Q3. Sixteen percent of Americans do not have health insurance. Suppose a simple random sample of 500 Americans is obtained. In a random sample of 500 Americans, what is the probability that more than 20% do not have health insurance?

Therefore, there is 0.0062 probability that more than 20% of Americans do not have health insurance.

Step-by-step explanation:

Sixteen percent of Americans do not have health insurance. Suppose a simple random sample of 500.

From the above information,

p = 16% = 0.16

n = 500

The mean is given by

The standard deviation is given by

What is the probability that more than 20% do not have health insurance?

We can use the Normal distribution as an approximation to the Binomial distribution since the following conditions are satisfied.

n×p ≥ 5  (satisfied)

n×(1 - p) ≥ 5   (satisfied)

So the probability is given by

500×0.20 = 100

We need to consider the continuity correction factor whenever we use continuous probability distribution (Normal distribution) to approximate discrete probability distribution (Binomial distribution).

The z-score corresponding to 2.50 is 0.9938

Therefore, there is 0.0062 probability that more than 20% of Americans do not have health insurance.

i dont know sorry

step-by-step explanation: