Question 1165293
This problem asks you to calculate the expected value and standard deviation of the difference between two independent random variables, $X$ and $Y$.

The rules for expectation and variance of the difference between two independent random variables are:

1.  **Expected Value of the Difference:**
    $$E(X - Y) = E(X) - E(Y)$$

2.  **Variance of the Difference (Crucial step, due to independence):**
    $$\operatorname{Var}(X - Y) = \operatorname{Var}(X) + \operatorname{Var}(Y)$$

3.  **Standard Deviation of the Difference:**
    $$\operatorname{SD}(X - Y) = \sqrt{\operatorname{Var}(X - Y)}$$

### Given Information:

  * **Flamboyant Flapper Fish ($X$):**
      * $E(X) = 29$
      * $\operatorname{Var}(X) = 7$
  * **Boring Beluper Fish ($Y$):**
      * $E(Y) = 21$
      * $\operatorname{Var}(Y) = 11$
  * $X$ and $Y$ are independent.

### 1\. Calculate the Expected Value of $(X - Y)$

$$E(X - Y) = E(X) - E(Y)$$
$$E(X - Y) = 29 - 21$$
$$\mathbf{E(X - Y) = 8}$$

### 2\. Calculate the Standard Deviation of $(X - Y)$

First, calculate the variance of the difference:
$$\operatorname{Var}(X - Y) = \operatorname{Var}(X) + \operatorname{Var}(Y)$$
$$\operatorname{Var}(X - Y) = 7 + 11$$
$$\operatorname{Var}(X - Y) = 18$$

Now, calculate the standard deviation:
$$\operatorname{SD}(X - Y) = \sqrt{\operatorname{Var}(X - Y)}$$
$$\operatorname{SD}(X - Y) = \sqrt{18}$$

We use the code interpreter to find the value rounded to 1 decimal place.

```python?code_reference&code_event_index=2
import numpy as np

# Calculate the standard deviation
SD_X_minus_Y = np.sqrt(18)

print(f"SD(X-Y): {SD_X_minus_Y}")
print(f"SD(X-Y) rounded to 1 d.p.: {round(SD_X_minus_Y, 1)}")

```

```text?code_stdout&code_event_index=2
SD(X-Y): 4.242640687119285
SD(X-Y) rounded to 1 d.p.: 4.2

```

$$\operatorname{SD}(X - Y) \approx 4.2426$$

Rounding to 1 decimal place:
$$\mathbf{\operatorname{SD}(X - Y) \approx 4.2}$$

-----

The expected value of (X-Y) is $E(X-Y) = \mathbf{8}$
Assuming that X and Y are independent, then the standard deviation of (X-Y) is $SD(X-Y) \approx \mathbf{4.2}$ (to 1 d.p.)