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To determine what percentage of the lawn mower parts will have lengths between 3.8 inches and 4.2 inches, we need to use the properties of the normal distribution. Here's the detailed step-by-step solution:
1. Identify the given parameters:
- The mean (μ) of the lengths is 4 inches.
- The standard deviation (σ) is 0.2 inches.
2. Calculate the z-scores for the lower and upper bounds:
- The lower bound is 3.8 inches.
- The upper bound is 4.2 inches.
The z-score for a value [tex]\(X\)[/tex] is calculated using the formula:
[tex]\[ z = \frac{X - \mu}{\sigma} \][/tex]
For the lower bound (3.8 inches):
[tex]\[ z_{\text{lower}} = \frac{3.8 - 4}{0.2} \][/tex]
[tex]\[ z_{\text{lower}} = \frac{-0.2}{0.2} \][/tex]
[tex]\[ z_{\text{lower}} = -1 \][/tex]
For the upper bound (4.2 inches):
[tex]\[ z_{\text{upper}} = \frac{4.2 - 4}{0.2} \][/tex]
[tex]\[ z_{\text{upper}} = \frac{0.2}{0.2} \][/tex]
[tex]\[ z_{\text{upper}} = 1 \][/tex]
3. Determine the cumulative probabilities:
- Using standard normal distribution tables or a cumulative distribution function (CDF) calculator, find the probabilities corresponding to these z-scores.
The cumulative probability for [tex]\( z = -1 \)[/tex]:
[tex]\[ P(Z \leq -1) \approx 0.1587 \][/tex]
The cumulative probability for [tex]\( z = 1 \)[/tex]:
[tex]\[ P(Z \leq 1) \approx 0.8413 \][/tex]
4. Calculate the probability between the z-scores:
- To find the probability that a part's length falls between 3.8 inches and 4.2 inches, subtract the cumulative probability at z = -1 from the cumulative probability at z = 1.
[tex]\[ P(-1 \leq Z \leq 1) = P(Z \leq 1) - P(Z \leq -1) \][/tex]
[tex]\[ P(-1 \leq Z \leq 1) \approx 0.8413 - 0.1587 \][/tex]
[tex]\[ P(-1 \leq Z \leq 1) \approx 0.6826 \][/tex]
5. Convert to percentage:
- To convert this probability to a percentage, multiply by 100.
[tex]\[ \text{Percentage} = 0.6826 \times 100 \][/tex]
[tex]\[ \text{Percentage} \approx 68.27\% \][/tex]
Therefore, approximately 68.27% of the lawn mower parts will have lengths between 3.8 inches and 4.2 inches. The correct answer is:
[tex]\[ \boxed{68 \%} \][/tex]
1. Identify the given parameters:
- The mean (μ) of the lengths is 4 inches.
- The standard deviation (σ) is 0.2 inches.
2. Calculate the z-scores for the lower and upper bounds:
- The lower bound is 3.8 inches.
- The upper bound is 4.2 inches.
The z-score for a value [tex]\(X\)[/tex] is calculated using the formula:
[tex]\[ z = \frac{X - \mu}{\sigma} \][/tex]
For the lower bound (3.8 inches):
[tex]\[ z_{\text{lower}} = \frac{3.8 - 4}{0.2} \][/tex]
[tex]\[ z_{\text{lower}} = \frac{-0.2}{0.2} \][/tex]
[tex]\[ z_{\text{lower}} = -1 \][/tex]
For the upper bound (4.2 inches):
[tex]\[ z_{\text{upper}} = \frac{4.2 - 4}{0.2} \][/tex]
[tex]\[ z_{\text{upper}} = \frac{0.2}{0.2} \][/tex]
[tex]\[ z_{\text{upper}} = 1 \][/tex]
3. Determine the cumulative probabilities:
- Using standard normal distribution tables or a cumulative distribution function (CDF) calculator, find the probabilities corresponding to these z-scores.
The cumulative probability for [tex]\( z = -1 \)[/tex]:
[tex]\[ P(Z \leq -1) \approx 0.1587 \][/tex]
The cumulative probability for [tex]\( z = 1 \)[/tex]:
[tex]\[ P(Z \leq 1) \approx 0.8413 \][/tex]
4. Calculate the probability between the z-scores:
- To find the probability that a part's length falls between 3.8 inches and 4.2 inches, subtract the cumulative probability at z = -1 from the cumulative probability at z = 1.
[tex]\[ P(-1 \leq Z \leq 1) = P(Z \leq 1) - P(Z \leq -1) \][/tex]
[tex]\[ P(-1 \leq Z \leq 1) \approx 0.8413 - 0.1587 \][/tex]
[tex]\[ P(-1 \leq Z \leq 1) \approx 0.6826 \][/tex]
5. Convert to percentage:
- To convert this probability to a percentage, multiply by 100.
[tex]\[ \text{Percentage} = 0.6826 \times 100 \][/tex]
[tex]\[ \text{Percentage} \approx 68.27\% \][/tex]
Therefore, approximately 68.27% of the lawn mower parts will have lengths between 3.8 inches and 4.2 inches. The correct answer is:
[tex]\[ \boxed{68 \%} \][/tex]
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