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Natasha2012 [34]
2 years ago
7

Using the following table determine the probability a person tests positive who does not actually have Tuberculosis.

Mathematics
2 answers:
Afina-wow [57]2 years ago
5 0

The probability a person tests positive who does not actually have Tuberculosis is 1/100

<h3>How to determine the probability a person tests positive who does not actually have Tuberculosis?</h3>

The probability is given as:

The probability a person tests positive who does not actually have Tuberculosis.

This is a conditional probability, and it can be calculated using

P = n(Tests positive | Does not have Tuberculosis)/n(Does not have Tuberculosis)

Using the given table of values, we have:

P = 98/9800

Evaluate the quotient

P = 1/100

Hence, the probability a person tests positive who does not actually have Tuberculosis is 1/100

Read more about probability at:

brainly.com/question/25870256

#SPJ1

Lady bird [3.3K]2 years ago
3 0

Using it's concept, the probability a person tests positive who does not actually have Tuberculosis is:

c. 49/148.

<h3>What is a probability?</h3>

A probability is given by the <u>number of desired outcomes divided by the number of total outcomes</u>.

In this problem, there are 198 + 98 = 296 people who test positive, which is the number of total outcomes, and of those, 98 do not have the disease, which is the number of desired outcomes. Hence the probability is given by:

p = 98/296 = 49/148.

Which means that option c is correct.

More can be learned about probabilities at brainly.com/question/14398287

#SPJ1

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In a clinical test with 2161 subjects, 1214 showed improvement from the treatment. Find the margin of error for the 95% confiden
Vlad [161]

Answer:

The margin of error for the 95% confidence interval used to estimate the population proportion is of 0.0209.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the z-score that has a p-value of 1 - \frac{\alpha}{2}.

The margin of error is of:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

In a clinical test with 2161 subjects, 1214 showed improvement from the treatment.

This means that n = 2161, \pi = \frac{1214}{2161} = 0.5618

95% confidence level

So \alpha = 0.05, z is the value of Z that has a p-value of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

Margin of error:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

M = 1.96\sqrt{\frac{0.5618*0.4382}{2161}}

M = 0.0209

The margin of error for the 95% confidence interval used to estimate the population proportion is of 0.0209.

4 0
3 years ago
The swim team has 41 people on their team they need new swim gear the swim caps cost $5 and the swim suit cost $ 16 dollars how
IRISSAK [1]

Answer:

The money they will have to pay is <u>$861</u>.

Step-by-step explanation:

Given:

The swim team has 41 people on their team.

They need new swim gear the swim caps cost $5 and the swim suit cost $16.

Now, to find the money they will have to pay.

Number of people on the swim team = 41.

Cost of swim caps = $5.

Cost of swim suit = $16.

Now, to get the total money they will have to pay:

<u><em>Number of people on the swim team× (cost of swim caps + cost of swim suit)</em></u>

=41\times (5 +16)

=41\times 21

=\$861.

Therefore, the money they will have to pay is $861.

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