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skelet666 [1.2K]
3 years ago
13

Restless Leg Syndrome and Fibromyalgia

Mathematics
1 answer:
Anna11 [10]3 years ago
3 0

Answer:

The probability that a person with restless leg syndrome has fibromyalgia is 0.183.

Step-by-step explanation:

Denote the events as follows:

<em>F</em> = a person with fibromyalgia

<em>R</em> = a person having restless leg syndrome

The information provided is as follows:

P (R | F) = 0.33

P (R | F') = 0.03

P (F) = 0.02

Consider the tree diagram attached below.

Compute the probability that a person with restless leg syndrome has fibromyalgia as follows:

P(F|R)=\frac{P(R|F)P(F)}{P(R|F)P(F)+P(R|F')P(F')}

            =\frac{(0.33\times 0.02)}{(0.33\times 0.02)+(0.03\times 0.98)}\\\\=\frac{0.0066}{0.0066+0.0294}\\\\=0.183333\\\\\approx 0.183

Thus, the probability that a person with restless leg syndrome has fibromyalgia is 0.183.

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In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by \alpha=1-0.99=0.01 and \alpha/2 =0.005. And the critical value would be given by:

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n=\frac{0.5(1-0.5)}{(\frac{0.009}{2.58})^2}=20544.44  

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