Question

For their statistics project, a group of students want to determine the proportion of LMC students who eat fast food frequently. They select a random sample of 50 students and find that 22 report eating fast food frequently (more than 3 times a week.)

Answer 1

Answer:

The number of successes is, x = 22.

The number of failures is, (n - x) = 28.

Yes, normality conditions are met.

Step-by-step explanation:

The complete question is:

For their statistics project, a group of students want to determine the proportion of LMC students who eat fast food frequently. They select a random sample of 50 students and find that 22 report eating fast food frequently (more than 3 times a week.)

What is the count of success? Options are: 22,28, 50

What is the count of failures? Options are: 22,28,50

Can the students use this data to calculate a 95% confidence interval? Options are: Yes, normality conditions are met or No, normality conditions are not met

Solution:

Let X denote the number of LMC students who eat fast food frequently.

A random sample of n = 50 students and find that 22 report eating fast food frequently.

The number of successes is, x = 22.

The number of failures is, (n - x) = 28.

The random variable X follows a Binomial distribution with parameters n = 50 and p = 22/50 = 0.44.

But the sample selected is too large and the probability of success is close to 0.50.

So a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:

1. np ≥ 10

2. n(1 - p) ≥ 10

Check the conditions as follows:

$np=50\times 0.44=22>10\\\\n(1-p)=50\times (1-0.44)=28>10$

Thus, a Normal approximation to binomial can be applied.

Yes, normality conditions are met.