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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?
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- 68.4%
- 99.75%
- 81.85%
- 86.64%
Correct Option: C
Given, mean, (μ) = 440 mm Standard deviation, σ = 1 mm
| Z = | σ |
lower limit,
| Z(x = 438) = | = - 2 | 1 |
| Z(x = 441 mm) = | = 1 | 1 |
Percentage of rods whose lengths lie between 438 mm and 441 mm.
= 0.3413 + (0.5 – 0.0228)
= 0.81854 = 81.854%
