Math Problem Statement

Suppose that the number of cars X that pass through a car wash both 4:00 pm and 5:00 pm on any Friday has the following probability distribution: X 4 5 6 7 8 9 P(X=x) 1/12 1/12 1/4 1/4 1/6 1/6 Let g(x)=2x-1 represents the amount of money in dollars, paid to the attendant by the manager. Find the attendants expected earning for this particular time period.

Solution

To find the attendant's expected earnings for this time period, we'll need to calculate the expected value of the function g(X)=2X1g(X) = 2X - 1. The expected value of g(X)g(X) is given by:

E[g(X)]=xg(x)P(X=x)E[g(X)] = \sum_{x} g(x) \cdot P(X = x)

Given:

  • XX represents the number of cars.
  • The probability distribution for XX is:

P(X=4)=112,P(X=5)=112,P(X=6)=14,P(X=7)=14,P(X=8)=16,P(X=9)=16P(X = 4) = \frac{1}{12}, \quad P(X = 5) = \frac{1}{12}, \quad P(X = 6) = \frac{1}{4}, \quad P(X = 7) = \frac{1}{4}, \quad P(X = 8) = \frac{1}{6}, \quad P(X = 9) = \frac{1}{6}

  • The function g(x)=2x1g(x) = 2x - 1 represents the amount of money paid.

Now, let's calculate g(x)g(x) for each value of xx:

g(4)=2(4)1=81=7g(4) = 2(4) - 1 = 8 - 1 = 7 g(5)=2(5)1=101=9g(5) = 2(5) - 1 = 10 - 1 = 9 g(6)=2(6)1=121=11g(6) = 2(6) - 1 = 12 - 1 = 11 g(7)=2(7)1=141=13g(7) = 2(7) - 1 = 14 - 1 = 13 g(8)=2(8)1=161=15g(8) = 2(8) - 1 = 16 - 1 = 15 g(9)=2(9)1=181=17g(9) = 2(9) - 1 = 18 - 1 = 17

Next, multiply each value of g(x)g(x) by the corresponding probability P(X=x)P(X = x) and sum the results to find the expected value E[g(X)]E[g(X)]:

E[g(X)]=7×112+9×112+11×14+13×14+15×16+17×16E[g(X)] = 7 \times \frac{1}{12} + 9 \times \frac{1}{12} + 11 \times \frac{1}{4} + 13 \times \frac{1}{4} + 15 \times \frac{1}{6} + 17 \times \frac{1}{6}

Calculate each term:

7×112=7127 \times \frac{1}{12} = \frac{7}{12} 9×112=9129 \times \frac{1}{12} = \frac{9}{12} 11×14=114=331211 \times \frac{1}{4} = \frac{11}{4} = \frac{33}{12} 13×14=134=391213 \times \frac{1}{4} = \frac{13}{4} = \frac{39}{12} 15×16=156=301215 \times \frac{1}{6} = \frac{15}{6} = \frac{30}{12} 17×16=176=341217 \times \frac{1}{6} = \frac{17}{6} = \frac{34}{12}

Now sum these values:

E[g(X)]=712+912+3312+3912+3012+3412=15212=12.67 dollarsE[g(X)] = \frac{7}{12} + \frac{9}{12} + \frac{33}{12} + \frac{39}{12} + \frac{30}{12} + \frac{34}{12} = \frac{152}{12} = 12.67 \text{ dollars}

Thus, the expected earnings for the attendant during this time period are approximately $12.67.

Would you like more details or have any other questions?

Here are 5 related questions:

  1. How do you calculate the expected value for a different function of XX?
  2. What if the probability distribution of XX changes? How would that affect the expected earnings?
  3. Can you explain the concept of expected value in general?
  4. How would you calculate the variance of the earnings g(X)g(X)?
  5. What is the significance of the function g(x)g(x) in practical applications?

Tip: When calculating expected value, always make sure to multiply the function values by their corresponding probabilities and sum them up.

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Math Problem Analysis

Mathematical Concepts

Probability Theory
Expected Value
Probability Distribution

Formulas

Expected Value formula: E[g(X)] = \sum_{x} g(x) \cdot P(X = x)

Theorems

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Suitable Grade Level

Advanced High School / College