The weights of pears are normally distributed with mean 118g and standard deviation 4.6g. 5 pears are chosen at random. Question:
The weights of pears are normally distributed with mean 118g and standard deviation 4.6g. 5 pears are chosen at random.
Find the probability that their combined weight is less than 570 grams.
I have solved these questions before, but I'm doing them after 1 year hence not being able to create a certain link. [Have Covid, suffering from apparent brain fog]
What I'm doing:
Consider one pear to be denoted by P, which is distributed as: P~N(118,4.6^2)
Hence 5P will have:
mean:-
118*5 = 590
variance:
5^2 * 4.6^2 = 529
Hence 5P distributed as: 5P~N(590, 529)
Finding probability using calculator function:
p = 0.192
Answer in the book:
0.0259
Additional information:
I'm pretty sure that my variance calculation is wrong. I am confused as to when is the coefficient of a combination of random variables supposed to be squared as well. In the book itself it has been explained that variance is a^2*Var(x), hence I have squared a (5) and squared the standard deviation to give the variance.
 A: A simple way to illustrate the issue is to consider the following situation.  Suppose $X$ is a Bernoulli random variable with probability mass function $$\Pr[X = 0] = \Pr[X = 1] = 1/2.$$  Then the expectation is $\operatorname{E}[X] = 0 \cdot \Pr[X = 0] + 1 \cdot \Pr[X = 1] = 0(1/2) + 1(1/2) = 1/2$, and $$\operatorname{Var}[X] = (0 - 1/2)^2 \Pr[X = 0] + (1 - 1/2)^2 \Pr[X = 1] = (1/4)(1/2) + (1/4)(1/2) = 1/4.$$  What is $\operatorname{Var}[5X]$?  The random variable $Y = 5X$ has probability mass function $$\Pr[Y = 0] = \Pr[Y = 5] = 1/2$$ and its support is $Y \in \{0, 5\}$.  So $$\operatorname{Var}[Y] = (0 - (5/2))^2 \Pr[Y = 0] + (5 - (5/2))^2 \Pr[Y = 5] = (5/2)^2 (1/2) + (5/2)^2 (1/2) = 25/4.$$  Therefore $$\operatorname{Var}[Y] = 5^2 \operatorname{Var}[X].$$
Now suppose we have $W = X_1 + X_2 + X_3 + X_4 + X_5$, where for each $i \in \{1, 2, \ldots, 5\}$, $X_i$ is independent and identically Bernoulli distributed as $X$; i.e., $\Pr[X_i = 0] = \Pr[X_i = 1] = 1/2$.  Then $$W \sim \operatorname{Binomial}(n = 5, p = 1/2)$$ with $$\Pr[W = w] = \binom{5}{w} \frac{1}{2^5}, \quad w \in \{0, 1, \ldots, 5\}$$ and $$\operatorname{Var}[W] = np(1-p) = 5(1/2)(1/2) = \frac{5}{4} \ne \operatorname{Var}[Y].$$
Why are these two variables different?  Well, you can immediately see they are different because they do not have the same support:  $Y \in \{0, 5\}$, whereas $W \in \{0, 1, 2, 3, 4, 5\}$.
Suppose you flip a fair coin.  If the coin lands heads, I pay you a dollar.  If it lands tails, you win nothing.  Then $X$ is a random variable that describes your winnings after a single coin toss.  $W$ would describe your winnings after five coin tosses.  $Y$ would describe your winnings if you only tossed the coin once, but instead of the payout for heads being one dollar, it is five dollars.
Since in this last scenario (with random variable $Y$) does not admit a possibility of winning $1, 2, 3$, or $4$ dollars, the variance of the winnings is larger than the case where you get to flip the coin five times and allowing outcomes in which there are some wins and some losses.
Now to translate this situation to your original question, we can immediately see that when we take a random sample of $5$ pears and weigh them, their total weight is the sum of five independent normally distributed random variables.  Some of the pears might be lighter than the others, and some may be heavier.  This is in contrast to a situation in which you might pick a single pear and multiply its weight by $5$.  Consequently, the variance of the total weight of five randomly selected pears is not the same as the variance of five times the weight of a single pear.
