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Exercise 4: We consider a random experiment that records a person's blood type a) Find the Ω, the sample space of this experiment b) If P(A+) = P(B+) = P(AB+) = 0.04, P(A-) = P(B-) = P(AB-) = 0.14 P(O+) = 2P(O-). Find the probabilities P(O+) and P(0-). c) Let X a variable that describes the sign of the blood type + or -. In sense that X = (+1 if the sign is + -1 if the sign is - (i) Show that X is a random variable over (Ω, Ρ(Ω)) (ii) Find the mass probability function of X (iii) Find E [2X + 3]and Var [2X + 3] Exercise 5: Let X be a discreet random variable such that X(Ω) = {-2,-1,0,1,2} has probability masse function 1. Find c P(-2) = P(1) = P(2) = 2c and P(-1) = P(0) = c 2. Find E[X] and Var[X] 3. Define random variable Y = 2X2 + 1. a. Find the mass probability function of Y b. Find the expectation of 2Y+2. c. Find the variance of 2Y+2

          Exercise 4:
We consider a random experiment that records a person's blood type
a) Find the Ω, the sample space of this experiment
b) If P(A+) = P(B+) = P(AB+) = 0.04, P(A-) = P(B-) = P(AB-) = 0.14
P(O+) = 2P(O-). Find the probabilities P(O+) and P(0-).
c) Let X a variable that describes the sign of the blood type + or -. In sense that
X =
(+1 if the sign is +
-1 if the sign is -
(i) Show that X is a random variable over (Ω, Ρ(Ω))
(ii) Find the mass probability function of X
(iii)
Find E [2X + 3]and Var [2X + 3]
Exercise 5:
Let X be a discreet random variable such that X(Ω) = {-2,-1,0,1,2} has probability masse
function
1. Find c
P(-2) = P(1) = P(2) = 2c and P(-1) = P(0) = c
2. Find E[X] and Var[X]
3. Define random variable Y = 2X2 + 1.
a. Find the mass probability function of Y
b. Find the expectation of 2Y+2.
c. Find the variance of 2Y+2

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exercise 4 we consider a random experiment that records a persons blood type a find the the sample space of this experiment b if pa pb pab 004 pa pb pab 014 po 2po find the probabilities po 13875

Added by Albert S.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

Instant Answer

Solved by Expert Sheryl Ezze

02/26/2024

Step 1

Since there are four blood types (A, B, AB, and O), and each blood type can be either positive or negative, the sample space is: Ω = {A+, A-, B+, B-, AB+, AB-, O+, O-} Show more…

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Exercise 4: We consider a random experiment that records a person's blood type a) Find the Ω, the sample space of this experiment b) If P(A+) = P(B+) = P(AB+) = 0.04, P(A-) = P(B-) = P(AB-) = 0.14 P(O+) = 2P(O-). Find the probabilities P(O+) and P(0-). c) Let X a variable that describes the sign of the blood type + or -. In sense that X = (+1 if the sign is + -1 if the sign is - (i) Show that X is a random variable over (Ω, Ρ(Ω)) (ii) Find the mass probability function of X (iii) Find E [2X + 3]and Var [2X + 3] Exercise 5: Let X be a discreet random variable such that X(Ω) = {-2,-1,0,1,2} has probability masse function 1. Find c P(-2) = P(1) = P(2) = 2c and P(-1) = P(0) = c 2. Find E[X] and Var[X] 3. Define random variable Y = 2X2 + 1. a. Find the mass probability function of Y b. Find the expectation of 2Y+2. c. Find the variance of 2Y+2

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00:01 So for your problem, you have quite a few different questions asked here, and i'm going to help you on just a set of them.

00:06 So if we look at number 3, we can see that you have a probability distribution.

00:11 It looks like it is x value of 1, 2, 3, and that the probability of x is equal to 0 .5, 0 .35, and 0 .15.

00:22 So notice that the sum of these probabilities does add up to 1.

00:27 And you want to determine what the mean is and the standard deviation is.

00:33 So that's going to be taking 1 times 0 .5 plus 2 times 0 .35 plus 3 times 0 .15.

00:41 I would also suggest that you go into your calculator and add in 1, 2, and 3 into list 1 and then these probabilities into list 2.

00:53 And then that way it will will allow you to find those values directly.

00:58 So i'm quickly doing that.

00:59 I'm up to putting the probabilities into list 2.

01:04 And so now i can go to one variable statistics on list 1 and frequency as list 2.

01:12 And so we find our mean comes out to be a value of 1 .65.

01:17 Now, the standard deviation is calculated long hand by taking the difference between each of these values squared times the corresponding probability and do that for all three categories.

01:33 And then you will add those up in square root.

01:36 And that value comes out to be 0 .7263.

01:42 Now, the next part of your question says to number four to deal with the binomial that has an n of 25 and a probability of 0 .4.

01:55 And you want to compute the probability that x is less than or equal to 24...

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