hba
  • hba
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
hba
  • hba
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hba
  • hba
Actually i know. \[Mean=\sum_{}^{}fx/\sum_{}^{}f\] I know, \[\sum_{}^{}f=2000\]
anonymous
  • anonymous
do you have the answer key is the answer 7.975 ???

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hba
  • hba
I also think that is the answer as i tried doing it.
hba
  • hba
But here x1,x2,x3 cannot be 8.5,7.5 and 8 because it is actually the mean
hba
  • hba
As mentioned in the ques
anonymous
  • anonymous
Okay give me a second.
hba
  • hba
Sure
anonymous
  • anonymous
Now suppose that we call \[ \sum fx \]The 'total'
anonymous
  • anonymous
We want the 'total' of each sub population.
anonymous
  • anonymous
Then we can add them
anonymous
  • anonymous
To get the 'total' of the whole population.
anonymous
  • anonymous
From there we can find the mean of the whole population.
ParthKohli
  • ParthKohli
We know that the sum is \(700 \times 8.5 + 800 \times 7.5 + 500 \times 8\)
hba
  • hba
No No No 8.5,7.5 and 8 cannot be x
hba
  • hba
They are the means
anonymous
  • anonymous
Yeah, but he's not using that formula.
ParthKohli
  • ParthKohli
Since mean is sum of observations divided by number of observations, we know that the sum of observations multiplied by the number of observations is the sum of observations.
anonymous
  • anonymous
He using this: \[ \sum fx = \frac{\sum fx}{\sum f}\times \sum f \]
ParthKohli
  • ParthKohli
Can you continue from this point?
anonymous
  • anonymous
i don't think there is any problem with taking them as x's . what you alreadyy did is correct 7.975 .
ParthKohli
  • ParthKohli
lunch... g2g
hba
  • hba
But how can you say mean of x is actually x?
anonymous
  • anonymous
(8.5*700 + 800*7.5 + 500*8)/2000
anonymous
  • anonymous
these are different random variables for the whole population . you can just use them in the formula .
hba
  • hba
@wio Please justify
hba
  • hba
It says that it is the mean @sami-21
anonymous
  • anonymous
Okay so if you have three means... suppose they are \(m_1, m_2, m_3\)
anonymous
  • anonymous
\[ m_1 = \frac{\sum_1fx}{\sum_1f} \]
anonymous
  • anonymous
yes it does says . and requires the mean for population . which should be taking the means of the sub populatiions .
hba
  • hba
One more thing,If that is not the formula,What is it?
hba
  • hba
@xoya Shu away
anonymous
  • anonymous
the mean of all three will be: \[ m_4 = \frac{\sum_4 fx}{\sum_4x} = \frac{\sum_1 fx + \sum_2 fx+ \sum_3 fx}{\sum_1 f + \sum_2 f + \sum_3 f} \]
anonymous
  • anonymous
This is because the total frequency \(\sum_4 f\) is the sum of the sub frequencies. The total weight \(\sum_4 fx\) is equal to the sum of all the weights.
anonymous
  • anonymous
Notice how \[ m_1 = \frac{\sum_1 fx}{\sum_1 f} \implies m_1\sum_1 f = \sum_1 f x \]
hba
  • hba
Okay so what would the formula basically?
hba
  • hba
be*
anonymous
  • anonymous
Given means... \(m_1, m_2, \dots \) with total frequencies \(f_1,f_2,...\) then the mean of the totals is: \[ \frac{\sum m_if_i}{\sum f_i} \]
hba
  • hba
Thanks a lot :D :D
anonymous
  • anonymous
It's not the same formula, they just happen to be the same though by coincidence.
hba
  • hba
I see.

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