uri
  • uri
Stats help Q.The observations 3.2,5.8,7.9 and 4.5 have frequencies x-2,x-3,and x+6 respectively with unknown value of x.If the mean of the frequency distribution is 4.816.Find the actual values of frequencies?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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nubeer
  • nubeer
well do u know how to find the mean??
uri
  • uri
We have been using this formula |dw:1378232957885:dw|
anonymous
  • anonymous
Add up the frequencies to get the total frequency first.

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anonymous
  • anonymous
multiply the observations and frequencies and add them up to get the total of the observations.
anonymous
  • anonymous
\[ \frac{3.2(x-2)+5.8(x-3)+4.5(x+6)}{(x-2)+(x-3)+(x+6)} = 4.816 \]
anonymous
  • anonymous
You want to solve for \(x\) first.
anonymous
  • anonymous
\[ \frac{\sum fx}{\sum f}=\frac{3.2(x-2)+5.8(x-3)+4.5(x+6)}{(x-2)+(x-3)+(x+6)} = 4.816 = \overline{x} \]
uri
  • uri
We make a table so i guess that's how i arrange them:|dw:1378233042858:dw|
uri
  • uri
Btw there are 4 observations and 3 frequencies.
anonymous
  • anonymous
No, there are 3 observations and then a mean frequency.
anonymous
  • anonymous
Means are calculated, not observed directly.
anonymous
  • anonymous
Try solving for \(x\) in the equation I wrote above.
uri
  • uri
don't get you.It says *The observations 3.2,5.8,7.9 and 4.5*
anonymous
  • anonymous
Actually I'll change it because it shouldn't be \(x\) for observations. \(\color{blue}{\text{Originally Posted by}}\) @wio \[ \frac{\sum fn}{\sum f}=\frac{3.2(x-2)+5.8(x-3)+4.5(x+6)}{(x-2)+(x-3)+(x+6)} = 4.816 = \overline{n} \] \(\color{blue}{\text{End of Quote}}\)
anonymous
  • anonymous
Okay, you're saying the problem is incorrect?
uri
  • uri
What about the observation 7.9? :o
anonymous
  • anonymous
I dunno, we don't know the frequency for it
uri
  • uri
lol we don't have frequency for 4.5 actually.
uri
  • uri
see the table i made.
uri
  • uri
@amistre64
amistre64
  • amistre64
3.2(x-2) 5.8(x-3) 7.9(x+6) 4.5(f) If the mean of the frequency distribution (weighted average) is 4.816.Find the actual values of frequencies? 3.2(x-2) + 5.8(x-3) + 7.9(x+6) + 4.5(f) ----------------------------------- = 4.816 (x-2) + (x-3) + (x+6) + (f) the 4.5 would have to account for the remainging "percentage"
amistre64
  • amistre64
x-2 + x-3 + x+6 + f 3x+1+f = 100 x = (99 - f)/3 or, f = 99-3x
uri
  • uri
Oh so i should add all the frequencies? :3 Here |dw:1378234507330:dw|
amistre64
  • amistre64
3.2(x-2) + 5.8(x-3) + 7.9(x+6) + 4.5(99-3x) ---------------------------------------- = 4.816 100 3.2(x-2) + 5.8(x-3) + 7.9(x+6) + 4.5(99-3x) = 481.6 solve for x
uri
  • uri
3.2x-6.4+5.8x-17.4+7.9x+47.4+445.5-13.5x=481.6 3.2x+5.8x+7.9x-13.5x-6.4-17.4+47.4+445.5=481.6 3.4x+469.1=481.6 3.4x=481.6-469.1 3.4x=12.5 x=12.5/3.4 x=3.67647
amistre64
  • amistre64
thats what i get .. assuming ove interpreted this stuff correctly
uri
  • uri
Ah so is that the ansur? :o
amistre64
  • amistre64
the mean is the expected value; which takes each value and assigns to it some part out of 100; then adds up all the calculations
uri
  • uri
Can you tell me how did you get 99-3x? :/
amistre64
  • amistre64
\[a(\frac{x-2}{100})+b(\frac{x-3}{100})+c(\frac{x+6}{100})+d(\frac{f}{100})=4.816\] the sum of the numerators would have to be 100 x-2+x-3+x+6+f = 100, solve for f
uri
  • uri
I've never been taught this...We just did some simple question using this formula... |dw:1378235723640:dw|
amistre64
  • amistre64
the only caveat i see is that the same frequency would have to result if say we used denominators of 1000
amistre64
  • amistre64
your trying to do a weighted average is all:\[\frac{af_1+bf_2+cf_3+df_4}{f_1+f_2+f_3+f_4}\] \[~~~a\frac{af_1}{f_1+f_2+f_3+f_4}\\+b\frac{f_2}{f_1+f_2+f_3+f_4}\\+c\frac{f_3}{f_1+f_2+f_3+f_4}\\+d\frac{f_4}{f_1+f_2+f_3+f_4}\]
amistre64
  • amistre64
bad little a up top ....
amistre64
  • amistre64
the missing frequency ... if it is missing .... is throwing a loop into this
uri
  • uri
I don't really understand but thankyouu :) I'll take a stats class tommorrow.
amistre64
  • amistre64
ok, im not that sure if equating it to 100 is a good idea .... but then I dont really have a better one to try out :/
amistre64
  • amistre64
maybe x is equal to the mean and they are presenting distances? i cant say fer sure
uri
  • uri
Hmm.I will do this tommorrow and tell you what my stats teacher says or explains :)
uri
  • uri
hello @amistre64 i asked him.There was a typo in the question.The frequency was *x-2,x,x-3,and x+6*
uri
  • uri
That way i was able to solve it.:)
amistre64
  • amistre64
i had a feeling one of them was "x" ... good job tho ;)
uri
  • uri
:)

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