Math Problem Statement

A large industrial firm purchases several new word processors at the end of each year, the exact number depending on the frequency of repairs in the previous year. Suppose that the number of word processors, X, purchased each year has the following probability distribution: X = [0, 1, 2, 3] with probabilities f(x) = [1/10, 3/10, 2/5, 1/5]. The cost of the desired model is $1200 per unit, and a refund of 50X^2 dollars is issued at the end of the year. How much can the firm expect to spend?

Solution

Let's break down the problem to calculate how much the firm can expect to spend on new word processors during the year.

1. Expected Number of Word Processors, E(X)E(X):

The number of word processors XX purchased each year follows a given probability distribution. To find the expected number of word processors, we calculate the expected value E(X)E(X), which is the weighted average of XX using the probabilities f(x)f(x).

E(X)=xxf(x)E(X) = \sum_{x} x \cdot f(x)

Using the values from the table:

  • f(0)=110,f(1)=310,f(2)=25,f(3)=15f(0) = \frac{1}{10}, f(1) = \frac{3}{10}, f(2) = \frac{2}{5}, f(3) = \frac{1}{5}

Thus:

E(X)=0×110+1×310+2×25+3×15E(X) = 0 \times \frac{1}{10} + 1 \times \frac{3}{10} + 2 \times \frac{2}{5} + 3 \times \frac{1}{5} E(X)=0+310+45+35E(X) = 0 + \frac{3}{10} + \frac{4}{5} + \frac{3}{5} Converting to a common denominator (10): E(X)=310+810+610=1710=1.7E(X) = \frac{3}{10} + \frac{8}{10} + \frac{6}{10} = \frac{17}{10} = 1.7 Thus, the expected number of word processors purchased is 1.7.

2. Expected Cost for the Year:

Each word processor costs $1200, but the firm also receives a refund of 50X250X^2 dollars at the end of the year. So, the net cost for the firm is:

Total Cost=1200X50X2\text{Total Cost} = 1200X - 50X^2

To find the expected cost, we compute the expected value of this expression, E(1200X50X2)E(1200X - 50X^2):

E(1200X50X2)=1200E(X)50E(X2)E(1200X - 50X^2) = 1200E(X) - 50E(X^2)

We already know E(X)=1.7E(X) = 1.7. Now, let's calculate E(X2)E(X^2).

3. Expected Value of X2X^2, E(X2)E(X^2):

We calculate E(X2)E(X^2) as:

E(X2)=xx2f(x)E(X^2) = \sum_{x} x^2 \cdot f(x)

So: E(X2)=02×110+12×310+22×25+32×15E(X^2) = 0^2 \times \frac{1}{10} + 1^2 \times \frac{3}{10} + 2^2 \times \frac{2}{5} + 3^2 \times \frac{1}{5} E(X2)=0+310+85+95E(X^2) = 0 + \frac{3}{10} + \frac{8}{5} + \frac{9}{5} Converting to a common denominator (10): E(X2)=310+1610+1810=3710=3.7E(X^2) = \frac{3}{10} + \frac{16}{10} + \frac{18}{10} = \frac{37}{10} = 3.7

4. Final Expected Cost Calculation:

Now, we can compute the expected total cost:

E(1200X50X2)=1200×1.750×3.7E(1200X - 50X^2) = 1200 \times 1.7 - 50 \times 3.7 =2040185=1855= 2040 - 185 = 1855

Conclusion:

The firm can expect to spend $1855 on new word processors during the year.

Would you like a more detailed breakdown or explanation of any part? Here are some related questions for further exploration:

  1. How is the concept of expected value used in other business scenarios?
  2. Can the firm minimize costs based on this probability model? If so, how?
  3. How would the total cost change if the refund formula was different (e.g., 100X2100X^2)?
  4. What if the distribution of XX changed? How would that affect expected spending?
  5. How can this model be extended to cover more complex purchasing decisions?

Tip: Always double-check your probability distributions and ensure the sum of probabilities equals 1 when working with expected values.

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Math Problem Analysis

Mathematical Concepts

Expected Value
Probability Distributions
Cost Function

Formulas

E(X) = Σ [x * f(x)]
E(X^2) = Σ [x^2 * f(x)]
Total Cost = 1200X - 50X^2
E(Total Cost) = 1200 * E(X) - 50 * E(X^2)

Theorems

Expected Value Theorem

Suitable Grade Level

University/Advanced High School