D

#### A quality control inspector routinely takes random samples of 200 cans of fruit cocktail produced in a packing plant and calculates the proportion \widehat{p} of cans from each sample with at least 3 cherries. Suppose that 98% of fruit cocktail cans produced in that packing plant contain at least 3 cherries.

405 viewed last edited 1 year ago Anonymous
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Which of the following distributions is the best approximation of the sampling distribution of \widehat{p}? Each distribution uses the same scale. Sahil Khan
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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