Question

The proportion of brown M&M’s in a milk chocolate packet is approximately 14% (Madison, 2013). Suppose a package of M&M’s typically contains 52 M&M’s. a.) State the random variable. b.) Find the mean. c.) Find the variance. d.) Find the standard deviation.

Answer 1

Answer: a) Let x be a random variable that represents the number of brown M&M’s in a milk chocolate packet.

b) Mean =7.28

c) Variance = 6.2608

d) Standard deviation = 2.5022

Step-by-step explanation:

Given : The proportion of brown M&M’s in a milk chocolate packet is approximately : p=14% =0.14

We have outcomes for every randomly selection : 1) its in a milk chocolate packet 2) its not.

a) Let x be a random variable that represents the number of brown M&M’s in a milk chocolate packet.

So x follows binomial distribution.

b) Sample size : n=52  (given)

Mean $= np = (0.14)(52) =7.28$

c) Variance = $n p (1-p)= 52 (0.14)(1-0.14) =7.28(0.86)=6.2608$

d) Standard deviation =$\sqrt{Variance}=\sqrt{6.2608}\approx2.5022$