Question

A manufacturing process produces semiconductor chips with a known failure rate of 7,4%. If a random sample of 295 chips is selected, approximate the probability that fewer than 22 will be defective. Use the normal approximation to the binomial with a correction for continuity.Round your answer to at least three decimal places. Do not round any intermediate steps.

Answer 1

To determine the probability using a normal approximation we need to determine the z-score of the sample. To do that we use the following formula:

$z=\frac{x-\mu}{\sigma}$

Where:

$\begin{gathered} z=\text{ z-score} \\ \mu=\text{ mean} \\ \sigma=\text{ standard deviation} \end{gathered}$

First, we need to calculate the mean. To do that we use the following formula:

$\mu=n\pi$

Where

$\begin{gathered} n=\text{ sample size} \\ \pi=\text{ failure rate in decimal form} \end{gathered}$

To determine the decimal form of the failure rate we divide the percentage by 100, like this:

$\pi=\frac{7.4}{100}=0.074$

The sample size, in this case, is 295. Substituting we get:

$\begin{gathered} \mu=(295)(0.074) \\ \mu=21.83 \end{gathered}$

Now, we need to calculate the standard deviation. We do this using the following formula:

$\sigma=\sqrt{n\pi(1-\pi)}$

Substituting the values we get:

$\sigma=\sqrt[]{(295)(0.074)(1-0.074)}$

Solving the operations:

$\sigma=4.49$

Since we are approximating to binomial we need to use a continuity correction factor of 0.5 to the number of chips we are considering. Let "x" be the number of chips, we have that:

$x=22-0.5=21.5$

Now we substitute in the formula for the z-score:

$z=\frac{21.5-21.83}{4.49}$

Now, we solve the operations:

$z=-0.073$

Therefore, in the binomial distribution, we need to determine the area for z = -0.073. Therefore, the probability is:

$P(x<22)=0.472$

Therefore, the probability is 0.472