Respuesta :
Answer:
[tex]P(X > 100.5) = 0.0062 \\\\[/tex]
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
[tex]\mu = n \times p \\\\\mu = 500 \times 0.16 \\\\\mu = 80[/tex]
The standard deviation is given by
[tex]\sigma = \sqrt{n \times p(1-p)} \\\\\sigma = \sqrt{500 \times 0.16(1-0.16)} \\\\\sigma = 8.197[/tex]
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
[tex]P(X > 100) = 1 - P(X < 100)[/tex]
We need to consider the continuity correction factor whenever we use continuous probability distribution (Normal distribution) to approximate discrete probability distribution (Binomial distribution).
[tex]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)\\\\[/tex]
The z-score corresponding to 2.50 is 0.9938
[tex]P(X > 100.5) = 1 - 0.9938\\\\P(X > 100.5) = 0.0062 \\\\P(X > 100.5) = 0.62\% \\\\[/tex]
Therefore, there is 0.0062 probability that more than 20% of Americans do not have health insurance.
