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STAT410- Homework #01 Solved

1. Grades on Fall 2021 STAT 410 Exam I were not very good *. Graphed, their distribution had a shape similar to the probability density function ** 

fx(x)              80,          zero elsewhere. c

a)                  Find the value of C that makes A (x) a valid probability density function.

b)                 Find the cumulative distribution function of X, F x(x) = P (X x).

"Hint":         To double-check your answer:           should be   Fx(80) 

 

* The probability distribution is fictional, the exam has not happened yet. Hopefully, the actual grades will be slightly better than these.

 ** Exam scores should have a discrete (instead of continuous) nature. A continuous probability distribution is used as an approximation, since the alternative would have been dealing with a discrete random variable with 65 possible values ( 16, 17, 18, 79, 80 ), which is not nearly as much fun as I am describing it here.

1.         (continued)

As a way of "curving" the results, the instructor announced that he would replace each person's grade, X, with anew grade, Y where g (x) = 8 x+20  

c)                  Find the support ( the range of possible values ) of the probability distribution of Y.

d)                 Use pan (b) and the c.d.f. approach to find the c.d.f. of Y, Fy(y).

"Hint":         Fy(y) =  

e)                  Use the change-of-variable technique to find the p.d.f. of Y,            (y ).

 "Hint": fy(y)  1 0')) —dx

dy

"Hint":         To double-check your answer:           should be fy(y) = FY(y).

f)                  Is Ply equal to g(gx)?   

2.       Consider a continuous random variable X with the probability density function

fx(x) zero elsewhere.

18

16

 Consider Find the probability distribution of Y.

2

x

You are welcome to use a computer and/or calculator on any problem to evaluate any integral. For the supporting work, you should include the full integral (with the function and the bounds) and the answer. For example,

 x                       3             4

J u 2 du — x                         x 2 y dy dx = 32,                                       dx dy    92  

3

0          0          

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