Question

a drug test is accurate 98% of the time. if the test is given to 2200 people who have not taken drugs, what is the probability that at least 47 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 = 0.352
give your answers to at least 3 decimal places.
question help: written example message instructor

Explanation:

Step1: Identify parameters for binomial distribution

We have a binomial distribution with \( n = 2200 \) (number of trials, i.e., number of people tested) and \( p \) is the probability that the test is positive (which is the probability that the test is inaccurate, since these people have not taken drugs). The test is accurate 98% of the time, so the probability of inaccuracy (false positive) is \( p=1 - 0.98=0.02 \).

The mean of the binomial distribution is \( \mu=np \) and the standard deviation is \( \sigma=\sqrt{np(1 - p)} \)

Calculate \( \mu \) and \( \sigma \):
\( \mu=2200\times0.02 = 44 \)
\( \sigma=\sqrt{2200\times0.02\times(1 - 0.02)}=\sqrt{2200\times0.02\times0.98}=\sqrt{43.12}\approx6.566 \)

Step2: Apply normal approximation to binomial

We want \( P(X\geq47) \). For normal approximation to binomial, we use continuity correction. So \( P(X\geq47)=P(X > 46.5) \)

First, we calculate the z - score: \( z=\frac{x-\mu}{\sigma} \), where \( x = 46.5 \), \( \mu = 44 \), \( \sigma\approx6.566 \)

\( z=\frac{46.5 - 44}{6.566}=\frac{2.5}{6.566}\approx0.38 \)

Step3: Find the probability using z - table

We know that \( P(X > 46.5)=1 - P(X\leq46.5) \). Looking up the z - score of \( 0.38 \) in the standard normal table, \( P(Z\leq0.38) = 0.6480 \)

So \( P(X\geq47)=1 - 0.6480 = 0.352 \) (Wait, but let's check the calculation again. Wait, maybe I made a mistake in the z - score. Wait, \( x = 47 \), continuity correction: \( P(X\geq47)=P(X>46.5) \). \( \mu = 44 \), \( \sigma=\sqrt{2200\times0.02\times0.98}=\sqrt{43.12}\approx6.566 \)

\( z=\frac{46.5 - 44}{6.566}=\frac{2.5}{6.566}\approx0.38 \). The area to the left of \( z = 0.38 \) is \( 0.6480 \), so the area to the right is \( 1 - 0.6480=0.352 \), which matches the given hint value. Wait, but maybe the question is to find the probability that at least 47 test positive. Let's confirm the steps.

Wait, the binomial distribution: \( n = 2200 \), \( p = 0.02 \) (probability of test being inaccurate, i.e., false positive). We use normal approximation \( N(np, np(1 - p))=N(44, 43.12) \)

For \( P(X\geq47) \), continuity correction: \( X\geq47 \) is equivalent to \( X>46.5 \) in normal distribution.

Z - score: \( z=\frac{46.5 - 44}{\sqrt{43.12}}=\frac{2.5}{6.566}\approx0.38 \)

\( P(Z > 0.38)=1 - P(Z\leq0.38) \). From standard normal table, \( P(Z\leq0.38)=0.6480 \), so \( P(Z > 0.38)=1 - 0.6480 = 0.352 \)

Answer:

\( 0.352 \)