The lengths of a large stock of titanium rods follow a normal distribution with a mean (μ) of 440 mm and a standard deviation (σ) of 1 mm. What is the percentage of rods whose lengths lie between 438 mm and 441 mm?
81.85%
To determine the percentage of titanium rods whose lengths fall between 438 mm and 441 mm, we need to use the properties of a normal distribution. We are given the mean (\(\mu\)) and standard deviation (\(\sigma\)) of the rod lengths.
Titanium Rod Lengths: Understanding the Normal Distribution
The lengths of the large stock of titanium rods are stated to follow a normal distribution. This is a common statistical distribution used to model many natural phenomena, including product dimensions in manufacturing.
- Mean (\(\mu\)): The average length of the titanium rods is 440 mm. This is the center point of our distribution.
- Standard Deviation (\(\sigma\)): The spread or variability of the rod lengths is 1 mm. A smaller standard deviation indicates that the data points are closer to the mean.
Rod Lengths: Calculating Z-Scores for Specific Values
To find the percentage of rods within a certain range, we first convert the given length values into Z-scores. A Z-score tells us how many standard deviations an element is from the mean. The formula for a Z-score (standard score) is:
\[Z = \frac{X - \mu}{\sigma}\]Where:
- \(X\) is the specific value from the distribution.
- \(\mu\) is the mean of the distribution.
- \(\sigma\) is the standard deviation of the distribution.
Z-Score for 438 mm:
For the lower limit, \(X_1 = 438\) mm:
\[Z_1 = \frac{438 - 440}{1} = \frac{-2}{1} = -2\]This means 438 mm is 2 standard deviations below the mean.
Z-Score for 441 mm:
For the upper limit, \(X_2 = 441\) mm:
\[Z_2 = \frac{441 - 440}{1} = \frac{1}{1} = 1\]This means 441 mm is 1 standard deviation above the mean.
Percentage of Rods: Finding Probability using Z-Scores
Once we have the Z-scores, we can use a standard normal distribution table (also known as a Z-table) or a calculator to find the probability (area under the curve) corresponding to these Z-scores. We are looking for the probability \(P(438 < X < 441)\), which is equivalent to \(P(-2 < Z < 1)\).
Using a standard normal distribution table (common approximate values):
- The probability \(P(Z < 1)\) (area to the left of Z = 1) is approximately 0.8413.
- The probability \(P(Z < -2)\) (area to the left of Z = -2) is approximately 0.0228.
The percentage of rods whose lengths lie between 438 mm and 441 mm is the difference between these two probabilities:
\[P(-2 < Z < 1) = P(Z < 1) - P(Z < -2)\] \[P(-2 < Z < 1) = 0.8413 - 0.0228\] \[P(-2 < Z < 1) = 0.8185\]To express this as a percentage, we multiply by 100:
\[\text{Percentage} = 0.8185 \times 100\% = 81.85\%\]Summary of Rod Length Calculation
Here's a concise summary of the steps involved in finding the percentage of titanium rods within the specified length range:
| Parameter | Value |
|---|---|
| Mean (\(\mu\)) | 440 mm |
| Standard Deviation (\(\sigma\)) | 1 mm |
| Lower Length (\(X_1\)) | 438 mm |
| Upper Length (\(X_2\)) | 441 mm |
| Z-score for \(X_1\) | \(Z_1 = -2\) |
| Z-score for \(X_2\) | \(Z_2 = 1\) |
| Probability \(P(Z < 1)\) | 0.8413 |
| Probability \(P(Z < -2)\) | 0.0228 |
| Desired Probability | \(0.8413 - 0.0228 = 0.8185\) |
| Percentage | 81.85% |
Therefore, 81.85% of the titanium rods have lengths between 438 mm and 441 mm.
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