Answer:
The correct option is (A).
Step-by-step explanation:
Let <em>X</em> = number of orange milk chocolate M&M’s.
The proportion of orange milk chocolate M&M’s is, <em>p</em> = 0.20.
The number of candies in a small bag of milk chocolate M&M’s is, <em>n</em> = 55.
The event of an milk chocolate M&M being orange is independent of the other candies.
The random variable <em>X</em> follows a Binomial distribution with parameter <em>n</em> = 55 and <em>p</em> = 0.20.
The expected value of a Binomial random variable is:

Compute the expected number of orange milk chocolate M&M’s in a bag of 55 candies as follows:


It is provided that in a randomly selected bag of milk chocolate M&M's there were 14 orange ones, i.e. the proportion of orange milk chocolate M&M's in a random bag was 25.5%.
This proportion is not surprising.
This is because the average number of orange milk chocolate M&M’s in a bag of 55 candies is expected to be 11. So, if a bag has 14 orange milk chocolate M&M’s it is not unusual at all.
All unusual events have a very low probability, i.e. less than 0.05.
Compute the probability of P (X ≥ 14) as follows:


The probability of having 14 or more orange candies in a bag of milk chocolate M&M’s is 0.1968.
This probability is quite larger than 0.05.
Thus, the correct option is (A).