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masya89 [10]
3 years ago
7

OMG PLEASE HELP IF YOU CAN 14 POINTS!!!!!! For each function what is the output for the given input? f(x) = 5x+1, find f(-4)

Mathematics
1 answer:
aleksandr82 [10.1K]3 years ago
3 0
<span>f(x) = 5x+1
f(-4) = 5(-4) + 1 = -20 + 1 = -19


hope that helps
</span>
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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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