Breakeven Point











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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!










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    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!










    share|cite|improve this question
























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      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!










      share|cite|improve this question













      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






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      asked Oct 14 '17 at 23:12









      Gill Dave

      174




      174






















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          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.






          share|cite|improve this answer





















          • Stupid question... but would I plug it in as: (20,000+4n)=17.89648?
            – Gill Dave
            Oct 15 '17 at 4:46












          • Almost that. Look again.
            – BruceET
            Oct 15 '17 at 6:27






          • 1




            Oh my goodness, thank you so much!
            – Gill Dave
            Oct 15 '17 at 14:30











          Your Answer





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          1 Answer
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          up vote
          0
          down vote













          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.






          share|cite|improve this answer





















          • Stupid question... but would I plug it in as: (20,000+4n)=17.89648?
            – Gill Dave
            Oct 15 '17 at 4:46












          • Almost that. Look again.
            – BruceET
            Oct 15 '17 at 6:27






          • 1




            Oh my goodness, thank you so much!
            – Gill Dave
            Oct 15 '17 at 14:30















          up vote
          0
          down vote













          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.






          share|cite|improve this answer





















          • Stupid question... but would I plug it in as: (20,000+4n)=17.89648?
            – Gill Dave
            Oct 15 '17 at 4:46












          • Almost that. Look again.
            – BruceET
            Oct 15 '17 at 6:27






          • 1




            Oh my goodness, thank you so much!
            – Gill Dave
            Oct 15 '17 at 14:30













          up vote
          0
          down vote










          up vote
          0
          down vote









          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.






          share|cite|improve this answer












          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Oct 15 '17 at 4:38









          BruceET

          35k71440




          35k71440












          • Stupid question... but would I plug it in as: (20,000+4n)=17.89648?
            – Gill Dave
            Oct 15 '17 at 4:46












          • Almost that. Look again.
            – BruceET
            Oct 15 '17 at 6:27






          • 1




            Oh my goodness, thank you so much!
            – Gill Dave
            Oct 15 '17 at 14:30


















          • Stupid question... but would I plug it in as: (20,000+4n)=17.89648?
            – Gill Dave
            Oct 15 '17 at 4:46












          • Almost that. Look again.
            – BruceET
            Oct 15 '17 at 6:27






          • 1




            Oh my goodness, thank you so much!
            – Gill Dave
            Oct 15 '17 at 14:30
















          Stupid question... but would I plug it in as: (20,000+4n)=17.89648?
          – Gill Dave
          Oct 15 '17 at 4:46






          Stupid question... but would I plug it in as: (20,000+4n)=17.89648?
          – Gill Dave
          Oct 15 '17 at 4:46














          Almost that. Look again.
          – BruceET
          Oct 15 '17 at 6:27




          Almost that. Look again.
          – BruceET
          Oct 15 '17 at 6:27




          1




          1




          Oh my goodness, thank you so much!
          – Gill Dave
          Oct 15 '17 at 14:30




          Oh my goodness, thank you so much!
          – Gill Dave
          Oct 15 '17 at 14:30


















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