Question

g 78% of all Millennials drink Starbucks coffee at least once a week. Suppose a random sample of 50 Millennials will be selected. What is the approximate distribution for the proportion in a sample who drink Starbucks coffee at least once a week (from all possible random samples of 50 Millennials)?

Answer 1

Answer:

We know that n = 50 and p =0.78.

We need to check the conditions in order to use the normal approximation.

$np=50*0.78=39 \geq 10$

$n(1-p)=50*(1-0.78)=11 \geq 10$

Since both conditions are satisfied we can use the normal approximation and the distribution for the proportion is given by:

$p \sim N (p, \sqrt{\frac{p(1-p)}{n}})$

With the following parameters:

$\mu_ p = 0.78$

$\sigma_p = \sqrt{\frac{0.78*(1-0.78)}{50}}= 0.0586$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

We know that n = 50 and p =0.78.

We need to check the conditions in order to use the normal approximation.

$np=50*0.78=39 \geq 10$

$n(1-p)=50*(1-0.78)=11 \geq 10$

Since both conditions are satisfied we can use the normal approximation and the distribution for the proportion is given by:

$p \sim N (p, \sqrt{\frac{p(1-p)}{n}})$

With the following parameters:

$\mu_ p = 0.78$

$\sigma_p = \sqrt{\frac{0.78*(1-0.78)}{50}}= 0.0586$