Let X be the temperature in at which a certain chemical reaction takes place, and let Y be the temperature in (so Y = 1.8X + 32). a. If the median of the X distribution is , show that 1.8 + 32 is the median of the Y distribution. b. How is the 90th percentile of the Y distribution related to the 90th percentile of the X distribution? Verify your conjecture. c. More generally, if Y = aX + b when a is non-zero, how is any particular percentile of the Y distribution related to the corresponding percentile of the X distribution? Distinguish the two cases when a is positive and when a is negative.

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Step-by-step explanation:

Solution:-

The random variable, Y be the temperature of chemical reaction in degree fahrenheit be a linear expression of a random variable X : The  temperature in at which a certain chemical reaction takes place.

                             Y = 1.8*X + 32

- The median of the random variate "X" is given to be equal to "η". We can mathematically express it as:

                             P ( X ≤ η ) = 0.5

- Then the median of "Y" distribution can be expressed with the help of the relation given:

                             P ( Y ≤ 1.8*η + 32 )

- The left hand side of the inequality can be replaced by the linear relation:

                             P ( 1.8*X + 32 ≤ 1.8*η + 32 )

                             P ( 1.8*X ≤ 1.8*η )   Cancel "1.8" on both sides.

                            P ( X ≤ η ) = 0.5 Proven

Hence,

- Through conjecture we proved that: (1.8*η + 32) has to be the median of distribution "Y".

b)

- Recall that the definition of proportion (p) of distribution that lie within the 90th percentile. It can be mathematically expressed as the probability of random variate "X" at 90th percentile :

                             P ( X ≤ p_.9 ) = 0.9 90th percentile

- Now use the conjecture given as a linear expression random variate "Y",

          P ( Y ≤ 1.8*p_0.9 + 32 ) = P ( 1.8*X + 32 ≤ 1.8*p_0.9 + 32 )

                                                 = P ( 1.8*X ≤ 1.8*p_0.9 )

                                                 = P ( X  ≤ p_0.9 )

                                                 = 0.9

- So from conjecture we saw that the 90th percentile of "X" distribution is also the 90th percentile of "Y" distribution.

c)

- The more general relation between two random variate "Y" and "X" is given:

                            Y = aX + b

Where, a : is either a positive or negative constant.

- Denote, (np) as the 100th percentile of the X distribution, so the corresponding 100th percentile of the Y distribution would be : (a*np + b).

- When a is positive,

                   P ( Y ≤ a*p_% + b ) = P ( a*X + b ≤ a*p_% + b )

                                                 = P ( a*X ≤ a*p_% )

                                                 = P ( X  ≤ p_% )

                                                 = np_%        

- When a is negative,

                   P ( Y ≤ a*p_% + b ) = P ( a*X + b ≤ a*p_% + b )

                                                 = P ( a*X ≤ a*p_% )

                                                 = P ( X  ≥ p_% )

                                                 = 1 - np_%        

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