To determine the percentage of lawn mower parts that will have lengths between 3.8 inches and 4.2 inches given that the lengths are approximately normally distributed with a mean ([tex]\(\mu\)[/tex]) of 4 inches and a standard deviation ([tex]\(\sigma\)[/tex]) of 0.2 inches, we should follow these steps:
1. Identify the key components of the problem:
- Mean ([tex]\(\mu\)[/tex]) = 4 inches
- Standard Deviation ([tex]\(\sigma\)[/tex]) = 0.2 inches
- Lower bound = 3.8 inches
- Upper bound = 4.2 inches
2. Calculate the z-scores for the lower and upper bounds:
- The z-score for a given value is calculated using the formula:
[tex]\[
z = \frac{X - \mu}{\sigma}
\][/tex]
- For the lower bound (3.8 inches):
[tex]\[
z_{\text{lower}} = \frac{3.8 - 4}{0.2} = -1
\][/tex]
- For the upper bound (4.2 inches):
[tex]\[
z_{\text{upper}} = \frac{4.2 - 4}{0.2} = 1
\][/tex]
3. Find the probabilities corresponding to these z-scores:
- Using the standard normal distribution table (or a calculator), we find:
- [tex]\( P(Z \leq z_{\text{lower}}) = P(Z \leq -1) \approx 0.1587 \)[/tex]
- [tex]\( P(Z \leq z_{\text{upper}}) = P(Z \leq 1) \approx 0.8413 \)[/tex]
4. Calculate the percentage of parts that fall between these bounds:
- The probability that a part's length falls between the lower and upper bounds is the difference between the two probabilities:
[tex]\[
P(3.8 \leq X \leq 4.2) = P(Z \leq 1) - P(Z \leq -1) = 0.8413 - 0.1587 = 0.6826
\][/tex]
5. Convert the probability to a percentage:
- [tex]\( 0.6826 \times 100\% \approx 68\% \)[/tex]
Therefore, approximately 68% of the lawn mower parts will have lengths between 3.8 inches and 4.2 inches.
Thus, the correct answer is [tex]\( 68\% \)[/tex].
