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 res

Answer:The correct option is (A).Step-by-step explanation:Let X = number of orange  milk chocolate M&M’s.The proportion of orange milk chocolate M&M’s is, p = 0.20.The number of candies in a small bag of milk chocolate M&M’s is, n = 55.The event of an milk chocolate M&M being orange is independent of the other candies.The random variable X follows a Binomial distribution with parameter n = 55 and p = 0.20.The expected value of a Binomial random variable is:[tex]E(X)=np[/tex]Compute the expected number of orange  milk chocolate M&M’s in a bag of 55 candies as follows:[tex]E(X)=np[/tex]          [tex]=55\times 0.20\\=11[/tex]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:[tex]P(X\geq 14)=\sum\limits^{55}_{x=14}{{55\choose x}0.20^{x}(1-0.20)^{55-x}}[/tex]                  [tex]=0.1968[/tex]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).