Calculate Variance of Tomato Crop Income: Steps & Answers

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Homework Help Overview

The discussion revolves around calculating the variance of income from a tomato crop based on its standard deviation and a linear relationship between crop yield and income. Participants are exploring the relationship between the standard deviation of the crop and the variance of income, as well as the Pearson correlation coefficient.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between standard deviation and variance, questioning whether squaring the standard deviation of the crop yields the variance of income. Some express confusion about how to apply the formula for variance in the context of a linear transformation of variables.

Discussion Status

There is ongoing exploration of the correct application of variance formulas and the relationship between the variables. Some participants have provided guidance on looking up relevant textbook information, while others have confirmed the application of the variance transformation formula.

Contextual Notes

Participants are working under the assumption that they have a standard deviation of 2 tons for the crop and are trying to determine the variance of income based on a linear equation. There is also a focus on the Pearson correlation coefficient, which remains unaddressed in terms of calculation.

omni
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is known that the Tomato crop (in ton) in some farm are Sampled for 10 years.
the Standard deviation of the crop was 2 ton.
the Income (Y) from the Tomato Depends on the crop (X)
according to following connection Y=3X-2 the Variance Income from the Tomato in this Sampled is 4?

if i know that the Standard deviation is 2 so i just need to do 2^2 to get the Variance yes?

and 1 more Question: the Value of Pearson product-moment correlation coefficient between the crop and the Income is?
how i can solve it?

thanks for help.
 
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omni said:
is known that the Tomato crop (in ton) in some farm are Sampled for 10 years.
the Standard deviation of the crop was 2 ton.
the Income (Y) from the Tomato Depends on the crop (X)
according to following connection Y=3X-2 the Variance Income from the Tomato in this Sampled is 4?

if i know that the Standard deviation is 2 so i just need to do 2^2 to get the Variance yes?
If you're referring to var(Y), then no, it isn't equal to 4.
and 1 more Question: the Value of Pearson product-moment correlation coefficient between the crop and the Income is?
how i can solve it?

thanks for help.
 
ok so how can i solve it?
and is told that the Standard deviation are 2
so for example if i know the Variance and it 4 so to find the Standard deviation i just need to do sqrt4.
and in this case i know the Standard deviation so to find the Variance i need to do 2^2 instead of sqrt4. no?

thank you.
 
ok so how can i solve it?
and is told that the Standard deviation are 2
so for example if i know the Variance and it 4 so to find the Standard deviation i just need to do sqrt4.
and in this case i know the Standard deviation so to find the Variance i need to do 2^2 instead of sqrt4. no?

thank you.
 
You're given stdev(X)=2 tons. You can't just square it to find the variance of Y, which is a different random variable.
 
so can you give me any way to solve it or give me direction ?

thanks
 
Look up in your textbook how the variances of X and Y are related.
 
well is will be correct to say:
if i use the formula var(aX+b)=a^2var(X) and then i will put numbers and get this:
9*4=36 and this is the Variance Income from the Tomato in this Sampled?

thanks.
 
Yes,that is correct. Are you interested in how that formula is derived?

For the variable x, the mean is given by
[tex]\mu_x= \frac{\sum x_i}{n}[/tex]
while, for the variable y, it is
[tex]\mu_y= \frac{\sum y_i}{n}[/tex].

If y= ax+ b, that is the same as
[tex]\mu_y= \frac{\sum (ax_i+ b)}{n}= \frac{a\sum x_i+ \sum b}{n}= a\frac{\sum x_i}{n}+ \frac{nb}{n}[/itex][itex]= a\mu_x+ b[/itex][/tex][itex].<br /> <br /> For the variable x, the variance is given by <br /> [tex]v_x= \frac{\sum (x_i- \mu_x)^2}{n}[/tex]<br /> while, for the variable y, it is <br /> [tex]v_y= \frac{\sum (y_i- \mu_y)^2}{n}[/tex].<br /> <br /> Again, if y= ax+ b, that is <br /> [tex]v_y= \frac{\sum (ax_i+ b- (a\mu_x+ b))^2}{n}=[/tex][/itex][tex][itex]\frac{\sum{(ax_i- a\mu_x)^2}{n}[/itex][/tex][itex] [tex]= \frac{a^2(x_i- \mu_x)^2}{n}= a^2\frac{(x_i- \mu_x)^2}{n}= a^2 v_x[/tex].[/itex]
 
  • #10
WOW.
thanks for show all this way.
 

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