Chapter 3 Questions. Question 3.1. Based on the nature of values that each random variable can take, we can have the following classifications:

Size: px

Start display at page:

Download "Chapter 3 Questions. Question 3.1. Based on the nature of values that each random variable can take, we can have the following classifications:"

Robert Wright
5 years ago
Views:

1 Chapter Questions Question. Based on the nature of values that each random variable can take, we can have the following classifications: X: Discrete; since X is essentially count data) Y: Continuous; since Y is measured data M: Continuous; since M is measured data N: Discrete; since N is count data P: Discrete; since P is count data Q: Continuous; since Q is measured data Question. The results could be summarized into the table below: Sample HHH HHT HTH HTT THH THT TTH TTT point w - = - = - = - = - - = - = - - = - - = - Question.5 The principal rule for solving this question is that all probabilities for all sample points within the sample space should add up to. We desire: c(x + 4) x= So: = c X [( + 4)+ ( + 4) + (4 + 4) + (9 + 4)] = c = Following the similar manner of computation: C[( ) ( ) + ( ) ( ) + ( )( )] =

2 C = + + = Question.8 We first compute the probability associated to each sample point, then add up the corresponding sample points to obtain probability associated with each value of W. P(W=-)= P(TTT) = / X / X / = /7 P(W=-) = P(HTT) + P(THT) + P(TTH) = X (/ X / X /) = /9 P(W=) = P(HHT) + P(HTH) + P(THH) = X (/ X / X /) = 4/9 P(W=) = P(HHH) = / X / X / = 8/7 Verify they sum up to one: /7 + /9 + 4/9 + 8/7 = Thus, above describes a valid probability distribution on discrete random variable W. Question.9 Integrate over the value of X to obtain the area under this part of the density function: P( < X < ) = (x+) dx = [ x +4x ] 5 = (+4)/5 Still use integration over density function to compute the probability: = P(/4 < X < /) = 5 = [ x +4x ] / 5 /4 / (x+) /4 5 = (/4+)/5 (/6 + )/5 = 9/8 dx

3 Question. First compute probability distribution of discrete random variable X: Note that the values X could take on are:,,. P(X=) = (5 ) P(X=) = (5 )( ) P(X=) = (5 )( ) ( 7 = /5 = /7 ) ( 7 ) = (X)/5 = 4/7 ( 7 ) = (5X)/5 = /7 Note that they sum up to, thus the above distribution is a valid probability distribution. The probability histogram is as following: Question. To obtain the cumulative distribution function, we just need to performing a cumulative running sum of the probability mass function on X. Thus, the cumulative distribution function is shown as following: X x < x < x < x < x < 4 x 4 F(x) = = = =.78

4 Note the ending value for the cumulative function always equals to. Question.4 We just need to directly plot in the value of x into F(x): With minutes =. hours, we have: First, compute probability density function: F(x =.5) = e 8. =.798 f(x) = F (x) = 8e 8x Then, compute the probability using integration over the density function:. p = f(x)dx. = 8 e 8x dx = 8 8 [e 8x ]. = e 8. =.798 First, find the cumulative distribution function F(x): Note that from question.: P(X=) = /7 P(X=) = 4/7 P(X=) = /7 Question.5 Then the cumulative function for F(x) is shown as following: x x < x < x < x F(x) /7 /7 + 4/7 = 6/7 6/7 + /7 = Then use F(x) for asked quantities.

5 P(X = ) = F(X = ) F(X = ) = 6/7 /7 = 4/7 (since there is no probability mass within (,)) P( < X ) = P(X ) P(X ) = F(X = ) F(X = ) = /7 = 5/7 The graph is shown as following: Question.6 Question.9 For this density function to be valid, it has to satisfy two conditions:. f(x) x

6 +. f(x)dx = Condition is easily verified as exponential function never takes on negative values. For condition verification: + f(x)dx Thus this density function is indeed valid. F(x) is computed by the integration of f(x): For x<, as f(x)= everywhere, F(x)=. Part (c) + = x 4 dx = [x ] + = F(X) = The probability can be computed using compliment: X f(a)da x = a 4 da = [a ] x = x (for x ) P = -F(4) = ( 4 ) = /64 Question.6 Given the questions asked are about probabilities of certain range, compute the cumulative distribution function first: F(x) = x f(a)da = ( a)da x

7 = [ a + a] x = x + x (if <= x <= ) If x < : trivially F(x) = If x > : F(x) = Use cumulative function: P(X /) = F(x = /) = ( ) + / = 5/9 Use compliment for computing the desired probability, with cumulative function: Part (c) P = -P(x<=.5) = -F(x=.5) = (.5 +.5) =.5 P(x <.75 x.5) = p(.5 x<.75) p(x.5) = F(x=.75) F(x=.5) F(x=.5) = (.5 +.5) (.5 +.5) = =.75

8 STA86 Problem Set Solutions Problem.8 p(x, Y = ) = p(x =, Y = ) + p(x =, Y = ) + p(x =, Y = ) = f(,) + f(,) + f(,) = = 5 p(x >, Y ) = p(x =, Y = ) + P(X =, Y = ) = f(,) + f(,) = = 7 Part (c) p(x > Y) = f(,) + f(,) + f(,) + f(,) + f(,) + f(,) ( + ) + ( + ) + ( + ) + ( + ) + ( + ) + ( + ) = = 5 Part (d) p(x + Y = 4) = f(,) + f(,) ( + ) + ( + ) = = 4 5

9 Problem.4 Integrate the joint density function over all values of Y to get marginal density of X: g(x) = f(x, y)dy = (x + y)dy = [xy + y ] = x + Integrate the joint density function over all values of X to get marginal density of Y: h(y) = f(x, y)dy = (x + y)dx = [x + xy] Part (c) = 4y + This event corresponds to the condition on random variable: X.5. As only the drive-through facility is in the picture of the problem, we only need to consider the marginal density function g(x). Integrate over the density function to get the desired event probability:.5 p(x.5) = g(x)dx.5 = x + dx

10 = [ x + x ].5 =.5 = 5 Problem.56 Note that for values of f(x,y), Y s range is actually defined based on x, we can have the following observation: Given a fixed value of Y =.5: Case : x =.. Then y < -x, and thus f(.,.5) = 6X. =.8 (Note that since it s density function, so a value greater than is possible). Case : x =.7. Then y > -x, and thus f(.7,.5) = Thus we have f(.,.5) f(.7,.5), meaning the probability density for y =.5 is dependent on x. Thus, X and Y are not independent. Note that for Y =.5, when X.5, y < -x doesn t hold and leads to probability density. Thus, we only need to consider domains of X where the density is non-zero. P(X >. Y =.5) = f(x,.5)dx..5 = f(x,.5)dx..5 = 6xdx. = [x ].5. =.48

11 Problem.57 The requirement for probability density function to be valid is that it would be integrated to over all domains. f(x, y, z)dxdydz = kxy zdxdydz = ky z dydz = kz 6 dz Thus, k =. = k = (by definition) Integrate the corresponding region with respect to each of the variable, we have: p(x < 4, Y >, < Z < ) 4 = f(x, y, z)dxdydz 4 = kxy zdxdydz = ky z dydz = 7kz 768 dz = [ 7kz 56 ]

12 = 6 56 = 5

Events A and B are said to be independent if the occurrence of A does not affect the probability of B.

Independent Events Events A and B are said to be independent if the occurrence of A does not affect the probability of B. Probability experiment of flipping a coin and rolling a dice. Sample Space: {(H,