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Need help with the Mean formula

Inferno

New member
I thought I understood the formula, but I can't figure out how to set it up for this problem. compute.jpg
 

stapel

Super Moderator
Staff member
I thought I understood the formula, but I can't figure out how to set it up for this problem.
Which formula? What did you plug in, where? Where did you get stuck? ;)
 

Inferno

New member
I am pretty sure I plugged the stuff in wrong. here's what I did for Set A

E20/6 = 5x10
E3.3 = 50
E=15.1

and set B

E20/6 = 50x10
E3.3=500
E=151.5
 

Subhotosh Khan

Super Moderator
I am pretty sure I plugged the stuff in wrong. here's what I did for Set A

E20/6 = 5x10
E3.3 = 50
E=15.1

and set B

E20/6 = 50x10
E3.3=500
E=151.5
Okay that is not the way to work with mean (or average).

Tell us the definition of mean (or average).
 

Inferno

New member
mean is when you add all the numbers up and divide by how many there are
 

Inferno

New member
10+20/6 = Set A is 5?

and 500+20/6= Set B is 86.7 ? are those right? I did that without the formula because the formula confuses me.
 

JeffM

New member
10+20/6 = Set A is 5?

and 500+20/6= Set B is 86.7 ? are those right? I did that without the formula because the formula confuses me.
The formula is not confusing. The problem is confusingly worded because it changes the definitions of the sets.

For set \(\displaystyle A = \{x_1,\ x_2,\ x_3,\ x_4,\ x_5\}.\)

\(\displaystyle \displaystyle \bar x_A = \dfrac{\displaystyle \sum_{i=1}^5x_i}{5} = 10 \implies\sum_{i=1}^5x_i = 5 * 10 = 50.\) Follow that?

Now let's define a new set \(\displaystyle H = \{x_1,\ x_2,\ x_3,\ x_4,\ x_5\, x_6\},\ where\ x_6 = 20.\)

So \(\displaystyle \displaystyle \bar x_H = \dfrac{\displaystyle \sum_{i=1}^6x_i}{6}.\) Still using the formula.

But what is the numerator in that formula equal to?

Here is the trick

\(\displaystyle \displaystyle \sum_{i=1}^6x_i = \left(\sum_{i=1}^5x_i\right) + x_6.\) Does that make sense?

And we know what the two terms on the right of the equation equal.

\(\displaystyle \displaystyle \sum_{i=1}^6x_i = 50 + 20 = 70 \implies \bar x_H = \dfrac{70}{6} \approx 11.67\)

Now try the second problem on your own, and let us know what you get.
 

Inferno

New member
50x10= 500

500+20= 520

520/6 = 86.7

still wrong? lol
 

Subhotosh Khan

Super Moderator
50x10= 500

500+20= 520

520/6 = 86.7

still wrong? lol
In the third line - why are you dividing by '6'?

You started 50 data points (n = 50) - you added one more. How many data points do you have now?
 

Inferno

New member
oh I guess I was looking at the 6 from Set A.

so

51x10= 510

510+20= 530

530/51 = 10.4?
 

Subhotosh Khan

Super Moderator
oh I guess I was looking at the 6 from Set A.

so

51x10= 510

510+20= 530

530/51 = 10.4?
Still wrong .... you are not thinking straight:

old sum → 50 * 10 = 500

new sum → 500 + 20 = 520

New average = 520/51 = 10.196
 
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