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 =
give your answers to at least 3 decimal places.
question help: written example message instructor
submit question

Explanation:

Step1: Identify distribution parameters

We have a binomial distribution with \( n = 2200 \) (number of trials) and \( p = 0.02 \) (probability of test being inaccurate, since accuracy is 98%, inaccuracy is \( 1 - 0.98 = 0.02 \)). We want \( P(X \geq 47) \), where \( X \) is the number of inaccurate tests (positive tests for non - drug users, which is inaccurate).

First, check if normal approximation is valid: \( np=2200\times0.02 = 44\) and \( n(1 - p)=2200\times0.98 = 2156\), both are greater than 5, so normal approximation is appropriate.

The mean of the binomial distribution \( \mu=np = 2200\times0.02=44 \)

The standard deviation \( \sigma=\sqrt{np(1 - p)}=\sqrt{2200\times0.02\times0.98}=\sqrt{43.12}\approx6.566 \)

Step2: Apply continuity correction

For \( P(X\geq47) \) in binomial, using continuity correction, we find \( P(X\geq47)=P(X > 46.5) \) in the normal approximation.

Step3: Calculate z - score

The z - score is calculated as \( z=\frac{x-\mu}{\sigma} \), where \( x = 46.5 \), \( \mu = 44 \), and \( \sigma\approx6.566 \)

\( z=\frac{46.5 - 44}{6.566}=\frac{2.5}{6.566}\approx0.381 \)

Step4: Find the probability

We know that \( P(X > 46.5)=1 - P(X\leq46.5) \). Looking up the z - score of \( 0.381 \) in the standard normal table, \( P(Z\leq0.381)\approx0.648 \)

So \( P(X\geq47)=1 - 0.648 = 0.352 \) (approximate value, more precise calculation:

Using a calculator for more precise z - score calculation:

\( z=\frac{46.5 - 44}{\sqrt{2200\times0.02\times0.98}}=\frac{2.5}{\sqrt{43.12}}\approx\frac{2.5}{6.566}\approx0.381 \)

Using the standard normal distribution, \( P(Z < 0.381)=0.6480 \) (from standard normal table or calculator). Then \( P(Z\geq0.381)=1 - 0.6480 = 0.352 \)

Answer:

\( 0.352 \) (or a more precise value around this, depending on the z - table or calculator used. If we use more precise calculation of z - score and standard normal distribution, the value may vary slightly, but approximately 0.352)