Let X be the temperature in °C at which a certain chemical reaction takes place, and let Y be the temperature in °F (so Y = 1.8X + 32).(a) If the median of the X distribution is p, show that 1.8 + 32 is the median of the Y distribution.P(Y s 1.8ũ + 32)= PY< 1.8 + 32= PSince i is the median of X, P(X sP) =0.50.5. Thus, 1.8ũ + 32 is the median of Y.(b) How is the 90th percentile of the Y distribution related to the 90th percentile of the X distribution?ny(.9) = ( 1.8n,(.9) + ( 32Verify your conjecture.P(Y s 1.87(.9) + 32)<1.87 .9) + 32= PlX

Here X is the temperature in 0C , and let Y be a the temperature in 0F, thus Y=1.8X+32 a.) If the…

Let X be the temperature in °C at which a certain chemical reaction takes place, and let Y be the temperature in °F (so Y = 1.8X + 32). (a) If the median of the X distribution is p, show that 1.8 + 32 is the median of the Y distribution. P(Y s 1.8ũ + 32) = PY S 1.8 + 32 = P X Since i is the median of X, P(X s p) = 0.5 0.5. Thus, 1.8 + 32 is the median of Y. (b) How is the 9oth percentile of the Y distribution related to the 90th percentile of the X distribution? n,(.9) = ( 1.8 + Verify your conjecture. P(Y s 1.8n(.9) + 32) = P 1 S 1.8n 1.9) + 32 (16% 5 Since n1.9) is the 90th percentile of X, P(X s n.9)) = 0.9 X . Thus, 1.87 4.9) + 32 is the 90th percentile of Y. (c) More generally, if Y = ax + b, how is any particular percentile of the Y distribution related to the rresponding percentile of the X distribution? ndo) - ( 32 1.8 This is a result of the following. P(Y s an Ap) + b) P Y s anlp) + b) = P Since n (p) is the (100p)th percentile of X, P(X s n (p)) = p. Thus, an (p) + b is the (100p)th percentile of Y.

Let X be the temperature in °C at which a certain chemical reaction takes place, and let Y be the temperature in °F (so Y = 1.8X + 32). (a) If the median of the X distribution is p, show that 1.8 + 32 is the median of the Y distribution. P(Y s 1.8ũ + 32) = PY S 1.8 + 32 = P X Since i is the median of X, P(X s p) = 0.5 0.5. Thus, 1.8 + 32 is the median of Y. (b) How is the 9oth percentile of the Y distribution related to the 90th percentile of the X distribution? n,(.9) = ( 1.8 + Verify your conjecture. P(Y s 1.8n(.9) + 32) = P 1 S 1.8n 1.9) + 32 (16% 5 Since n1.9) is the 90th percentile of X, P(X s n.9)) = 0.9 X . Thus, 1.87 4.9) + 32 is the 90th percentile of Y. (c) More generally, if Y = ax + b, how is any particular percentile of the Y distribution related to the rresponding percentile of the X distribution? ndo) - ( 32 1.8 This is a result of the following. P(Y s an Ap) + b) P Y s anlp) + b) = P Since n (p) is the (100p)th percentile of X, P(X s n (p)) = p. Thus, an (p) + b is the (100p)th percentile of Y.

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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