Suppose that 10% of all steel shafts produced by a certain process are nonconforming but can be reworked (rather than having to be scrapped). Consider a random sample of 200 shafts, and let X denote the number among these that are nonconforming and can be reworked. (Round your answers to four decimal places.)

(a) What is the (approximate) probability that X is at most 30?
(b) What is the (approximate) probability that X is less than 30?
(c) What is the (approximate) probability that X is between 15 and 25 (inclusive)?


Detailed Answer:

Calculating Probabilities for Nonconforming Steel Shafts: A Binomial Distribution Problem

In manufacturing, quality control is crucial. Understanding the likelihood of defective products, or in this case, nonconforming steel shafts, allows manufacturers to optimize processes and reduce waste. In this blog, we’ll calculate the probabilities for different outcomes using the binomial distribution and normal approximation, focusing on nonconforming steel shafts.

Problem Overview

We are told that 10% of steel shafts produced are nonconforming but can be reworked. A random sample of 200 shafts is selected, and we are asked to determine the probability of specific outcomes, where X denotes the number of nonconforming shafts that can be reworked.

Given Data:

• Probability of nonconformance $(p)=0.1$
• Sample size $(n)=200$

Our task is to calculate:

• (a) The probability that X is at most 30.
• (b) The probability that X is less than 30.
• (c) The probability that X is between 15 and 25 (inclusive).

Step-by-Step Solution

Since we are dealing with a large sample size ($n=200$), we can use the normal approximation to the binomial distribution to calculate these probabilities.

1. Calculating the Mean and Standard Deviation

The mean $(\mu )$ is given by:

$\mu =n\times p$

$=200\times 0.1$

$=20$

The standard deviation $(\sigma )$ is:

$\sigma =\sqrt{n\times p\times (1–p)}$

$=\sqrt{200\times 0.1\times (1–0.1)}$

$=\sqrt{200\times 0.1\times 0.9}$

$=4.243$

2. A) Probability that X is at Most 30

To find the probability that $X$ is at most 30, we apply the normal approximation:

$P(X_{binomial}\le 30)\approx P(X_{normal}\le 30+0.5)$

Using the continuity correction factor:

$P(X_{normal}\le 30.5)$

We now calculate the Z-score:

$Z=\frac{X_{normal}–\mu }{\sigma }$

$=\frac{30.5–20}{4.243}$

$\approx 2.47$

Now, we use the Z-table to find the probability:

$P(Z\le 2.47)\approx 0.9932$

Thus, the probability that X is at most 30 is approximately 0.9932.

3. B) Probability that X is Less than 30

Next, we calculate the probability that $X$ is less than 30. The process is similar to part (a), as we use the normal approximation and continuity correction:

$P(X_{binomial}<30)=P(X_{normal}<30+0.5)$

$=P(X_{normal}\le 30.5)$

Using the same Z-score:

$Z=\frac{30.5–20}{4.243}\approx 2.47$

From the Z-table:

$P(Z\le 2.47)\approx 0.9932$

Therefore, the probability that X is less than 30 is also approximately 0.9932.

4. C) Probability that X is Between 15 and 25 (Inclusive)

Finally, we calculate the probability that $X$ is between 15 and 25 (inclusive). We again use the normal approximation and continuity correction:

$P(15\le X_{binomial}\le 25)\approx P(14.5\le X_{normal}\le 25.5)$

We calculate the Z-scores for both 14.5 and 25.5:

For $X=14.5$:

$Z=\frac{14.5–20}{4.243}=\frac{-5.5}{4.243}\approx -1.30$

For $X=25.5$:

$Z=\frac{25.5–20}{4.243}=\frac{5.5}{4.243}\approx 1.30$

Now, we find the probabilities from the Z-table:

$P(Z\le 1.30)\approx 0.9032$

$P(Z\le -1.30)\approx 0.0968$

Now, subtract the probabilities:

$P(-1.30\le Z\le 1.30)=P(Z\le 1.30)–P(Z\le -1.30)$

$=0.9032–0.0968$

$=0.8064$

Thus, the probability that X is between 15 and 25 (inclusive) is approximately 0.8064.

Conclusion

In this example, we used the normal approximation to the binomial distribution to calculate the probabilities for nonconforming steel shafts that can be reworked. The probabilities for the different scenarios are:

• The probability that X is at most 30 is approximately 0.9932.
• The probability that X is less than 30 is also approximately 0.9932.
• The probability that X is between 15 and 25 is approximately 0.8064.

Understanding how to use the normal approximation for binomial distributions is crucial in quality control and many other fields where large sample sizes are common.