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Data for costs of production for a firm every month from January 2001 to December 2002. Data is denoted by $d_1, d_2,...,d_{24}$. The following info is calculated from the data set. Mean cost = \$2500, Range of Costs = \$7500, Sample Standard Deviation of Costs = \$5500. Relationship between Cost and Revenue is given by $R_i = -7d_i + 1000, i = 1,2,...,24$.

Answer the following:

  1. The average revenue for our data set is ....
  2. The Range of the Revenue is ....
  3. The variance in revenues of the data set is ...

Answers:

  1. -\$16,500
  2. \$52,500
  3. $38500^2$

Can someone help explain this to me how they got these numbers, I have already been trying to come up with a solution.

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1 Answer 1

I didnt understand the part (b). However part (a) and (c) are straightforward.

$(a)$ For $24$ data you are given the mean cost that is $2500$\$. If you want to calculate the average Revenue, then you can write it in a formula: $$\frac{1}{24}\sum_i R_i$$ so $1000\frac{1}{24}*24=1000$ and for the first term you have $-7\sum_i d_i/24$ here you are given what $\sum_i d_i/24$ is; that is the mean cost which is $2500$\$. So you can do the rest.

(b) Here you are asked to calculate the variance. If you multiply a data with a number which is $7$ here, then what happens to its variance? So if you subtract some value from you each element of your data then the variance will not be affected. Becase it doesnt afffect how your data $changes$. However if you multiply each element of your data then what happens? you have the sample variance as follows $$\frac{1}{24}\sum_i (d_i-\mu)^2.$$ If you multiply with $7$, then you get $$\frac{1}{24}\sum_i (7(d_i-\mu))^2=7^2\frac{1}{24}\sum_i (d_i-\mu)^2.$$ Since you are given the standard deviation $$\sqrt{\frac{1}{24}\sum_i (d_i-\mu)^2}=5500\$$$ then when you plug in this result to the above given equation, you get $$7^25500^2=(7\times5500)^2$$

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Part b) is just a rescaling and shifting of the range of Costs with the formula that gives the relation between costs and revenues. –  Raskolnikov Feb 5 '13 at 11:23
    
@Raskolnikov isn't it just the scaling? Now I guess the range is where the data lies. When the data is multiplied by $7$ then the range is also multiplied by $7$, indep. of the shifting I guess. –  Seyhmus Güngören Feb 5 '13 at 11:34
    
Yes, you're right. I actually had the lower and upper bound of the range in mind when I made the comment. But if we're just talking about the width, scaling is enough. –  Raskolnikov Feb 5 '13 at 12:05
    
Can you explain your a) and b) please? I still do not see it –  user61049 Feb 5 '13 at 15:08
    
@user61049 in part (a) you will multiply $2500$ with $-7$ and then add $1000$. In part (b) you will multiply the range $7500$ with $7$ and in part (c) you will multiply the standard deviation $5500$ with $7$ and then take the square of it. Tell me the unclear parts if there are and then i think about another way of explanation. –  Seyhmus Güngören Feb 5 '13 at 18:44

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