Answered

\begin{tabular}{|c|c|c|}
\hline Number of Bagels & Money Collected [tex]$x$[/tex] & Probability [tex]$P(z)$[/tex] \\
\hline 1 & [tex]$0.80$[/tex] & 0.5 \\
\hline 2 & [tex]$1.60$[/tex] & 0.1 \\
\hline 3 & [tex]$2.40$[/tex] & 0.1 \\
\hline 6 & [tex]$4.80$[/tex] & 0.1 \\
\hline 12 & [tex]$9.60$[/tex] & 0.1 \\
\hline
\end{tabular}

The expected value of [tex]$X$[/tex], the amount of money Bob can expect to make per customer, is [tex]$E(X)=\Sigma X P(X) = \$[/tex]3.03[tex]$. What is the standard deviation?

\begin{tabular}{|c|c|c|c|}
\hline Number of Bagels & $[/tex](X-\mu)^2[tex]$ (Rounded to nearest hundredth) & $[/tex]P(x)[tex]$ & $[/tex](x-\mu)^2 \cdot P(x)[tex]$ \\
\hline 1 & $[/tex](0.80-3.03)^2 \approx 4.97[tex]$ & 0.5 & 2.49 \\
\hline 2 & $[/tex](1.60-3.03)^2 \approx 2.05[tex]$ & 0.1 & 0.21 \\
\hline 3 & $[/tex](2.40-3.03)^2 \approx 0.40[tex]$ & 0.1 & 0.04 \\
\hline 6 & $[/tex](4.80-3.03)^2 \approx 3.13[tex]$ & 0.1 & 0.31 \\
\hline 12 & $[/tex](9.60-3.03)^2 \approx 43.24[tex]$ & 0.1 & 4.32 \\
\hline \text{Sum} & & & 7.37 \\
\hline \text{Standard Deviation} & & & $[/tex]\sqrt{7.37} \approx 2.71[tex]$ \\
\hline
\end{tabular}

The standard deviation is, therefore, approximately $[/tex]2.71$.



Answer :

GinnyAnswer
Let's construct the step-by-step solution to find the expected value, variance, and standard deviation based on the given data.

### Step-by-Step Solution

#### 1. Complete the Table for [tex]\( x \)[/tex] (Money Collected)
Given that Bob collects $0.80 per bagel, we can calculate [tex]\( x \)[/tex] for different numbers of bagels:

| Number of Bagels | [tex]\( x \)[/tex] = Money Collected (in dollars) | [tex]\( P(z) \)[/tex] = Probability |
|------------------|---------------------------------------|--------------------------|
| 1 | 0.80 | 0.5 |
| 2 | 1.60 | 0.1 |
| 3 | 2.40 | 0.1 |
| 6 | 4.80 | 0.1 |
| 12 | 9.6 | 0.1 |

#### 2. Calculate the Expected Value [tex]\( E(X) \)[/tex]
The expected value [tex]\( E(X) \)[/tex] is defined as:

[tex]\[ E(X) = \sum (x_i \cdot P(x_i)) \][/tex]

Using the values of [tex]\( x \)[/tex] and [tex]\( P(x) \)[/tex]:

[tex]\[ E(X) = (0.80 \cdot 0.5) + (1.60 \cdot 0.1) + (2.40 \cdot 0.1) + (4.80 \cdot 0.1) + (9.6 \cdot 0.1) \][/tex]
[tex]\[ E(X) = 0.40 + 0.16 + 0.24 + 0.48 + 0.96 \][/tex]
[tex]\[ E(X) = 2.24 \][/tex]

So, the expected value [tex]\( E(X) \)[/tex] is 2.24.

#### 3. Calculate the Variance [tex]\( \text{Var}(X) \)[/tex]
Variance [tex]\( \text{Var}(X) \)[/tex] is defined as:

[tex]\[ \text{Var}(X) = \sum \left( (x_i - \mu)^2 \cdot P(x_i) \right) \][/tex]

Where [tex]\( \mu \)[/tex] is the expected value. First, we find [tex]\( (x_i - \mu)^2 \)[/tex] for each [tex]\( x \)[/tex]:

[tex]\[ \begin{align*} (0.80 - 2.24)^2 & = 2.07 \\ (1.60 - 2.24)^2 & = 0.4096 \\ (2.40 - 2.24)^2 & = 0.0256 \\ (4.80 - 2.24)^2 & = 6.5536 \\ (9.60 - 2.24)^2 & = 54.0544 \\ \end{align*} \][/tex]

Next, we multiply by their respective probabilities [tex]\( P(x) \)[/tex]:

[tex]\[ \begin{align*} 2.07 \cdot 0.5 & = 1.035 \\ 0.4096 \cdot 0.1 & = 0.04096 \\ 0.0256 \cdot 0.1 & = 0.00256 \\ 6.5536 \cdot 0.1 & = 0.65536 \\ 54.0544 \cdot 0.1 & = 5.40544 \\ \end{align*} \][/tex]

Finally, summing these values gives the variance:

[tex]\[ \text{Var}(X) = 1.035 + 0.04096 + 0.00256 + 0.65536 + 5.40544 = 7.13932 \][/tex]

#### 4. Calculate the Standard Deviation [tex]\( \sigma \)[/tex]

The standard deviation is the square root of the variance:

[tex]\[ \sigma = \sqrt{\text{Var}(X)} \][/tex]

[tex]\[ \sigma = \sqrt{7.13932} \approx 2.6744 \][/tex]

Thus, the standard deviation is approximately 2.67 (rounded to the nearest hundredth).

### Conclusion
In conclusion:

- The expected value [tex]\( E(X) \)[/tex] is 2.24.
- The variance [tex]\( \text{Var}(X) \)[/tex] is 7.13932.
- The standard deviation [tex]\( \sigma \)[/tex] is 2.67.

Other Questions