A bakery sells rolls in units of a dozen. The demand X (in 1000 units) for rolls has a gamma distribution with parameters α = 3, θ = 0.5, where θ is in units of days per 1000 units of rolls. It costs $ 2 to make a unit that sells for $ 5 on the first day when the rolls are fresh. Any leftover units are sold on the second day for $ 1. How many units should be made to maximize the expected value of the profit?

Respuesta :

Answer:

The value  is  [tex]E(X) =  \$ 1.7067 [/tex]

Step-by-step explanation:

From the question we are told that

   The  parameters  are  α = 3, θ = 0.5

    The cost of making a unit on the first day  is  c = $2

    The selling price of a  unit on the first day is  s = $5

    The selling price of a leftover unit on the second day is  v  = $ 1

Generally the profit of a unit on the first day is

        [tex]p_1 = 5 - 2[/tex]

           [tex]p_1 = \$3 [/tex]

The profit of a unit on the second day is

       [tex]p_2 = 1 - 2[/tex]

=>     [tex]p_2 = - \$1 [/tex]

Generally the probability of making profit greater than $ 1 is mathematically represented as

    [tex]P(X >  1 ) = Gamma (X ,\alpha , \theta)[/tex]

=>   [tex]P(X >  1 ) = Gamma (1 ,3 , 0.5)[/tex]

Now from the gamma distribution table  we have that

    [tex]P(X >  1 ) =  0.67668[/tex]

Generally the probability of making profit less than or  equal to  $ 1 is mathematically represented as

       [tex] P(X \le  1 ) = 1 - P(X >  1 )[/tex]

=>     [tex] P(X \le  1 ) = 1 - 0.67668[/tex]

=>     [tex] P(X \le  1 ) = 0.32332[/tex]    

So  the probability of making  $3  is    [tex]P(X >  1 ) =  0.67668[/tex]

and  the probability of making  -$1  is   [tex] P(X \le  1 ) = 0.32332[/tex]  

Generally the value of profit per day is mathematically represented as

      [tex]E(X) =  3 *  P(X >  1 )   +   (-1  *  P(X \le 1 ) )[/tex]

=>     [tex]E(X) =  3 * 0.67668   +   (-1  *  0.32332 )[/tex]

=>     [tex]E(X) =  \$ 1.7067 [/tex]