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It is known that 35 percent of members of a population suffer from at least one...

It is known that 35 percent of members of a population suffer from at least one chronic disease. What is the probability that in a random sample of 200 individuals 100 or more do not have chronic diseases?  

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Answer #1

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First, we need to find mean and standard deviation for the given distribution

we have p = 35% = 35/100 = 0.35 and sample size n = 200

Mean = n*p = 0.35*200 = 70 and standard deviation = уп *p * (1-p) V/200 * 0.35 * (1-0.35-7.75

we have to find the probability that 100 or more do not have chronic disease. So, we will find probability of more than 99.

So, using the formula P(X>x) = P(z>(ar{x}-mu)/(sigma))

where we have  ar{x} = 99, mu = 70, n =200, sigma = 7.75

setting the given values, we get

P(X > 99) P(z > (99-70)/(7.75))-P(z > (29/7.75))

on solving, we get

P(X > 99) P(z > (3.74))

Using identity P(z>a) =1-P(z<a)

we can write it as

P(z > 3.74) 1- Pz<3.74)

using z distribution table(looking 3.0 in first column on left side and 0.74 on top most row) , we get

P(z > 3.74) 1- Pz<3.74) = 1 - 0.9999 = 0.0001

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