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a. If a normal distribution has 30 and 5, what is the 91st

Probability and Statistics for Engineering and the Sciences (with Student Suite Online) | 7th Edition | ISBN: 9780495382171 | Authors: Jay L. Devore ISBN: 9780495382171 196

Solution for problem 4.47 Chapter 4

Probability and Statistics for Engineering and the Sciences (with Student Suite Online) | 7th Edition

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Probability and Statistics for Engineering and the Sciences (with Student Suite Online) | 7th Edition | ISBN: 9780495382171 | Authors: Jay L. Devore

Probability and Statistics for Engineering and the Sciences (with Student Suite Online) | 7th Edition

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Problem 4.47

a. If a normal distribution has 30 and 5, what is the 91st percentile of the distribution? b. What is the 6th percentile of the distribution? c. The width of a line etched on an integrated circuit chip is normally distributed with mean 3.000 m and standard deviation .140. What width value separates the widest 10% of all such lines from the other 90%? 4

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----- Lecture 10/23/17 ----- → Section 5.3 ← x ~ N( μ, ��) → Z using Z = (x - μ )/�)� (Z ~ N(0,1)), using this to find probability (area) using:  Z - table,  Symmetry  Complement rules,  etc., Example: Mean = 174 Standard Deviation = 20 X = 200 → P(x < 200) = x ~ N(174, 200) → P((x - μ)/(�) < (200 - 174)/(20)) → P(z < 1.3) = 0.9032 Ex. 2.1 Mean = 174, Standard Deviation = 20 Find ‘A’ given that P(x < a) = 0.9505 → P(Z < (a - 174)/(20)) = 0.9505 → P(Z < + 1.65) = 0.9505 → (a - 174)/20 = 1.65 → a = 174 + (20 • 1.65) = 207. Ex 2.2 Find ‘A’ given that P(x > a) = 0.8888 NOTE: Z is Negative in this case. → P(Z > (a - 174)/(20)) = 0.8888 → P(Z > - 1.22) = 0.8888 → (a - 174)/20 = - 1.22 → a = 174 + (20 • - 1.22) = 149.6 ~ 150. Ex. 2.

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Textbook: Probability and Statistics for Engineering and the Sciences (with Student Suite Online)
Edition: 7
Author: Jay L. Devore
ISBN: 9780495382171

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a. If a normal distribution has 30 and 5, what is the 91st