Stats help required

- hba

Stats help required

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- hba

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- hba

Actually i know.
\[Mean=\sum_{}^{}fx/\sum_{}^{}f\]
I know,
\[\sum_{}^{}f=2000\]

- anonymous

do you have the answer key
is the answer 7.975 ???

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## More answers

- hba

I also think that is the answer as i tried doing it.

- hba

But here x1,x2,x3 cannot be 8.5,7.5 and 8 because it is actually the mean

- hba

As mentioned in the ques

- anonymous

Okay give me a second.

- hba

Sure

- anonymous

Now suppose that we call \[
\sum fx
\]The 'total'

- anonymous

We want the 'total' of each sub population.

- anonymous

Then we can add them

- anonymous

To get the 'total' of the whole population.

- anonymous

From there we can find the mean of the whole population.

- ParthKohli

We know that the sum is \(700 \times 8.5 + 800 \times 7.5 + 500 \times 8\)

- hba

No No No 8.5,7.5 and 8 cannot be x

- hba

They are the means

- anonymous

Yeah, but he's not using that formula.

- 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

He using this: \[
\sum fx = \frac{\sum fx}{\sum f}\times \sum f
\]

- ParthKohli

Can you continue from this point?

- anonymous

i don't think there is any problem with taking them as x's .
what you alreadyy did is correct
7.975 .

- ParthKohli

lunch... g2g

- hba

But how can you say mean of x is actually x?

- anonymous

(8.5*700 + 800*7.5 + 500*8)/2000

- anonymous

these are different random variables for the whole population . you can just use them in the formula .

- hba

@wio Please justify

- hba

It says that it is the mean @sami-21

- anonymous

Okay so if you have three means... suppose they are \(m_1, m_2, m_3\)

- anonymous

\[
m_1 = \frac{\sum_1fx}{\sum_1f}
\]

- anonymous

yes it does says . and requires the mean for population . which should be taking the means of the sub populatiions .

- hba

One more thing,If that is not the formula,What is it?

- hba

@xoya Shu away

- 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

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

Notice how \[
m_1 = \frac{\sum_1 fx}{\sum_1 f} \implies m_1\sum_1 f = \sum_1 f x
\]

- hba

Okay so what would the formula basically?

- hba

be*

- 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

Thanks a lot :D :D

- anonymous

It's not the same formula, they just happen to be the same though by coincidence.

- hba

I see.

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