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BaLLatris [955]
3 years ago
8

A new medical test provides a false positive result for hepatitis 2% of the time that is a perfectly healthy subject being teste

d for hepatitis will test as being infected 2% of the time. And research, the test is given to 30 healthy (not having hepatitis) subjects. Let X be the number of subjects who test positive for the disease
A. What is the probability that all 30 subjects will appropriately test as not being infected?
B. What are the mean and standard deviation of X?
C. To what extent do you think this is a viable test to use in the field of medicine?
Mathematics
1 answer:
zloy xaker [14]3 years ago
4 0
A) You need to use the binomial distribution, for which the probability of an event X is given by:
P(X) = \frac{n!}{k!(n-k)!}  p^{k} (1-p)^{n-k}
where:
n = total number of events
k = number of success we want
p = probability of success

Therefore, since the problem tells you that <span>X is the number of subjects who test positive for the disease, you will have:

</span><span>P(X) = \frac{30!}{0!(30-0)!} 0.02^{0} (1-0.02)^{30-0}

= 1 </span>· 1 · 0.98³⁰
= 0.5455

Hence, the probability of none of the 30 subjects testing positive to the desease is 54.55%


B) In a binomial distribution, the mean is given by the formula:
μ = n · p
   = 30 · 0.02
   = 0.6

And the standard deviation is given by the formula:
σ = √[n·p·(1-p)]
   = √[30·0.02·0.98]
   = √0.588
   = 0.77

Hence, the mean is 0.6 and the standard deviation is 0.77


<span>C) This test is not very viable: 30 subjects are a sample too small compared to the population (millions of people who need to be tested), the probability of finding that all the 30 subjects are healty is only a little bit over 50%, the standard deviation is too high compared to the mean, and 2% of false positive is a percentage too high to consider the test viable.</span>
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Step-by-step explanation:

Given - A normally distributed random variable with mean 4.5 and standard deviation 7.6 is sampled to get two independent values, X1 and X2. The mean is estimated using the formula (3X1 + 4X2)/8.

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Bias for the estimator = E(μ bar) - μ

                                    = E( \frac{3X_{1} + 4X_{2}  }{8}) - 4.5

                                    = \frac{3E(X_{1}) + 4E(X_{2})}{8} - 4.5

                                    = \frac{3(4.5) + 4(4.5)}{8} - 4.5

                                    = \frac{13.5 + 18}{8} - 4.5

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Bias for the estimator = -0.56

Now,

Mean Square Error for the estimator = E[(μ bar - μ)²]

                                                             = Var(μ bar) + [Bias(μ bar, μ)]²

                                                             = Var( \frac{3X_{1} + 4X_{2}  }{8}) + 0.3136

                                                             = \frac{1}{64} Var( {3X_{1} + 4X_{2}  }) + 0.3136

                                                             = \frac{1}{64} ( [{3Var(X_{1}) + 4Var(X_{2})]  }) + 0.3136

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Mean Square Error for the estimator = 6.6311

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