Statistics Question: Revenue & Cost












0












$begingroup$


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.






statistics standard-deviation







share|cite|improve this question











$endgroup$












  • $begingroup$
    I'd like to know how they got those numbers too. I don't mean the numbers in the answers (those are easy), I mean the numbers in the question: the mean, range, and sample standard deviation of cost. I do not see how the values given for those three numbers can possibly occur together.
    $endgroup$
    – David K
    Oct 13 '15 at 5:19


















0












$begingroup$


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.






statistics standard-deviation







share|cite|improve this question











$endgroup$












  • $begingroup$
    I'd like to know how they got those numbers too. I don't mean the numbers in the answers (those are easy), I mean the numbers in the question: the mean, range, and sample standard deviation of cost. I do not see how the values given for those three numbers can possibly occur together.
    $endgroup$
    – David K
    Oct 13 '15 at 5:19
















0












0








0





$begingroup$


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.






statistics standard-deviation







share|cite|improve this question











$endgroup$




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.






statistics standard-deviation




statistics standard-deviation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 5 '13 at 11:21









Raskolnikov

12.5k23571




12.5k23571










asked Feb 5 '13 at 8:32









user61049user61049

135




135












  • $begingroup$
    I'd like to know how they got those numbers too. I don't mean the numbers in the answers (those are easy), I mean the numbers in the question: the mean, range, and sample standard deviation of cost. I do not see how the values given for those three numbers can possibly occur together.
    $endgroup$
    – David K
    Oct 13 '15 at 5:19




















  • $begingroup$
    I'd like to know how they got those numbers too. I don't mean the numbers in the answers (those are easy), I mean the numbers in the question: the mean, range, and sample standard deviation of cost. I do not see how the values given for those three numbers can possibly occur together.
    $endgroup$
    – David K
    Oct 13 '15 at 5:19


















$begingroup$
I'd like to know how they got those numbers too. I don't mean the numbers in the answers (those are easy), I mean the numbers in the question: the mean, range, and sample standard deviation of cost. I do not see how the values given for those three numbers can possibly occur together.
$endgroup$
– David K
Oct 13 '15 at 5:19






$begingroup$
I'd like to know how they got those numbers too. I don't mean the numbers in the answers (those are easy), I mean the numbers in the question: the mean, range, and sample standard deviation of cost. I do not see how the values given for those three numbers can possibly occur together.
$endgroup$
– David K
Oct 13 '15 at 5:19












1 Answer
1






active

oldest

votes


















0












$begingroup$

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$$ so $1000frac{1}{24}*24=1000$ and for the first term you have $-7sum_i d_i/24$ here you are given what $sum_i d_i/24$ is; that is the mean cost which is $2500$$. 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 $$frac{1}{24}sum_i (7(d_i-mu))^2=7^2frac{1}{24}sum_i (d_i-mu)^2.$$ 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 $$7^25500^2=(7times5500)^2$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Part b) is just a rescaling and shifting of the range of Costs with the formula that gives the relation between costs and revenues.
    $endgroup$
    – Raskolnikov
    Feb 5 '13 at 11:23










  • $begingroup$
    @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.
    $endgroup$
    – Seyhmus Güngören
    Feb 5 '13 at 11:34












  • $begingroup$
    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.
    $endgroup$
    – Raskolnikov
    Feb 5 '13 at 12:05










  • $begingroup$
    Can you explain your a) and b) please? I still do not see it
    $endgroup$
    – user61049
    Feb 5 '13 at 15:08










  • $begingroup$
    @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.
    $endgroup$
    – Seyhmus Güngören
    Feb 5 '13 at 18:44













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0












$begingroup$

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$$ so $1000frac{1}{24}*24=1000$ and for the first term you have $-7sum_i d_i/24$ here you are given what $sum_i d_i/24$ is; that is the mean cost which is $2500$$. 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 $$frac{1}{24}sum_i (7(d_i-mu))^2=7^2frac{1}{24}sum_i (d_i-mu)^2.$$ 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 $$7^25500^2=(7times5500)^2$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Part b) is just a rescaling and shifting of the range of Costs with the formula that gives the relation between costs and revenues.
    $endgroup$
    – Raskolnikov
    Feb 5 '13 at 11:23










  • $begingroup$
    @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.
    $endgroup$
    – Seyhmus Güngören
    Feb 5 '13 at 11:34












  • $begingroup$
    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.
    $endgroup$
    – Raskolnikov
    Feb 5 '13 at 12:05










  • $begingroup$
    Can you explain your a) and b) please? I still do not see it
    $endgroup$
    – user61049
    Feb 5 '13 at 15:08










  • $begingroup$
    @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.
    $endgroup$
    – Seyhmus Güngören
    Feb 5 '13 at 18:44


















0












$begingroup$

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$$ so $1000frac{1}{24}*24=1000$ and for the first term you have $-7sum_i d_i/24$ here you are given what $sum_i d_i/24$ is; that is the mean cost which is $2500$$. 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 $$frac{1}{24}sum_i (7(d_i-mu))^2=7^2frac{1}{24}sum_i (d_i-mu)^2.$$ 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 $$7^25500^2=(7times5500)^2$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Part b) is just a rescaling and shifting of the range of Costs with the formula that gives the relation between costs and revenues.
    $endgroup$
    – Raskolnikov
    Feb 5 '13 at 11:23










  • $begingroup$
    @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.
    $endgroup$
    – Seyhmus Güngören
    Feb 5 '13 at 11:34












  • $begingroup$
    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.
    $endgroup$
    – Raskolnikov
    Feb 5 '13 at 12:05










  • $begingroup$
    Can you explain your a) and b) please? I still do not see it
    $endgroup$
    – user61049
    Feb 5 '13 at 15:08










  • $begingroup$
    @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.
    $endgroup$
    – Seyhmus Güngören
    Feb 5 '13 at 18:44
















0












0








0





$begingroup$

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$$ so $1000frac{1}{24}*24=1000$ and for the first term you have $-7sum_i d_i/24$ here you are given what $sum_i d_i/24$ is; that is the mean cost which is $2500$$. 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 $$frac{1}{24}sum_i (7(d_i-mu))^2=7^2frac{1}{24}sum_i (d_i-mu)^2.$$ 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 $$7^25500^2=(7times5500)^2$$






share|cite|improve this answer











$endgroup$



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$$ so $1000frac{1}{24}*24=1000$ and for the first term you have $-7sum_i d_i/24$ here you are given what $sum_i d_i/24$ is; that is the mean cost which is $2500$$. 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 $$frac{1}{24}sum_i (7(d_i-mu))^2=7^2frac{1}{24}sum_i (d_i-mu)^2.$$ 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 $$7^25500^2=(7times5500)^2$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Feb 5 '13 at 11:24

























answered Feb 5 '13 at 11:16









Seyhmus GüngörenSeyhmus Güngören

5,57131736




5,57131736












  • $begingroup$
    Part b) is just a rescaling and shifting of the range of Costs with the formula that gives the relation between costs and revenues.
    $endgroup$
    – Raskolnikov
    Feb 5 '13 at 11:23










  • $begingroup$
    @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.
    $endgroup$
    – Seyhmus Güngören
    Feb 5 '13 at 11:34












  • $begingroup$
    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.
    $endgroup$
    – Raskolnikov
    Feb 5 '13 at 12:05










  • $begingroup$
    Can you explain your a) and b) please? I still do not see it
    $endgroup$
    – user61049
    Feb 5 '13 at 15:08










  • $begingroup$
    @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.
    $endgroup$
    – Seyhmus Güngören
    Feb 5 '13 at 18:44




















  • $begingroup$
    Part b) is just a rescaling and shifting of the range of Costs with the formula that gives the relation between costs and revenues.
    $endgroup$
    – Raskolnikov
    Feb 5 '13 at 11:23










  • $begingroup$
    @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.
    $endgroup$
    – Seyhmus Güngören
    Feb 5 '13 at 11:34












  • $begingroup$
    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.
    $endgroup$
    – Raskolnikov
    Feb 5 '13 at 12:05










  • $begingroup$
    Can you explain your a) and b) please? I still do not see it
    $endgroup$
    – user61049
    Feb 5 '13 at 15:08










  • $begingroup$
    @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.
    $endgroup$
    – Seyhmus Güngören
    Feb 5 '13 at 18:44


















$begingroup$
Part b) is just a rescaling and shifting of the range of Costs with the formula that gives the relation between costs and revenues.
$endgroup$
– Raskolnikov
Feb 5 '13 at 11:23




$begingroup$
Part b) is just a rescaling and shifting of the range of Costs with the formula that gives the relation between costs and revenues.
$endgroup$
– Raskolnikov
Feb 5 '13 at 11:23












$begingroup$
@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.
$endgroup$
– Seyhmus Güngören
Feb 5 '13 at 11:34






$begingroup$
@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.
$endgroup$
– Seyhmus Güngören
Feb 5 '13 at 11:34














$begingroup$
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.
$endgroup$
– Raskolnikov
Feb 5 '13 at 12:05




$begingroup$
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.
$endgroup$
– Raskolnikov
Feb 5 '13 at 12:05












$begingroup$
Can you explain your a) and b) please? I still do not see it
$endgroup$
– user61049
Feb 5 '13 at 15:08




$begingroup$
Can you explain your a) and b) please? I still do not see it
$endgroup$
– user61049
Feb 5 '13 at 15:08












$begingroup$
@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.
$endgroup$
– Seyhmus Güngören
Feb 5 '13 at 18:44






$begingroup$
@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.
$endgroup$
– Seyhmus Güngören
Feb 5 '13 at 18:44




















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Has there ever been an instance of an active nuclear power plant within or near a war zone?

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36 3 The question Has there ever been a military combat going on around an active (running or in outage but not decommissioned) commercial nuclear power plant? Or dangerously near it? If yes, how did the plant and its personnel fare in such situation? Why I ask Let's assume for the scope of this question modern commercial nuclear power plants (CANDU, PWR, BWR of generation II plus all of gen III and newer) are usually reasonably safe to operate given that the engineering has been done right and there is a whole cohort of very well trained personnel on-site at all times, well rested and with considerable resources on their hands. Things hard to get with an armed conflict like the recent Syrian war raging around. I can't seem to find any reference or a comprehensive article about this, hence this question.
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