Math Problem Statement

Suppose that a grocery store purchases 5 cartons of skim milk at the wholesale price of $1.20 per carton and retails the milk at $1.65 per carton. After the expiration date, the unsold milk is removed from the shelf and the grocer receives a credit from the distributor equal to three-fourths of the wholesale price. If the probability distribution of the random variable X, the number of cartons that are sold from this lot, is in the picture provided. find the expected profit

Suppose that a grocery store purchases 5 cartons of skim milk at the wholesale price of $1.20 per carton and retails the milk at $1.65 per carton. After the expiration date, the unsold milk is removed from the shelf and the grocer receives a credit from the distributor equal to three-fourths of the wholesale price. If the probability distribution of the random variable X, the number of cartons that are sold from this lot, is in the picture provided, find the expected profit.

Solution

To find the expected profit based on the provided probability distribution, we need to calculate the profit for each possible value of XX, the number of cartons sold, and then use the expected value formula for a discrete random variable.

Step-by-step process:

  1. Wholesale cost: Each carton costs $1.20 to the grocer.
  2. Retail price: Each carton is sold for $1.65.
  3. Credit for unsold cartons: For each unsold carton, the grocer receives a credit of 34×1.20=0.90\frac{3}{4} \times 1.20 = 0.90.

Profit for selling XX cartons:

  • If XX cartons are sold, the grocer earns a profit from those cartons, and also gets a credit for the unsold 5X5 - X cartons.

The profit function can be written as: Profit(X)=X×(1.651.20)+(5X)×0.90\text{Profit}(X) = X \times (1.65 - 1.20) + (5 - X) \times 0.90 Profit(X)=X×0.45+(5X)×0.90\text{Profit}(X) = X \times 0.45 + (5 - X) \times 0.90

Now, let's compute the profit for each possible value of XX:

  • For X=0X = 0: Profit(0)=0×0.45+(50)×0.90=5×0.90=4.50\text{Profit}(0) = 0 \times 0.45 + (5 - 0) \times 0.90 = 5 \times 0.90 = 4.50

  • For X=1X = 1: Profit(1)=1×0.45+(51)×0.90=0.45+4×0.90=0.45+3.60=4.05\text{Profit}(1) = 1 \times 0.45 + (5 - 1) \times 0.90 = 0.45 + 4 \times 0.90 = 0.45 + 3.60 = 4.05

  • For X=2X = 2: Profit(2)=2×0.45+(52)×0.90=0.90+3×0.90=0.90+2.70=3.60\text{Profit}(2) = 2 \times 0.45 + (5 - 2) \times 0.90 = 0.90 + 3 \times 0.90 = 0.90 + 2.70 = 3.60

  • For X=3X = 3: Profit(3)=3×0.45+(53)×0.90=1.35+2×0.90=1.35+1.80=3.15\text{Profit}(3) = 3 \times 0.45 + (5 - 3) \times 0.90 = 1.35 + 2 \times 0.90 = 1.35 + 1.80 = 3.15

  • For X=4X = 4: Profit(4)=4×0.45+(54)×0.90=1.80+1×0.90=1.80+0.90=2.70\text{Profit}(4) = 4 \times 0.45 + (5 - 4) \times 0.90 = 1.80 + 1 \times 0.90 = 1.80 + 0.90 = 2.70

  • For X=5X = 5: Profit(5)=5×0.45+(55)×0.90=2.25+0=2.25\text{Profit}(5) = 5 \times 0.45 + (5 - 5) \times 0.90 = 2.25 + 0 = 2.25

Expected profit:

The expected value E[X]E[X] of the random variable XX is given by the formula: E[Profit]=xProfit(x)P(X=x)E[\text{Profit}] = \sum_{x} \text{Profit}(x) \cdot P(X = x)

From the table in the image, we can use the probabilities for each XX:

E[Profit]=4.50×115+4.05×215+3.60×215+3.15×315+2.70×415+2.25×315E[\text{Profit}] = 4.50 \times \frac{1}{15} + 4.05 \times \frac{2}{15} + 3.60 \times \frac{2}{15} + 3.15 \times \frac{3}{15} + 2.70 \times \frac{4}{15} + 2.25 \times \frac{3}{15}

Let's compute this sum step by step:

E[Profit]=4.5015+8.1015+7.2015+9.4515+10.8015+6.7515E[\text{Profit}] = \frac{4.50}{15} + \frac{8.10}{15} + \frac{7.20}{15} + \frac{9.45}{15} + \frac{10.80}{15} + \frac{6.75}{15} E[Profit]=46.8015=3.12E[\text{Profit}] = \frac{46.80}{15} = 3.12

Thus, the expected profit is $3.12.


Would you like further clarification or details on this process?

Related Questions:

  1. What would the expected profit be if the retail price changed to $1.75 per carton?
  2. How does changing the wholesale price affect the expected profit?
  3. What is the variance of the profit based on this probability distribution?
  4. How would the expected profit change if the credit for unsold cartons was half the wholesale price?
  5. How can we use this profit distribution to model different scenarios for milk sales?

Tip:

The expected value is a great tool for making decisions in uncertain situations. It helps to average out all possible outcomes weighted by their likelihood, giving a clearer picture of potential results.

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Math Problem Analysis

Mathematical Concepts

Probability
Expected Value
Profit Calculation

Formulas

Profit(X) = X × (1.65 - 1.20) + (5 - X) × 0.90
E[Profit] = Σ Profit(X) × P(X = x)

Theorems

Expected Value Theorem

Suitable Grade Level

Grades 9-12