ECMT quiz 2 (1 Viewer)

[s.m.i.t.h]

is there gonna be a practice set of questions like last time?
when are they gonna go up on blckboard?

 

[04er]

Exactly what I was going to ask. Also, if my work shop is on Friday, does that mean I have both assessments (quiz and assignment) on next Friday?

 

[absolution*]

farrrk.

i hope so.

ecmt is pooh.

bring back tig.

 

[Lainee]

The questions should be up on Blackboard on Monday.

And I think everyone has to do the quiz as well as hand in the assignment in next week's workshop...

 

[kow_dude]

There is an announcement that the possible questions will be posted on blackboard soon

 

[stazi]

9am Monday morning it will be up. I just wonder what chapters theyll test us on.

 

[stazi]

Got a question:
For Promblems of Section 5.2:
Q.5.7:
It gives you a probability dristribution for variables X and Y:
P(XiYi) X Y
0.4 100 200
0.6 200 100

Compute:
a.E(X)

How do we compute the expected value of X, as I'm not sure how to find out what the probability of just X is or how to work it out.

 

[sarevok]

i think P(XiYi) just means the probabiltiy of x or y occuring are both that value

*shrug* seems to work for q 5.8

 

[stazi]

Hmm it shoudn't be:
P(XiYi) = P(Xi) times P(Yi)

 

[04er]

There is an announcement that the possible questions will be posted on blackboard soon

YES! good to hear

 

[stazi]

havent gotten an answer for the last Q I asked yet, but I have another one.
Week 7 workshop sheet. Question 1.
Consider a population consisting of 0,4,12
x ...........0............4.............12
P(x)........0.3.......0.2............0.5
Random sample of n=2 measurements is selected from the population

b) Calculate the probability for each possible sample
To do this I'm pretty sure we just times the probabilities together, correct me if I'm wrong.
eg. P{0,0}=0.3*0.3=0.09

b) This is the one I'm not sure how to approach:
Calculate the mean for each possible sample.
Is it just mean{0,1}= (X1+X2)/N = 0+1/2 = 0.5?
Or is it something more complex as probabilities are involved

 

[jpr333]

Mean is the equivalent of expected value.

 

[stazi]

so it would be E(X1)+E(X2) divided by 2 etc?
eg.
Mean{0,0}=(0.3*0.3)/2 = 0.3?

 

[jpr333]

No expected value is the sum of X1*PX1, X2*PX2, X3*PX3 etc. So in this case for a sample (0,4) expected value (or mean) for the sample would be (0*0.3)+(4*0.2)

 

[stazi]

its up now. from first glance, doesnt seem difficult. However, there are A LOT!!!! of questions

 

[absolution*]

homorific

 

[absolution*]

Dont laugh at me. Im crap and havnt been to any workshops/lectures since like wk.6. Most of these will be horribly wrong. I dont understand the normal distribution table.

1.
a) ? 4.2? wtf. i dont get it.
b) 0.9367
c) 0.3497
d) z = -2.2
e) P(z<-2.27)

2.
a)c=58.25
b)a=40.2, b=59.8
c) 0.9192
d) i dont get it. does the question make sense?

3
a) 0.062
b)?
c) 62 batteries
d) 0.4761
e)?
f) 351.69< Xbar < 398
g) i) 21 ii) 161/141 - i cant tell what i wrote i was on the train
h) 0.18298 < p < 0.357
i)$127.28 < p < $128.71
j) 3.0008 < Xbar < 3.019911
k) I looked this up in the book but have no idea what it means even after reading the explanation
l) ?

 

[011]

Can someone explain the 4.2 thing?

 

[stazi]

I'll come to the rescue tomorrow and post up answers. maybe in part today.
GO TEAM STAS!

 

[sarevok]

If any one is confused about 3g)iii as I was, the variance of the sum of two random variables when they're independent is just the sum of their variances (i.e. no need to compute the covariance)

see: http://cnx.rice.edu/content/m10656/latest/ under 'properties of variance'