It is given that our sample size is 200 cans, and that 98% of the population contains at least 3 cherries.

So, n = 200 and p = 98% = 0.98

Expected successes : np = 200(0.98) = 196

Expected failures : n(1-p) = 200( 1- 0.98) = 4

Since the expected failures, 4 \lt 10, the sample proportions will not be normally distributed.

Since the probability of getting a can with at least 3 cherries is relatively high, we expect a large percentage of most of the samples to have at least 3 cherries. Some samples however could have a relatively lower percentage of cans with at least 3

cherries.

So the proportion of \widehat{p} will be high in most samples and skewed towards the lower proportions.

We expect only a left-skewed sampling distribution when the population proportion is so extreme relative to the sample size that there are fewer than 10 expected failures per sample.

So, option A is the correct choice.

Here is more on sampling distributions

https://www.khanacademy.org/math/ap-statistics/sampling-distribution-ap/sampling-distribution-proportion/e/normal-condition-sample-proportions?modal=1