Need help with the Mean formula

I

Inferno

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Sep 11, 2013
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I thought I understood the formula, but I can't figure out how to set it up for this problem. compute.jpg
 
Inferno said:
I thought I understood the formula, but I can't figure out how to set it up for this problem.
Click to expand...
Which formula? What did you plug in, where? Where did you get stuck? ;)
 
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
 
Inferno said:
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
Click to expand...

Okay that is not the way to work with mean (or average).

Tell us the definition of mean (or average).
 
mean is when you add all the numbers up and divide by how many there are
 
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.
 
Inferno said:
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.
Click to expand...
The formula is not confusing. The problem is confusingly worded because it changes the definitions of the sets.

For set A={x1, x2, x3, x4, x5}.\displaystyle A = \{x_1,\ x_2,\ x_3,\ x_4,\ x_5\}.

xˉA=∑i=15xi5=10  ⟹  ∑i=15xi=5∗10=50.\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 H={x1, x2, x3, x4, x5 x6}, where x6=20.\displaystyle H = \{x_1,\ x_2,\ x_3,\ x_4,\ x_5\, x_6\},\ where\ x_6 = 20.

So xˉH=∑i=16xi6.\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

∑i=16xi=(∑i=15xi)+x6.\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.

∑i=16xi=50+20=70  ⟹  xˉH=706≈11.67\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.
 
attachment.php
Inferno said:
50x10= 500

500+20= 520

520/6 = 86.7

still wrong? lol
Click to expand...

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

so

51x10= 510

510+20= 530

530/51 = 10.4?
 
Inferno said:
oh I guess I was looking at the 6 from Set A.

so

51x10= 510

510+20= 530

530/51 = 10.4?
Click to expand...

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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