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Given that $X$ and $Y$ are independent random variables with normal distributions, what is the mean and variance of the distribution of $2X - Y$ when the corresponding distributions are $X\sim \mathcal{N}(3, 4)$ and $Y \sim \mathcal{N}(1, 4)$?

Given that $X$ and $Y$ are independent random variables with normal distributions, and the corresponding distributions are $X\sim \mathcal{N}(3, 4)$ and $Y \sim \mathcal{N}(1, 4)$, what is the mean of the distribution of $2X - Y$?

Given that $X$ and $Y$ are independent random variables with normal distributions, and the corresponding distributions are $X\sim \mathcal{N}(3, 4)$ and $Y \sim \mathcal{N}(1, 4)$, what is the mean of the distribution of $2X - Y$?

Distribution of 2X - Y I

Given that $X$ and $Y$ are independent random variables with normal distributions, and the corresponding distributions are $X\sim \mathcal{N}(3, 4)$ and $Y \sim \mathcal{N}(1, 4)$, what is the variance of the distribution of $2X - Y$?

Given that $X$ and $Y$ are independent random variables with normal distributions, and the corresponding distributions are $X\sim \mathcal{N}(3, 4)$ and $Y \sim \mathcal{N}(1, 4)$, what is the variance of the distribution of $2X - Y$?

Distribution of 2X - Y II

Distribution of 2X - Y

Write a function <code>compute_deviation</code> that takes in a list of dictionaries with a key and list of integers and returns a dictionary with the standard deviation of each list.

Note: This should be done without using the NumPy built-in functions.

Example:

Input:

<pre tabindex="0" class="chroma"><code>input = [
 {
 &#39;key&#39;: &#39;list1&#39;,
 &#39;values&#39;: [4,5,2,3,4,5,2,3],
 },
 {
 &#39;key&#39;: &#39;list2&#39;,
 &#39;values&#39;: [1,1,34,12,40,3,9,7],
 }
]
</code></pre>

Output:

<pre tabindex="0" class="chroma"><code> output = {&#39;list1&#39;: 1.12, &#39;list2&#39;: 14.19}
</code></pre>

Compute Deviation

Let’s say that your company is running a standard control and variant AB test on a feature to increase conversion rates on the landing page. The PM checks the results and finds a .04 p-value.

How would you assess the validity of the result?

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