Solution to the Middle Term

Sept 26, Fang Wang


1. P (Complement of A) = 1-P (A) =1-0.4 =0.6 ______(C)

  1. By definition of Independent, P (A|B)=P(A). So P(A|B)=0.4______(B)
  1. Define event A as “the card is a king” , event B as “the card is a spade”. Notice A and B are not mutually exclusive, and P (A and B)=1/52. P(A)=4/52 , P(B)=13/52 , So P(A or B)=P (A)+P (B)-P (A and B)=4/52+13/52-1/52=16/52______(A)
  1. Variance (5-2X)= 2^2*Variance (X)=4*10=40______(D) Notice: Variance is always >=0.
  1. P= 1/(Possible ways of draw 4 randomly from those 10 socks)=1/210______( B)
  1. Remember the Van Diagram we draw before? The area of the intersection of two events (The shaded part) must be less than or equal to the area of any of the two circles. That is the reason why we have

P (A and B)<= P (A); P (A and B)<= P (B)______(C)

  1. E(X^2+5)=E(X^2)+5= (0^2)*0.4+(1^2)*0.3+(2^2)*0.3+5=0+0.3+1.2+5=6.5 ______(A)

8. Q1=12

Q2 (Median)=13.5

Q3=15

This is a distribution pretty skewed to the right.

  1. This problem is almost the same as the one I give to you on the review problem set. The heart part is you have to realize that the number of blue chips we get in this experiment follows a Binomial distribution. The reasons are (Check those two assumptions about Binomial Distribution): 1. In an experiment of drawing withreplacement, the P (get a blue chip) in every drawing is the same 6/10

2. The processes are independent (Which you can see from the irritating of the problem).

For a Binomial distribution, E (X)=n * p. which is 10*0.6=6. (Be careful: The n=10 here is the total number of the repeated processes. It is not the total number of chips inside the bowl which is 10 as well)

The second part of this problem is quite easy as long as you know your random variable follows a binomial distribution. P (X>=7,n=10,p=0.6)=P (X’<=3,n=10,p’=1-0.6=0.4)=0.3823(Look it up in the Binomial table.) And by the way, some students used Poisson approximation distribution, which is not right. Because when you use the Poisson approximation, the certain kinds of Binomial distribution must satisfy some conditions: The n is big enough (Which is very important in all the approximation you have learned so far and those you are going to learn), and the “p” is very small (Typically n>=20,p<=0.5 or n>=100,np<=0.1) If it is the case that your ”p” is very big, it can still work, but you have to make a transformation to approximate the complementary event. Then the conditions I just talked about apply to p’=(1-p) instead of p.

  1. 1) Poisson Distribution. When we considering about those 100-yard cable, the average number of defects on it is (2/1000)*100=0.2. This means the intensity parameter Lambda here, as well as E(X), is 0.2.

2) P (X=0)=0.819

3) Suppose X is the number of defects on a typical 100-yards cable. Z is the profit random variable. The problem says:

When X=0, Z=20-10=10, So P (Z=10)=P(X=0)=0.819

When X>=1, Z=20-20=0 (because the company only replace the cable once). P (Z=0)=1-0.819=0.181

E(Z)=10*0.819+0*0.181=8.19

  1. 1) P (Test Positive)= P(Test Positive |cancer)*P(Cancer)+P(Test Positive |No Cancer)*P(No Cancer)

=0.001*0.9 + 0.999*0.05=0.05085

2) P (Healthy | Test Positive)=P (Test Positive and Healthy)/ P (Test Positive)

= P (Test Positive |Healthy)*P (Healthy)/ P (Test Positive)

= 0.999*0.05/0.05085

=0.9823

  1. 1. P(X>Y)=P(X=0,Y=-1)+P(X=1,Y=-1)+P(X=1,Y=0)=0.2
  1. Compute the marginal distribution of X and Y first.

P (X=-1)= 0.2+0.1+0.1=0.4 (The summation of the certain row)

P (X=0)=0+0.2+0=0.2

P (X=1)=0.1+0.1+0.2=0.4

! Check whether your probability sums to 1!

P (Y=-1)=0.2+0+0.1=0.3 (The summation of the certain column)

P (Y=0)=0.1+0.2+0.1=0.4

P (Y=1)=0.1+0.2=0.3

! Check whether your probability sums to 1!

E (X)=0; E (Y)=0;

E (XY)=(-1)*(-1)*0.2+(0)*(-1)*0+(1)*(-1)*0.1+(-1)*(0)*0.1+(0)*(0)*(0.2)+(1*0)*0.1+

(-1)*(1)*0.1+(0*1)*0+(1*1)*0.2=0.2

Covariance (X, Y)=E (XY)-(EX)*(EY)=0.2-0*0=0.2

  1. Not independent. Covariance(X,Y) doesn’t equal 0.

Covariance only measures a certain kind of linear relationship between two random variables. If two random variables are independent,their covariance must be 0. But Covariance equals 0 doesn’t necessarily mean the two random variables are independent(You may image that there may be some kinds of nonlinear relationships between two random variables).The safest way to check whether tworandom variables are independent or not is: go check whether f(X,Y)=f(X)*f(Y) stands for all pairs of X and Y. If this is the case,X and Y are independent. Otherwise X and Y are not independent.