To solve this problem using the normal approximation to the binomial distribution, we will first calculate the mean and standard deviation and then proceed to the probability questions using the appropriate z-scores.

We are given:

• Defective rate $p = 0.066$ (6.6%),
• Number of strings $n = 130130$.

Part a: Mean and Standard Deviation

The formulas for the mean $\mu$ and standard deviation $\sigma$ of a binomial distribution are:

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

Let's calculate these values.

1. Mean ($\mu$):

$\mu = 130130 \times 0.066$

2. Standard deviation ($\sigma$):

$\sigma = \sqrt{130130 \times 0.066 \times (1 - 0.066)}$

I'll compute these now.The mean ($\mu$) is approximately 8588.5800, and the standard deviation ($\sigma$) is approximately 89.5641.

Part b: Probability that fewer than 44 strings will be defective

For this part, we use the normal approximation. Since we're dealing with a binomial distribution, the normal approximation requires a continuity correction. So, for "fewer than 44," we consider $X = 43.5$.

The z-score is calculated as follows:

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

For $X = 43.5$, we substitute the values of $\mu$ and $\sigma$ that we calculated earlier:

$z = \frac{43.5 - 8588.58}{89.5641}$

I'll compute this z-score and find the corresponding probability.The z-score for $X = 43.5$ is approximately $-95.4074$, which is extremely low. This results in a probability of essentially 0. Therefore, the probability that fewer than 44 strings will be defective is approximately 0.

Part c: Probability that exactly 99 strings will be defective

For this, we also use the normal approximation with continuity correction. To approximate the probability for exactly 99 defective strings, we calculate the z-scores for $X = 98.5$ and $X = 99.5$.

The probabilities are found by determining the area between these two z-scores:

$z_1 = \frac{98.5 - \mu}{\sigma}, \quad z_2 = \frac{99.5 - \mu}{\sigma}$

I'll compute the z-scores and the probability now.It seems like I can’t do more advanced data analysis right now. Please try again later. If you'd like, you can provide a new request or continue with questions based on the current explanations.

Would you like to proceed with further questions or explore another part of the problem?

Here are five relative questions:

1. How does the continuity correction improve normal approximation in binomial problems?
2. What conditions are required for the normal approximation to the binomial distribution to be valid?
3. How would you calculate the z-score for other defective counts, such as 150 or 200?
4. Why do probabilities for extremely low z-scores approach zero?
5. How can you determine whether to use a one-tailed or two-tailed probability in binomial problems?

Tip: For large sample sizes, the binomial distribution can often be approximated by the normal distribution when $np$ and $n(1-p)$ are both greater than 5.