Statistics Practice Review For Midterm (Need answer, and Method for study)

Question

Andy is a restaurant owner. He believes 82% of his customers are satisfied with the food quality of his restaurant. From a random sample of 99 customers, what are the following probabilities? (Round your answers to four decimal places, if needed.)

(a) What is the probability that less than 82 customers are satisfied with the food quality?

(b) What is the probability that at least 82 customers are satisfied with the food quality?

 

(c) What is the probability that the sample proportion of customers who are satisfied with the food quality is between 80% and 86%?

 

Answer 1

Samuel,

I'm going to make some assumptions as not all information present is enough, but let's forge ahead.

Let's first of all say that customers can only be satisfied or dissatisfied so that we have a binary response. Also, let's say that Andy's belief is p and the "true" model we want to operate under (similar to a null hypothesis). There's lots of reasons in experiments to assume "truth" like this for example: assuming a coin is fair until I find proof otherwise.

Now, when we have repeated binary or Bernoulli sampling a Binomial distribution is the best model for this total/aggreation. To be more rigorous if the Bernoulli distribution for one person is defined as:

P(X = 1) = 0.82

P(X = 0) = 0.18

This gives us the Binomial distribution that (where in our case n = 99):

P(X = x) = C(x, 99) p^x(1-p)^99-x

This is a great model however it can be computationally intensive. For example, to answer (a) we'd need:

P(X <= 81) = P(X=1) + P(X=2) + ... + P(X=81)

But let's do it anyways with a computer:

from scipy.stats import binom

dist = binom(99, 0.82)

print(dist.cdf(81))

We get P(X < 82) = P(X <= 81) ≈ 0.5222. With that Now (b) is as easy as:

P(X>=82) = 1 - P(X <= 81) ≈ 1- 0.5222 ≈ 0.4778

Now as a nifty trick we can use the central limit theorem (sometimes it will just be called normal approximation for binomial distribution). Now you have to be careful with this since the central limit theorem here is on the Bernoulli distribution.

The central limit theorem allows us to approximate the binomial distribution with:

µ = np =σ = 82 and σ²= np(1-p) ≈ 14.612 => σ ≈ 3.822

(Note for this to be appropriate approximation n needs to be relatively large and p needs to be towards the middle of [0, 1])

This allows us to use z-scores to go back and answer (a) and (b):

z = (X - μ) / σ = (82 - 82) / 3.822 = 0

a) P(X < 82) ≈ P(z < 0) = 0.5

b) 1 - 0.5 = 0.5

However because we are estimating a discrete distribution with the a continuous distribution, it is common practice to adjust for this (a discrete bar centered on 82 really goes from 81.5 to 82.5). So adjusting for this we get:

a) P(X < 82.5) ≈ P(z < 0.5/3.8222) ≈ 0.552

b) 1 - 0.552 ≈ 0.448

For part c), let's consider the same thing and use a little algebra (change proportion back to our normal problem):

P( 0.8 < X/n < 0.86)

= P( 0.8*88 < X < 0.86*99)

= P(79.2 < X < 85.14)

= P((79.2-82)/3.822 < z < (85.14 -82)/3.822)

≈ P(-0.733 < z < 0.822)

≈ P(z <= 0.822) - P(z <= -0.733)

(we want to use CDF to calculate this, area between is larger area - smaller area)

and using some kind of table or in my case code we get:

from scipy.stats import norm

z = norm(0, 1)

print(z.cdf(0.822) - z.cdf(-0.733))

and that gives ≈ 0.563

All in all, be careful with your nomenclature here. But essentially I laid out both a model for the problem and showed you two methods to solve for the answers to questions based on the model.

To reiterate, we assumed a Bernoulli distribution (satisfied/dissatisfied). Repeated Bernoulli sampling (i.e our 99 customers) gives us a Binomial distribution. Using the Central Limit Theorem / Normal Approximation for a Binomial, we answered the probability questions using z-scores.