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Vlad1618 [11]
1 year ago
10

a certain virus infects one in every 600 people. a test used to detect the virus in a person is positive 90% of the time if the

person has the virus and 10% of the time if the person does not have the virus. (a) find the probability that a person has the virus given that they have tested positive. $.0148026316$correct (b) find the probability that a person does not have the virus given that they have tested negative.
Mathematics
1 answer:
Alona [7]1 year ago
6 0
  1. the probability that a person has the virus given that they have tested positive is  0.0151.
  2. the probability that a person does not have the virus given that they have tested negative is  0.9999

P(A) = 1/600 = 0.0017

P(B) = 0.9 * 0.0017 + 0.1 * (1 - 0.0017) = 0.1014

A) P (has the virus | tested positive) = P (tested positive | has the virus) ×  

                                                          P (has the virus)/ P (tested positive)

                                                         = 0.9 × 0.0017/0.1014  

                                                         = 0.0151

B) P (does not have the virus | tested negative) = P (tested negative | does not have the virus) × P (does not have the virus)/ P (tested negative)

= (1 - 0.1) *× (1 - 0.0017)/ (1 - 0.1014)  

= 0.9999

Probability is the department of mathematics regarding numerical descriptions of ways likely an occasion is to occur, or how possibly it's far that a proposition is genuine. The possibility of an occasion varies between zero and 1, wherein, roughly speaking, 0 suggests the impossibility of the occasion and 1 shows certainty. The better the possibility of an event, the more likely it is that the event will arise.  

A simple instance is the tossing of an honest (unbiased) coin. since the coin is truthful, the 2 results ("heads" and "tails") are both equally likely; the possibility of "heads" equals the chance of "tails"; and considering the fact that no different results are feasible, the possibility of both "heads" or "tails" is 1/2 (that could additionally be written as 0.5 or 50%).

To learn more about Probability visit here:

brainly.com/question/11234923

#SPJ4

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Historically, the proportion of people who trade in their old car to a car dealer when purchasing a new car is 48%. Over the pre
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Answer:

z=\frac{0.4 -0.48}{\sqrt{\frac{0.48(1-0.48)}{115}}}=-1.717  

p_v =P(z  

So the p value obtained was a very low value and using the significance level given \alpha=0.1 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of people that have traded in their old car is lower than 0.48 or 48%.  

Step-by-step explanation:

Data given and notation

n=115 represent the random sample taken

X=46 represent the number of people that have traded in their old car.

\hat p=\frac{46}{115}=0.4 estimated proportion of people that have traded in their old car

p_o=0.48 is the value that we want to test

\alpha=0.1 represent the significance level

Confidence=90% or 0.9

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the proportion is less than 0.48.:  

Null hypothesis:p\geq 0.48  

Alternative hypothesis:p < 0.48  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.4 -0.48}{\sqrt{\frac{0.48(1-0.48)}{115}}}=-1.717  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided \alpha=0.1. The next step would be calculate the p value for this test.  

Since is a left tailed test the p value would be:  

p_v =P(z  

So the p value obtained was a very low value and using the significance level given \alpha=0.1 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of people that have traded in their old car is lower than 0.48 or 48%.  

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