Calculate standard deviation of normal distribution from sample of another normal distribution with known mean and standard deviation

Question

The question is as follows:

Suppose the weight of avocados in a large crate has a distribution with mean 197 grams and standard deviation 7.6 grams. Consider the process of picking 7 avocados at random from the crate and putting them in a bag. Let X be the total weight of the bag. The standard deviation of X is.....

I have attempted to solve using R code: X = sample(sum(rnorm(7,197,7.6)), 999999, replace=TRUE) then sd(X) which takes many samples from the sum of a selection of 7 avocados from the normal distribution, to no avail.

EDIT: I have solved it using $\bar{X}$ is $\sigma_\bar{X}=\frac{\sigma_X}{\sqrt{n}}$, where $\sigma_X$ is the standard deviation of $X$ and $n$ is the sample size (sourced from a different answer) but I don't understand what it actually means...

Answer 1

Let $X_i$, where $i \in \{1, 2, \ldots, 7\}$, represent the random weight of the $i^{\rm th}$ avocado picked. So the $X_i$ are independent and identically distributed random variables, each with mean $\mu = 197$ and standard deviation $\sigma = 7.6$. The total weight of the bag is $$X = X_1 + X_2 + \cdots + X_7.$$ Then, because each of the $X_i$ are independent, the variance of $X$ is equal to the sum of the variances of each $X_i$: $$\operatorname{Var}[X] \overset{\text{ind}}{=} \operatorname{Var}[X_1] + \operatorname{Var}[X_2] + \cdots + \operatorname{Var}[X_7],$$ where the symbol $\overset{\text{ind}}{=}$ means that the equality is true only if the independence assumption is met.

But the variance of each $X_i$ is the same, because they are also identically distributed random variables: $$\operatorname{Var}[X_i] = \sigma^2,$$ the square of the standard deviation. So we have

$$\operatorname{Var}[X] = 7\sigma^2.$$ And so the standard deviation of the total weight is $$\sigma_{\text{tot}} = \sqrt{7 \sigma^2} = \sigma \sqrt{7} \approx (7.6)(2.64575) \approx 20.1077 \text{ grams}.$$

Note that the formula you referred to, $$\sigma_{\bar X} = \frac{\sigma}{\sqrt{n}},$$ gives the standard deviation for the average weight of a sample of $n$ avocados, not the total weight.