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.
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 |
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