Question

m&m's chocolate candies come in six different colours. Suppose the colour distribution is as follows:

Blue = 0.13 Brown = 0.21 Green = 0.15 Orange = 0.17 Red = 0.21 Yellow = 0.13

a) If you randomly select 175 m&m's, what is the approximate probability that at least 20% of them are orange?
b) If you randomly select 250 m&m's, what is the approximate probability that between 20% and 26% of them are red?

Answer 1

This can be solved by using the normal approximation to the binomial distribution where\[mean=np=175\times0.17=29.75\]
\[standard\ deviation=\sqrt{np(1-p)}=\sqrt{175\times0.17\times0.83}=4.969\]
20% of 175 = 35
Applying the continuity correction, you need to find the z-score for 35.5. When you have the z-score, use a standard normal distribution table to find the cumulative probability. Then subtract that value of probability from 1 to find the required probability.

Answer 2

For (a) you will need to calculate the expected number of orange M&M's you expect to see

Expected # of orange = (175) x (0.17).

Then from here you'll need to calculate the z-score given by

z= (expected # of orange)(observed # of orange)/(standard deviation).

Where the standard deviation is given by

\[\sigma = \sqrt{npq}\]

where

n = total # of M&M's

p = probability of getting an orange = 0.17

q = probability of not getting an orange = 0.83

Answer 3

Haha, what @kropot72 said

Answer 4

Sorry, my bad. You need to find the probability that at least 20% are orange. Therefore applying the continuity correction, you need to find the z-score for 34.5.