Orange M&M’s: The M&M’s web site says that 20% of milk chocolate M&M’s are orange. Let’s assume this is true and set

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Orange M&M’s: The M&M’s web site says that 20% of milk chocolate M&M’s are orange. Let’s assume this is true and set

up a simulation to mimic buying 200 small bags of milk chocolate M&M’s. Each bag contains 55 candies. We made this dotplot of the results. Normal sampling distribution of proportion of orange candies Now suppose that we buy a small bag of M&M’s. We find that 25.5% (14 of the 55) of the M&M’s are orange. What can we conclude? Group of answer choices This result is not surprising because we expect to see many samples with 14 or more orange candies. This result is surprising because we expect the orange candies to make up no more than 20% of the candies in a packet. This result is surprising because it is unlikely that we will select a random sample with 25.5% or more orange candies if 20% of milk chocolate M&M’s are orange.

Mathematics

1 answer:

Zina [12.3K] · Answer 1 · 2026-04-08T15:25:34+03:00

Answer:

The answer is option (A).

Step-by-step explanation:

Let X represent the count of orange milk chocolate M&M’s.

The proportion of orange milk chocolate M&M’s is defined as p = 0.20.

In a small bag containing milk chocolate M&M’s, n is established at 55.The occurrence of an orange milk chocolate M&M is independent of the remaining candies.

The random variable X conforms to a Binomial distribution characterized by parameters n = 55 and p = 0.20.

Calculating the average number of orange milk chocolate M&M’s in a package of 55 candies proceeds as follows:

$E(X)=np$

It is noted that in a randomly selected pack of milk chocolate M&M's there were 14 orange M&M's, indicating that the ratio of orange milk chocolate M&M's in that selection was 25.5%.

This ratio is not unexpected.

This is due to the expected count of orange milk chocolate M&M’s in a 55 candy bag being estimated at 11. Therefore, encountering 14 orange milk chocolate M&M’s does not seem unusual.

Unusual instances typically have exceedingly low probabilities, specifically below 0.05.

To calculate the probability of having P (X ≥ 14), follow the subsequent calculation:

$P(X\geq 14)=\sum\limits^{55}_{x=14}{{55\choose x}0.20^{x}(1-0.20)^{55-x}}$

The likelihood of obtaining 14 or more orange candies in a milk chocolate M&M’s bag is 0.1968.

This probability is considerably higher than 0.05.

Thus, the correct answer is option (A).