Answer:
The distribution of the sample proportion ~ N(0.4, 0.07302)
Step-by-step explanation:
The distribution of the data can be modeled with a binomial distribution with probability p = 0.4 of finding a person that consume her or his daily recommended intake of fiber, and q = 0.6
If n is the sample size, we must check if np ≥ 10 and nq ≥ 10, so we can approximate the Binomial with a Normal distribution.
The sample size is 45
45*0.4 = 18 > 10 and 45*0.6 = 27 > 10.
The distribution of the sample proportion ~ the Normal with mean p and standard deviation
[tex]\bf \sqrt{\frac{pq}{n}}=\sqrt{\frac{0.4*0.6}{45}}=0.07302[/tex]
and
The distribution of the sample proportion ~ N(0.4, 0.07302)
