Question

Suppose that 65% of high school students eat breakfast before going to school. A random sample of 500 high school students were surveyed. Let X = the number of high school students in a random sample of size 500 who eat breakfast before going to school.
1. Explain why X can be modeled by a binomial distribution even though the sample was selected without replacement.



2. Use a binomial distribution to estimate the probability that 300 or fewer high school students in the sample eat breakfast before going to school.



3. Justify why X can be approximated by a Normal distribution.



4. Use a Normal distribution to estimate the probability that 300 or fewer high school students in the sample eat breakfast before going to school.

Answer 1

Answer:

1.) very large sample size

2.) 0.11326

3.) np and n(1-p) should be ≥ 10

4.) 0.010809

Step-by-step explanation:

Given :

p = 0.65 ; n = 500

2.)

P(x ≤ 300)

P(x =x) = nCx * p^x * (1 - p)^(n - x)

Using a binomial probability calculator :

P(x ≤ 300) = 0.11326

To justify why if it can be approximated using a normal distributon :

np and n(1-p) should be ≥ 10

np = 500 * 0.65 = 325

n(1-p) = 500(1 - 0.65) = 175

4.) probability using normal distribution estimate:

Mean, m = np = 500 * 0.65 = 325

Standard deviation, s = sqrt(n*p*(1-p)) = sqrt(500*0.65*(1 - 0.65)) = sqrt(113.75) = 10.665

x ≤ 300

Correction:

x = x + 0.5 = 300 - 0.5 = 300.5

P(x ≤ 300.5)

Zscore = (x - m) / s

Zscore = (300.5 - 325) / 10.665

Zscore = - 24.5 / 10.665

Zscore = - 2.297

P(Z ≤ - 2.297) = 0.010809