to solve
Transcribed Image Text:The height of a certain type of palm tree is normally distributed with mean 45 feet and standard deviation 10 feet. A random sample of fifteen of these palm trees are observed. It is of interest to determine the probability that a single palm tree has a height of more than 50 feet. Identify the correct formula for solving. P (X > 50) = P ( z (50–45) 10 V15 P (X > 50) = P Z > (45–50) 10 V15 Р (X > 50) — Р (Z> (50–45) 10 V15 O P(X > 50) = P ( Z (45–50) 10 (50–45) P (X > 50) = P (z Z > 100 V15 O P(X > 50) = P ( Z (50–45) 100 O P(X > 50) = P ( Z (50–45) 10 O None of the above / Unable to solve
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