Answer :
Certainly! Let's go through the detailed solution for calculating the expected value, variance, and standard deviation of the amount of money Bob collects from a randomly-selected customer.
### Step 1: Understanding the problem and probabilities
Firstly, we are provided with a probability distribution for the number of bagels sold to a randomly-selected customer:
[tex]\[ \begin{array}{|c|c|} \hline \text{Number of Bagels} & \text{Probability} \\ \hline 1 & 0.5 \\ \hline 2 & 0.1 \\ \hline 3 & 0.1 \\ \hline 6 & 0.1 \\ \hline 12 & 0.2 \\ \hline \end{array} \][/tex]
Bob sells bagels at:
- \[tex]$0.50 per bagel if fewer than 12 are bought. - \$[/tex]9.00 for a dozen.
### Step 2: Calculating the money collected for each possible number of bagels
The amount of money collected, [tex]\( x \)[/tex], for each number of bagels is calculated as follows:
- For 1 bagel: [tex]\( x = 1 \times 0.50 = \$0.50 \)[/tex]
- For 2 bagels: [tex]\( x = 2 \times 0.50 = \$1.00 \)[/tex]
- For 3 bagels: [tex]\( x = 3 \times 0.50 = \$1.50 \)[/tex]
- For 6 bagels: [tex]\( x = 6 \times 0.50 = \$3.00 \)[/tex]
- For 12 bagels: [tex]\( x = \$9.00 \)[/tex] (discounted price for a dozen)
So, the table including the amount of money collected looks like this:
[tex]\[ \begin{array}{|c|c|c|} \hline \text{Number of Bagels} & \text{Money Collected} & \text{Probability} \\ \hline 1 & \$0.50 & 0.5 \\ \hline 2 & \$1.00 & 0.1 \\ \hline 3 & \$1.50 & 0.1 \\ \hline 6 & \$3.00 & 0.1 \\ \hline 12 & \$9.00 & 0.2 \\ \hline \end{array} \][/tex]
### Step 3: Finding the expected value [tex]\( E(X) \)[/tex]
The expected value [tex]\( E(X) \)[/tex] is the sum of each value multiplied by its probability:
[tex]\[ E(X) = \sum_{i} (x_i \cdot P(x_i)) \][/tex]
Calculating this:
[tex]\[ E(X) = (0.50 \cdot 0.5) + (1.00 \cdot 0.1) + (1.50 \cdot 0.1) + (3.00 \cdot 0.1) + (9.00 \cdot 0.2) \][/tex]
[tex]\[ E(X) = 0.25 + 0.10 + 0.15 + 0.30 + 1.80 = 2.6 \][/tex]
Thus, the expected value [tex]\( E(X) \)[/tex] is \[tex]$2.60. ### Step 4: Finding the variance \( \text{Var}(X) \) The variance is calculated using the formula: \[ \text{Var}(X) = E(X^2) - [E(X)]^2 \] First, we find \( E(X^2) \): \[ E(X^2) = \sum_{i} (x_i^2 \cdot P(x_i)) \] Calculating \( x_i^2 \) for each \( x_i \): - For 1 bagel: \( (0.50)^2 = 0.25 \) - For 2 bagels: \( (1.00)^2 = 1.00 \) - For 3 bagels: \( (1.50)^2 = 2.25 \) - For 6 bagels: \( (3.00)^2 = 9.00 \) - For 12 bagels: \( (9.00)^2 = 81.00 \) Now, calculating \( E(X^2) \): \[ E(X^2) = (0.25 \cdot 0.5) + (1.00 \cdot 0.1) + (2.25 \cdot 0.1) + (9.00 \cdot 0.1) + (81.00 \cdot 0.2) \] \[ E(X^2) = 0.125 + 0.100 + 0.225 + 0.900 + 16.200 = 17.55 \] Finally, the variance: \[ \text{Var}(X) = 17.55 - (2.6)^2 = 17.55 - 6.76 = 10.79 \] Thus, the variance \( \text{Var}(X) \) is 10.79. ### Step 5: Finding the standard deviation \( \sigma(X) \) The standard deviation is the square root of the variance: \[ \sigma(X) = \sqrt{\text{Var}(X)} = \sqrt{10.79} \approx 3.285 \] Thus, the standard deviation \( \sigma(X) \) is approximately 3.285. ### Interpretation - The expected value, \( E(X) = \$[/tex]2.60 \), represents the average amount of money Bob can expect to collect from a randomly-selected customer over time.
- The standard deviation, [tex]\( \sigma(X) \approx 3.285 \)[/tex], indicates the variability or spread of the amount of money Bob collects around this average value.
In summary:
- Expected value, [tex]\( E(X) \)[/tex]: \$2.60
- Variance, [tex]\( \text{Var}(X) \)[/tex]: 10.79
- Standard deviation, [tex]\( \sigma(X) \)[/tex]: 3.285
### Step 1: Understanding the problem and probabilities
Firstly, we are provided with a probability distribution for the number of bagels sold to a randomly-selected customer:
[tex]\[ \begin{array}{|c|c|} \hline \text{Number of Bagels} & \text{Probability} \\ \hline 1 & 0.5 \\ \hline 2 & 0.1 \\ \hline 3 & 0.1 \\ \hline 6 & 0.1 \\ \hline 12 & 0.2 \\ \hline \end{array} \][/tex]
Bob sells bagels at:
- \[tex]$0.50 per bagel if fewer than 12 are bought. - \$[/tex]9.00 for a dozen.
### Step 2: Calculating the money collected for each possible number of bagels
The amount of money collected, [tex]\( x \)[/tex], for each number of bagels is calculated as follows:
- For 1 bagel: [tex]\( x = 1 \times 0.50 = \$0.50 \)[/tex]
- For 2 bagels: [tex]\( x = 2 \times 0.50 = \$1.00 \)[/tex]
- For 3 bagels: [tex]\( x = 3 \times 0.50 = \$1.50 \)[/tex]
- For 6 bagels: [tex]\( x = 6 \times 0.50 = \$3.00 \)[/tex]
- For 12 bagels: [tex]\( x = \$9.00 \)[/tex] (discounted price for a dozen)
So, the table including the amount of money collected looks like this:
[tex]\[ \begin{array}{|c|c|c|} \hline \text{Number of Bagels} & \text{Money Collected} & \text{Probability} \\ \hline 1 & \$0.50 & 0.5 \\ \hline 2 & \$1.00 & 0.1 \\ \hline 3 & \$1.50 & 0.1 \\ \hline 6 & \$3.00 & 0.1 \\ \hline 12 & \$9.00 & 0.2 \\ \hline \end{array} \][/tex]
### Step 3: Finding the expected value [tex]\( E(X) \)[/tex]
The expected value [tex]\( E(X) \)[/tex] is the sum of each value multiplied by its probability:
[tex]\[ E(X) = \sum_{i} (x_i \cdot P(x_i)) \][/tex]
Calculating this:
[tex]\[ E(X) = (0.50 \cdot 0.5) + (1.00 \cdot 0.1) + (1.50 \cdot 0.1) + (3.00 \cdot 0.1) + (9.00 \cdot 0.2) \][/tex]
[tex]\[ E(X) = 0.25 + 0.10 + 0.15 + 0.30 + 1.80 = 2.6 \][/tex]
Thus, the expected value [tex]\( E(X) \)[/tex] is \[tex]$2.60. ### Step 4: Finding the variance \( \text{Var}(X) \) The variance is calculated using the formula: \[ \text{Var}(X) = E(X^2) - [E(X)]^2 \] First, we find \( E(X^2) \): \[ E(X^2) = \sum_{i} (x_i^2 \cdot P(x_i)) \] Calculating \( x_i^2 \) for each \( x_i \): - For 1 bagel: \( (0.50)^2 = 0.25 \) - For 2 bagels: \( (1.00)^2 = 1.00 \) - For 3 bagels: \( (1.50)^2 = 2.25 \) - For 6 bagels: \( (3.00)^2 = 9.00 \) - For 12 bagels: \( (9.00)^2 = 81.00 \) Now, calculating \( E(X^2) \): \[ E(X^2) = (0.25 \cdot 0.5) + (1.00 \cdot 0.1) + (2.25 \cdot 0.1) + (9.00 \cdot 0.1) + (81.00 \cdot 0.2) \] \[ E(X^2) = 0.125 + 0.100 + 0.225 + 0.900 + 16.200 = 17.55 \] Finally, the variance: \[ \text{Var}(X) = 17.55 - (2.6)^2 = 17.55 - 6.76 = 10.79 \] Thus, the variance \( \text{Var}(X) \) is 10.79. ### Step 5: Finding the standard deviation \( \sigma(X) \) The standard deviation is the square root of the variance: \[ \sigma(X) = \sqrt{\text{Var}(X)} = \sqrt{10.79} \approx 3.285 \] Thus, the standard deviation \( \sigma(X) \) is approximately 3.285. ### Interpretation - The expected value, \( E(X) = \$[/tex]2.60 \), represents the average amount of money Bob can expect to collect from a randomly-selected customer over time.
- The standard deviation, [tex]\( \sigma(X) \approx 3.285 \)[/tex], indicates the variability or spread of the amount of money Bob collects around this average value.
In summary:
- Expected value, [tex]\( E(X) \)[/tex]: \$2.60
- Variance, [tex]\( \text{Var}(X) \)[/tex]: 10.79
- Standard deviation, [tex]\( \sigma(X) \)[/tex]: 3.285
