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yuradex [85]
3 years ago
10

Suppose that 5% of the general population has a disease and the test for the disease is accurate 80% of the time. What is the pr

obability of testing positive for the disease?
Mathematics
2 answers:
Aleonysh [2.5K]3 years ago
6 0

Answer:

Step-by-step explanation:

Say we are testing 100 people

we test 100 times

80 of the 100 tests result in 5 people having the disease.

so 5 people of 80 of 100 test have this disease

therefore, 5/80/100

.065

aivan3 [116]3 years ago
4 0
<h2>Answer:</h2>

The probability of testing positive for the disease is:

                                23%

<h2>Step-by-step explanation:</h2>

The population who have disease = 5%

Hence, the population who do not have disease= 95%

( since 100%-55=95%)

Also, the percent that the test is accurate= 80%

and the percent that the test is not accurate= 100%-80%=20%

Hence, the  probability of testing positive for the disease is :

The person actually has disease×Test is positive+Person do not have disease×test is not accurate

=  0.05×0.80+0.95×0.20

= 0.23

Hence, the probability is: 23%

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The Town of Hertfordshire clerk knows that 23% of dogs in the town have completed emotional support training. Hertfordshire plan
Nataly_w [17]

Answer:

95.64% probability that under 30% of the dogs are emotional support trained

Step-by-step explanation:

For each dog, there are only two possible outcomes. Either they have completed emotional support training, or they have not. So we use the binomial probability distribution to solve this problem.

However, we are working with samples that are considerably big. So i am going to aproximate this binomial distribution to the normal.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

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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)}.

In this problem, we have that:

\mu = E(X) = np = 100*0.23 = 23

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{100*0.23*0.77} = 4.21

What is the probability that under 30% of the dogs are emotional support trained?

30% of 100 is 0.3*100 = 30

So this is the pvalue of Z when X = 30.

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

Z = \frac{30 - 23}{4.1}

Z = 1.71

Z = 1.71 has a pvalue of 0.9564.

So there is a 95.64% probability that under 30% of the dogs are emotional support trained

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