Question

1- A single serving of milk chocolate M&M candies is 50 pieces. In a standard bag of M&Ms there are six colors (brown, yellow, green, red, orange, and blue) and assume equal probabilities for the color of a randomly selected M&M.

a) What is the mean value for the number of red M&Ms in a single serving of randomly selected M&Ms?

b)What is the variance of the number of red M&Ms in a single serving?

c) What is the probability that a single serving of M&Ms will contain more than 10 red M&Ms?

d)What is the probability that a single serving of M&Ms will contain exactly 8 blue M&Ms? Calculate this probability using the binomial distribution and by using the normal approximation to the binomial distribution and compare the results.

Answer 1

as each color occurs with equal probability ;therefore probability of each color p=1/6

a)

mean value for the number of red M&Ms=np=50*(1/6)=8.33

b)

variance =np(1-p)=50*(1/6)*(1-1/6)=6.94

from binomial distribution:

c)

probability that a single serving of M&Ms will contain more than 10 red M&Ms

=P(X>10)=1-P(X<=10)=1- =1-7986 =0.2014

d)

probability that a single serving of M&Ms will contain exactly 8 blue M&Ms =

=0.1510

from normal approximation:

mean =8.333

std deviaiton =sqrt(np(1-p))=2.635

also as z score=(X-mean)/std deviation

c) probability that a single serving of M&Ms will contain more than 10 red M&Ms:

probability =

P(X>10.5)

=

P(Z>0.82)=

1-P(Z<0.82)=

1-0.7939=

0.2061

d)

probability that a single serving of M&Ms will contain exactly 8 blue M&Ms:

probability =

P(7.5<X<8.5)

=

P(-0.32<Z<0.06)=

0.5239-0.3745=

0.1494