SOLUTION: A lake in foreign country contains two types of fish. Let X be the random variable for the length (in cm) of the Flamboyant Flapper fish. The expected value of X is E(X) = 29 and t

Question 1165293: A lake in foreign country contains two types of fish. Let X be the random variable for the length (in cm) of the Flamboyant Flapper fish. The expected value of X is E(X) = 29 and the variance Var(X) is 7.
Let Y be the random variable for the length (in cm) of the Boring Beluper fish. The expected value of Y is E(Y) = 21 and the variance Var(Y) is 11.
You and your friend go fishing. You catch a Flamboyant Flapper fish and your friend catchs a Boring Beluper. What is the probability that your friend's Boring Flapper is longer than your Flamboyant Flapper? To be able to work this out you need to know the distribution of the random variable (X-Y).
Calculate the expectation (X-Y), E(X-Y), and the standard deviation of (X-Y), SD(X-Y), assuming that X and Y are independent, and enter your answers below. The expected value of (X-Y) is E(X-Y) = ?
Assuming that X and Y are independent, then the standard deviation of (X-Y) is SD(X-Y) = ? (to 1 d.p.)

Answer by CPhill(2189) (Show Source): You can put this solution on YOUR website!
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.)
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