We are looking at which conditions does the sampling distributions of the sample proportions, looks, skewed to the left, skewed to the right, approximately normal, or uniform.

Our sample size n = 0,09 and p = 80.

Expected successes np = 0.09(80) = 7.2

Expected failures n(1-p) = 80( 1- 0.09) = 80(0.91) = 72.8

Since np\lt 10, and n(1-p) \geq 10, the shape of the sampling proportion is not normal.

Now to find in which direction the shape of the sampling distribution is skewed.

We would expect only a right-skewed sampling distribution when the population proportion is so extreme relative to the sample size that there are fewer than 10 expected successes per sample. So, since our expected successes is less than 10, we have a right-skewed sampling distribution.

Do have a look at the below discussion to understand the shape of a right-skewed sampling distribution.

https://www.qalaxia.com/viewDiscussion?messageId=606489b2f4d26b0f37dbd958

I found an answer from www.khanacademy.org

**Normal conditions for sampling distributions of sample proportions** ...

**AP**.**STATS**: UNC‑3 (EU). ,. UNC‑3.L (LO). ,. UNC‑3.L.1 (EK). About Transcript.
**Conditions** for roughly **normal sampling distribution of sample proportions**.

For more information, see **Normal conditions for sampling distributions of sample proportions** ...