Breakeven Point
$begingroup$
To examine the effectiveness of its four annual advertising promotions, a mail order company has sent a questionnaire to each of its customers, asking how many of the previous year's promotions prompted orders that would not have otherwise been made. The accompanying table lists the probabilities that were derived from the questionnaire, where X is the random variable representing the number of promotions that prompted orders. If we assume that overall customer behavior next year will be the same as last year, what is the expected number of promotions that each customer will take advantage of next year by ordering goods that otherwise would not be purchased?
X: 0 1 2 3 4
P(X) 0.078 0.242 0.337 0.16 0.183
A previous analysis of historical records found that the mean value of orders for promotional goods is 29 dollars, with the company earning a gross profit of 29% on each order. Calculate the expected value of the profit contribution next year. Answer is 17.89648
The fixed cost of conducting the four promotions is estimated to be 20000 dollars with a variable cost of 4 dollars per customer for mailing and handling costs. What is the minimum number of customers required by the company in order to cover the cost of promotions? (Round your answer to the next highest integer.)
Breakeven point:?????
I have tried to do 20000 = (# of customers) (29-4)
But got it wrong.... I don't know what method to use to find the breakeven point.
Thank you!
statistics
asked Oct 14 '17 at 23:12
Gill DaveGill Dave
176
$endgroup$
add a comment |
$begingroup$
To examine the effectiveness of its four annual advertising promotions, a mail order company has sent a questionnaire to each of its customers, asking how many of the previous year's promotions prompted orders that would not have otherwise been made. The accompanying table lists the probabilities that were derived from the questionnaire, where X is the random variable representing the number of promotions that prompted orders. If we assume that overall customer behavior next year will be the same as last year, what is the expected number of promotions that each customer will take advantage of next year by ordering goods that otherwise would not be purchased?
X: 0 1 2 3 4
P(X) 0.078 0.242 0.337 0.16 0.183
A previous analysis of historical records found that the mean value of orders for promotional goods is 29 dollars, with the company earning a gross profit of 29% on each order. Calculate the expected value of the profit contribution next year. Answer is 17.89648
The fixed cost of conducting the four promotions is estimated to be 20000 dollars with a variable cost of 4 dollars per customer for mailing and handling costs. What is the minimum number of customers required by the company in order to cover the cost of promotions? (Round your answer to the next highest integer.)
Breakeven point:?????
I have tried to do 20000 = (# of customers) (29-4)
But got it wrong.... I don't know what method to use to find the breakeven point.
Thank you!
statistics
asked Oct 14 '17 at 23:12
Gill DaveGill Dave
176
$endgroup$
add a comment |
$begingroup$
To examine the effectiveness of its four annual advertising promotions, a mail order company has sent a questionnaire to each of its customers, asking how many of the previous year's promotions prompted orders that would not have otherwise been made. The accompanying table lists the probabilities that were derived from the questionnaire, where X is the random variable representing the number of promotions that prompted orders. If we assume that overall customer behavior next year will be the same as last year, what is the expected number of promotions that each customer will take advantage of next year by ordering goods that otherwise would not be purchased?
X: 0 1 2 3 4
P(X) 0.078 0.242 0.337 0.16 0.183
A previous analysis of historical records found that the mean value of orders for promotional goods is 29 dollars, with the company earning a gross profit of 29% on each order. Calculate the expected value of the profit contribution next year. Answer is 17.89648
The fixed cost of conducting the four promotions is estimated to be 20000 dollars with a variable cost of 4 dollars per customer for mailing and handling costs. What is the minimum number of customers required by the company in order to cover the cost of promotions? (Round your answer to the next highest integer.)
Breakeven point:?????
I have tried to do 20000 = (# of customers) (29-4)
But got it wrong.... I don't know what method to use to find the breakeven point.
Thank you!
statistics
asked Oct 14 '17 at 23:12
Gill DaveGill Dave
176
$endgroup$
To examine the effectiveness of its four annual advertising promotions, a mail order company has sent a questionnaire to each of its customers, asking how many of the previous year's promotions prompted orders that would not have otherwise been made. The accompanying table lists the probabilities that were derived from the questionnaire, where X is the random variable representing the number of promotions that prompted orders. If we assume that overall customer behavior next year will be the same as last year, what is the expected number of promotions that each customer will take advantage of next year by ordering goods that otherwise would not be purchased?
X: 0 1 2 3 4
P(X) 0.078 0.242 0.337 0.16 0.183
A previous analysis of historical records found that the mean value of orders for promotional goods is 29 dollars, with the company earning a gross profit of 29% on each order. Calculate the expected value of the profit contribution next year. Answer is 17.89648
The fixed cost of conducting the four promotions is estimated to be 20000 dollars with a variable cost of 4 dollars per customer for mailing and handling costs. What is the minimum number of customers required by the company in order to cover the cost of promotions? (Round your answer to the next highest integer.)
Breakeven point:?????
I have tried to do 20000 = (# of customers) (29-4)
But got it wrong.... I don't know what method to use to find the breakeven point.
Thank you!
statistics
statistics
asked Oct 14 '17 at 23:12
Gill DaveGill Dave
176
asked Oct 14 '17 at 23:12
Gill DaveGill Dave
176
asked Oct 14 '17 at 23:12
Gill DaveGill Dave
176
asked Oct 14 '17 at 23:12
Gill DaveGill Dave
176
asked Oct 14 '17 at 23:12
Gill DaveGill Dave
176
176
add a comment |
add a comment |
1 Answer
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$begingroup$
In somewhat improved notation, the PDF table for the random variable $X$ is
k: 0 1 2 3 4
P(X=k): 0.078 0.242 0.337 0.160 0.183
First, notice that $sum_{k=0}^4 P(X=k) = 1,$ as required for a PDF. Then $$E(X) = sum_{k=0}^4 kP(X = k) = 2.128.$$
The profit per order is $29(.29) = $8.41.$ So the profit on the expected $2.128$ orders per customer
is $$17.89648.$
The profit for $n$ customers is $$17.89648n.$ From what you say, I suppose the cost of a campaign reaching $n$ customers
to be $$(20,000 + 4n).$ I will leave it to you to find the number $n$ of
customers to break even, as required.
answered Oct 15 '17 at 4:38
BruceETBruceET
35.9k71440
$endgroup$
$begingroup$
Stupid question... but would I plug it in as: (20,000+4n)=17.89648?
$endgroup$
– Gill Dave
Oct 15 '17 at 4:46
$begingroup$
Almost that. Look again.
$endgroup$
– BruceET
Oct 15 '17 at 6:27
1
$begingroup$
Oh my goodness, thank you so much!
$endgroup$
– Gill Dave
Oct 15 '17 at 14:30
add a comment |
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1 Answer
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$begingroup$
In somewhat improved notation, the PDF table for the random variable $X$ is
k: 0 1 2 3 4
P(X=k): 0.078 0.242 0.337 0.160 0.183
First, notice that $sum_{k=0}^4 P(X=k) = 1,$ as required for a PDF. Then $$E(X) = sum_{k=0}^4 kP(X = k) = 2.128.$$
The profit per order is $29(.29) = $8.41.$ So the profit on the expected $2.128$ orders per customer
is $$17.89648.$
The profit for $n$ customers is $$17.89648n.$ From what you say, I suppose the cost of a campaign reaching $n$ customers
to be $$(20,000 + 4n).$ I will leave it to you to find the number $n$ of
customers to break even, as required.
answered Oct 15 '17 at 4:38
BruceETBruceET
35.9k71440
$endgroup$
$begingroup$
Stupid question... but would I plug it in as: (20,000+4n)=17.89648?
$endgroup$
– Gill Dave
Oct 15 '17 at 4:46
$begingroup$
Almost that. Look again.
$endgroup$
– BruceET
Oct 15 '17 at 6:27
1
$begingroup$
Oh my goodness, thank you so much!
$endgroup$
– Gill Dave
Oct 15 '17 at 14:30
add a comment |
$begingroup$
In somewhat improved notation, the PDF table for the random variable $X$ is
k: 0 1 2 3 4
P(X=k): 0.078 0.242 0.337 0.160 0.183
First, notice that $sum_{k=0}^4 P(X=k) = 1,$ as required for a PDF. Then $$E(X) = sum_{k=0}^4 kP(X = k) = 2.128.$$
The profit per order is $29(.29) = $8.41.$ So the profit on the expected $2.128$ orders per customer
is $$17.89648.$
The profit for $n$ customers is $$17.89648n.$ From what you say, I suppose the cost of a campaign reaching $n$ customers
to be $$(20,000 + 4n).$ I will leave it to you to find the number $n$ of
customers to break even, as required.
answered Oct 15 '17 at 4:38
BruceETBruceET
35.9k71440
$endgroup$
$begingroup$
Stupid question... but would I plug it in as: (20,000+4n)=17.89648?
$endgroup$
– Gill Dave
Oct 15 '17 at 4:46
$begingroup$
Almost that. Look again.
$endgroup$
– BruceET
Oct 15 '17 at 6:27
1
$begingroup$
Oh my goodness, thank you so much!
$endgroup$
– Gill Dave
Oct 15 '17 at 14:30
add a comment |
$begingroup$
In somewhat improved notation, the PDF table for the random variable $X$ is
k: 0 1 2 3 4
P(X=k): 0.078 0.242 0.337 0.160 0.183
First, notice that $sum_{k=0}^4 P(X=k) = 1,$ as required for a PDF. Then $$E(X) = sum_{k=0}^4 kP(X = k) = 2.128.$$
The profit per order is $29(.29) = $8.41.$ So the profit on the expected $2.128$ orders per customer
is $$17.89648.$
The profit for $n$ customers is $$17.89648n.$ From what you say, I suppose the cost of a campaign reaching $n$ customers
to be $$(20,000 + 4n).$ I will leave it to you to find the number $n$ of
customers to break even, as required.
answered Oct 15 '17 at 4:38
BruceETBruceET
35.9k71440
$endgroup$
In somewhat improved notation, the PDF table for the random variable $X$ is
k: 0 1 2 3 4
P(X=k): 0.078 0.242 0.337 0.160 0.183
First, notice that $sum_{k=0}^4 P(X=k) = 1,$ as required for a PDF. Then $$E(X) = sum_{k=0}^4 kP(X = k) = 2.128.$$
The profit per order is $29(.29) = $8.41.$ So the profit on the expected $2.128$ orders per customer
is $$17.89648.$
The profit for $n$ customers is $$17.89648n.$ From what you say, I suppose the cost of a campaign reaching $n$ customers
to be $$(20,000 + 4n).$ I will leave it to you to find the number $n$ of
customers to break even, as required.
answered Oct 15 '17 at 4:38
BruceETBruceET
35.9k71440
answered Oct 15 '17 at 4:38
BruceETBruceET
35.9k71440
answered Oct 15 '17 at 4:38
BruceETBruceET
35.9k71440
answered Oct 15 '17 at 4:38
BruceETBruceET
35.9k71440
35.9k71440
$begingroup$
Stupid question... but would I plug it in as: (20,000+4n)=17.89648?
$endgroup$
– Gill Dave
Oct 15 '17 at 4:46
$begingroup$
Almost that. Look again.
$endgroup$
– BruceET
Oct 15 '17 at 6:27
1
$begingroup$
Oh my goodness, thank you so much!
$endgroup$
– Gill Dave
Oct 15 '17 at 14:30
add a comment |
$begingroup$
Stupid question... but would I plug it in as: (20,000+4n)=17.89648?
$endgroup$
– Gill Dave
Oct 15 '17 at 4:46
$begingroup$
Almost that. Look again.
$endgroup$
– BruceET
Oct 15 '17 at 6:27
1
$begingroup$
Oh my goodness, thank you so much!
$endgroup$
– Gill Dave
Oct 15 '17 at 14:30
$begingroup$
Stupid question... but would I plug it in as: (20,000+4n)=17.89648?
$endgroup$
– Gill Dave
Oct 15 '17 at 4:46
$begingroup$
Stupid question... but would I plug it in as: (20,000+4n)=17.89648?
$endgroup$
– Gill Dave
Oct 15 '17 at 4:46
$begingroup$
Almost that. Look again.
$endgroup$
– BruceET
Oct 15 '17 at 6:27
$begingroup$
Almost that. Look again.
$endgroup$
– BruceET
Oct 15 '17 at 6:27
1
1
$begingroup$
Oh my goodness, thank you so much!
$endgroup$
– Gill Dave
Oct 15 '17 at 14:30
$begingroup$
Oh my goodness, thank you so much!
$endgroup$
– Gill Dave
Oct 15 '17 at 14:30
add a comment |
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