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a drug test is accurate 97% of the time. if the test is given to 1700 people who have not taken drugs, what
is the probability that at least 55 will test positive (use the normal approximation to the binomial)?
hint: since the people have not taken drugs, if they test positive that means the test was inaccurate.
what is the probability the test is not accurate? use that value for p.
probability =
enter an integer or decimal number more...
give your answers to at least 3 decimal places.
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Explanation:

Step1: Calculate \(p\) and \(n\)

Since the test is accurate \(97\%\) of the time, the probability of an inaccurate test ( \(p\)) is \(p = 1 - 0.97=0.03\), and \(n = 1700\).

Step2: Calculate the mean \(\mu\) and standard deviation \(\sigma\)

The mean of a binomial distribution is \(\mu=np\). So, \(\mu = 1700\times0.03 = 51\).
The standard deviation of a binomial distribution is \(\sigma=\sqrt{np(1 - p)}\). Substitute \(n = 1700\) and \(p = 0.03\) into the formula: \(\sigma=\sqrt{1700\times0.03\times(1 - 0.03)}=\sqrt{1700\times0.03\times0.97}=\sqrt{49.47}\approx7.03\).

Step3: Use the continuity correction

For \(P(X\geq55)\) (using the normal approximation to the binomial), with continuity correction, we find \(P(X > 54.5)\).
The \(z\)-score is \(z=\frac{x-\mu}{\sigma}\). Substitute \(x = 54.5\), \(\mu = 51\), and \(\sigma\approx7.03\) into the formula: \(z=\frac{54.5 - 51}{7.03}=\frac{3.5}{7.03}\approx0.50\).

Step4: Calculate the probability

We want \(P(Z>0.50)\). Since \(P(Z > z)=1 - P(Z\leq z)\), and from the standard - normal table \(P(Z\leq0.50)=0.6915\). So \(P(Z>0.50)=1 - 0.6915 = 0.3085\).

Answer:

\(0.309\)