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.

Respuesta :

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:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

Where:

[tex]\begin{gathered} z=\text{ z-score} \\ \mu=\text{ mean} \\ \sigma=\text{ standard deviation} \end{gathered}[/tex]

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

[tex]\mu=n\pi[/tex]

Where

[tex]\begin{gathered} n=\text{ sample size} \\ \pi=\text{ failure rate in decimal form} \end{gathered}[/tex]

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

[tex]\pi=\frac{7.4}{100}=0.074[/tex]

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

[tex]\begin{gathered} \mu=(295)(0.074) \\ \mu=21.83 \end{gathered}[/tex]

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

[tex]\sigma=\sqrt{n\pi(1-\pi)}[/tex]

Substituting the values we get:

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

Solving the operations:

[tex]\sigma=4.49[/tex]

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:

[tex]x=22-0.5=21.5[/tex]

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

[tex]z=\frac{21.5-21.83}{4.49}[/tex]

Now, we solve the operations:

[tex]z=-0.073[/tex]

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

[tex]P(x<22)=0.472[/tex]

Therefore, the probability is 0.472

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