Find the 80th percentile for sample proportions of red M&M's in samples of size 46.

Question

The proportion of red M&M's made is 0.25. Suppose one took a random sample of 46 M&M's.

 

What is the probability that an SRS of 46 M&M's has as its proportion of red M&M's 0.24 or below?

 

Find the 80th percentile for sample proportions of red M&M's in samples of size 46.

Answer 1

Hi Jason,

 

Think of your first question as a binomial experiment with getting red considered as "success". The event you're interested in is that you get 24% or fewer reds out of 46 M&Ms. Since 0.24 × 46 = 11.04, this is the same as getting at most 11 reds (you can't get 0.04 M&Ms!). The probability of this occurring is P (exactly 0 red) + P (exactly 1 red) + ... + P (exactly 11 red).

 

Computing each of those individual probabilities and summing can be tedious so we approximate the above probability by a "normal approximation to the binomial distribution". Let 

 

n = 46

p = 0.25

phat = sample proportion of red M&Ms (this is random because it varies from sample to sample)

 

There's a rule of thumb that the normal approximation is valid if n ≥ 30, n × p ≥ 5, and n × (1 - p) ≥ 5. You can check that these hold for your problem.

 

P (phat ≤ 0.24)

= P ((phat - p)/√(p × (1 - p)/n) ≤ (0.24 - p) /√(p × (1 - p)/n))

 ≈ P (Z ≤ -0.1566)

 

where Z is a standard normally distributed random variable and -0.1566 is computed by substituting in the values for n and p.

Now you need to make use of a normal distribution table and your answer should be greater than 16% but less that 50%.

 

For your second question you have all the basic ingredients. You need to solve for phat in the equation

 

z_0.2 = (phat - p)/√(p × (1 - p)/n)

 

where z_0.2 is the 80th percentile of the standard normal distribution and can be found from a table (z_0.2 will be a positive value slightly less than 1).