The proportion of red M&M's made is 0.24. Suppose one took a random sample of 51 M&M's.

(a) What is the probability that an SRS of 51 M&M's has as its proportion of red M&M's 0.23 or below?
(b) Find the 80th percentile for sample proportions of red M&M's in samples of size 51.

To solve these questions, we will use the sampling distribution of sample proportions. The sampling distribution follows a normal distribution if certain conditions are met: random sample, independence, and a sufficiently large sample size.

(a) To find the probability that the proportion of red M&M's in a sample of 51 is 0.23 or below, we need to calculate the z-score and use the standard normal distribution.

Step 1: Calculate the sample mean (x̄) and standard deviation (σ) for the sampling distribution.
x̄ = 51 * 0.24 = 12.24
σ = sqrt[(0.24 * (1 - 0.24)) / 51] = 0.0496

Step 2: Calculate the z-score:
z = (x - x̄) / σ
z = (0.23 - 0.24) / 0.0496 ≈ -0.202

Step 3: Use the z-score to find the probability using a standard normal distribution table or calculator. For a z-score of -0.202, the corresponding probability is approximately 0.419.

Therefore, the probability that an SRS of 51 M&M's has a proportion of red M&M's 0.23 or below is approximately 0.419.

(b) To find the 80th percentile for sample proportions of red M&M's in samples of size 51, we need to find the z-score that corresponds to the 80th percentile.

Step 1: Find the z-score for the 80th percentile using a standard normal distribution table or calculator. For the 80th percentile, the z-score is approximately 0.842.

Step 2: Calculate the sample mean (x̄) and standard deviation (σ) for the sampling distribution (same as in part a).

Step 3: Use the z-score to find the corresponding sample proportion using the formula:
x = x̄ + z * σ
x = 12.24 + 0.842 * 0.0496 ≈ 12.282

Therefore, the 80th percentile for sample proportions of red M&M's in samples of size 51 is approximately 0.12282 (or 12.282%).