Jessie estimated the weight of his cat to be 12 pounds. The actual weight of the cat is 15 pounds.

According to a 2014 Gallup poll, 56% of uninsured Americans who plan to get health insurance say they will do so through a gover

Airida [17]

Answer:

a) 24.27% probability that in a random sample of 10 people exactly 6 plan to get health insurance through a government health insurance exchange

b) 0.1% probability that in a random sample of 1000 people exactly 600 plan to get health insurance through a government health insurance exchange

c) Expected value is 560, variance is 246.4

d) 99.34% probability that less than 600 people plan to get health insurance through a government health insurance exchange

Step-by-step explanation:

To solve this question, we need to understand the binomial probability distribution and the binomial approximation to the normal.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

The expected value of the binomial distribution is:

$E(X) = np$

The variance of the binomial distribution is:

$V(X) = np(1-p)$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

56% of uninsured Americans who plan to get health insurance say they will do so through a government health insurance exchange.

This means that $p = 0.56$

a. What is the probability that in a random sample of 10 people exactly 6 plan to get health insurance through a government health insurance exchange?

This is P(X = 6) when n = 10. So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 6) = C_{10,6}.(0.56)^{6}.(0.44)^{4} = 0.2427$

24.27% probability that in a random sample of 10 people exactly 6 plan to get health insurance through a government health insurance exchange

b. What is the probability that in a random sample of 1000 people exactly 600 plan to get health insurance through a government health insurance exchange?

This is P(X = 600) when n = 1000. So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 600) = C_{1000,600}.(0.56)^{600}.(0.44)^{400} = 0.001$

0.1% probability that in a random sample of 1000 people exactly 600 plan to get health insurance through a government health insurance exchange

c. What are the expected value and the variance of X?

$E(X) = np = 1000*0.56 = 560$

$V(X) = np(1-p) = 1000*0.56*0.44 = 246.4$

d. What is the probability that less than 600 people plan to get health insurance through a government health insurance exchange?

Using the approximation to the normal

$\mu = 560, \sigma = \sqrt{246.4} = 15.70$

This is the pvalue of Z when X = 600-1 = 599. Subtract by 1 because it is less, and not less or equal.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{599 - 560}{15.70}$

$Z = 2.48$

$Z = 2.48$ has a pvalue of 0.9934

99.34% probability that less than 600 people plan to get health insurance through a government health insurance exchange

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Step-by-step explanation:

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