Statistics question on interpreting my results for sampling distribution

The weights of pennies minted after 1982 are approximately normally distributed with mean 2.46 grams and standard deviation 0.02 grams.

What is the probability that in a simple random sample of 10 pennies minted after 1982, we obtain a sample mean of at least 2.465 grams?

I found out that the answer is 0.2148, but I don't know how to interpret my result. I know that if I take 100 of these pennies about 21 of them will result in a sample mean of at least 2.465, but when I say this it makes little sense. Am I saying that I am averaging 21 of these pennies to be at least 2.465 grams? Are they averaged randomly?

What's a better interpretation? The probability that a randomly selected penny will have a sample mean of 2.465 grams is 21.48%? When I say this, I still can't make sense of the wording.

It seems you're missing the crucial information in the very text you typed. (And it sounds like you have been taught a way of thinking about probability that is leading you into confusion - one where you interpret probabilities as small-sample proportions. While it might sort of work on the simplest problems, it can easily mislead you. Drop that way of thinking, it is serving you badly here because - among other things - it's leading you to introduce additional numbers that distract you from the actual numbers.)

What is the probability that in a simple random sample of 10 pennies minted after 1982, we obtain a sample mean of at least 2.465 grams?

• So you take 10 pennies. Not 100. Not 21. 10 pennies. It's right there in your question!

• You average the weight of those 10 pennies and record the number of grams for that average.

• Now do it again. And again, and again and again*, an extremely large (notionally infinite) number of times.

• you now have a 'population' of average-weights from samples of ten coins.

• If you count what fraction of those average-weights exceed 2.465$g$, you find that (approximately**) a proportion of 0.2148 of those average weights exceed the mark.

* this is the long-run understanding of probability common in the frequentist interpretation of probability, which is the understanding you would bring to this sort of problem

** (averages of samples of size 10 from an approximately normal parent distribution will themselves still only be approximately normally distributed.)