Question

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?

Answer 1

Answer:

$P(X > 100.5) = 0.0062$

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

$\mu = n \times p \\\\\mu = 500 \times 0.16 \\\\\mu = 80$

The standard deviation is given by

$\sigma = \sqrt{n \times p(1-p)} \\\\\sigma = \sqrt{500 \times 0.16(1-0.16)} \\\\\sigma = 8.197$

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

$P(X > 100) = 1 - P(X < 100)$

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

$P(X > 100.5) = 1 - P(X < 100.5)\\\\P(X > 100.5) = 1 - P(Z < \frac{x - \mu}{\sigma} )\\\\P(X > 100.5) = 1 - P(Z < \frac{100.5 - 80}{8.197} )\\\\P(X > 100.5) = 1 - P(Z < \frac{20.5}{8.197} )\\\\P(X > 100.5) = 1 - P(Z < 2.50)$

The z-score corresponding to 2.50 is 0.9938

$P(X > 100.5) = 1 - 0.9938\\\\P(X > 100.5) = 0.0062 \\\\P(X > 100.5) = 0.62\%$

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