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A small grocery store sells fresh produce, which it obtains from a local farmer. During the strawberry season, demand for fresh strawberries can be reasonably approximated using a normal distribution with a mean of 40 quarts per day and a standard deviation of 6 quarts per day. Excess costs run 35 cents per quart. The grocer orders 49 quarts per day.

a. What is the implied cost of shortage per quart?

b.Why might this be a reasonable figure?

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Answer:

(a)The implied cost of shortage per quart is = $4.75

(b) This could be viewed as reasonable figure, because is (approximately) equal to the loss per quart of strawberry.

Explanation:

Solution

Given that:

Mean =μ = 40

Standard deviation =σ = 6

Excess cost= Ce =$0.35

The amount ordered =S₀= 49

Thus

Z =(49 -40)/6

=1.5

Now

From the Table Z, we have the service level which is,

P(X <49 ) = P(Z < 1.5)

= 0.9332

Since we know that,

Service level (SL) =Cs/Cs+Ce

So,

0,9332 =Cs/Cs+0.35

Thus

0.9332Cs + 0.35* 0.9332 =Cs

0.0668Cs =0.32662

Hence

Cs = $4.75

(a) The implied cost of shortage per quart is = $4.75

(b) Therefore,this could be regarded as reasonable figure, because is (approximately) equal to the loss per quart of strawberry.

Cricetus

(a) "$4.75" would be the implied cost of shortage per quart.

(b) Just because is equivalent to the loss of strawberry, this could be the reasonable figure.

According to the question,

Mean,

  • [tex]\mu = 40[/tex]

Standard deviation,

  • [tex]\sigma = 6[/tex]

Excess cost,

  • [tex]C_e = 0.35[/tex]

Amount ordered,

  • [tex]S_o = 49[/tex]

Now,

→ [tex]Z = \frac{S_o-\mu}{\sigma}[/tex]

      [tex]= \frac{49-40}{6}[/tex]

      [tex]= 1.5[/tex]

With the help of Z-table, we get

→ [tex]P(X < 49) = P(Z < 1.5)[/tex]

                    [tex]= 0.9332[/tex]

As we know,

→ Service level,         [tex]SL = \frac{C_s}{C_s+C_e}[/tex]

By substituting the values, we get

                             [tex]0.9332 = \frac{C_s}{C_s+0.35}[/tex]

[tex]0.9332 C_s +0.35\times 0.9332 = C_s[/tex]                                

                         [tex]0.0668 C_s = 0.32662[/tex]

                                   [tex]C_s = \frac{0.32662}{0.0668}[/tex]

                                        [tex]= 4.75[/tex] ($)      

Thus the above response is correct.

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