As far as the method of calculating the mean is concerned the two questions are the same.

By the way you should not use E in place of .

Results 1 to 6 of 6

- September 15th 2012, 09:53 PM #1

- Joined
- Mar 2012
- From
- LA
- Posts
- 68

## Question.

I am somewhat confused on these 2. I want to know what the difference is in finding the mean. For one means, I did 4(0.0537)+5(0.1124)+6(.1473) and etc. For the other problem, to find he mean, you have to multiply, it was more like 0(0.0776);1(.103); 2(.047) and etc. I am just wondering the term I have to really look out for when I look for the mean in these 2 problems. Meaning, when are the times I have to add abd tge other times I have to multiply.

P.S. I know I can just input this in my TI84 and get the answers quickly, but I am more curious by putting it on paper.

I also know the rule is u=E[X.P(X)]

- September 16th 2012, 12:01 AM #2

- Joined
- Jan 2008
- From
- UK
- Posts
- 484
- Thanks
- 66

- September 16th 2012, 12:53 PM #3

- Joined
- Mar 2012
- From
- LA
- Posts
- 68

- September 16th 2012, 01:06 PM #4

- Joined
- Jan 2008
- From
- UK
- Posts
- 484
- Thanks
- 66

- September 16th 2012, 01:48 PM #5

- Joined
- Mar 2012
- From
- LA
- Posts
- 68

## Re: Question.

See, the first one, has 4.0.0537

**+**5. 0.1124 and then does 6+ etc, 7+... The second problem there is no plus, it's all multiplication until you add all of them up. Because it looks like 1.0.103= 0.103.. Then it's 2 . etc... There is no addition between those numbers. Get what I am saying? Sorry if it sounds confusing.

- September 16th 2012, 02:26 PM #6

- Joined
- Jan 2008
- From
- UK
- Posts
- 484
- Thanks
- 66