Q. 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?
(A) 81.85% (B) 68.4% (C) 99.75% (D) 86.64%
Ans: 81.85%
Sol:
Given, mean, (μ) = 440 mm Standard deviation, σ = 1 mm
lower limit,
Percentage of rods whose lengths lie between 438 mm and 441 mm.
= 0.3413 + (0.5 – 0.0228)
= 0.81854 = 81.854%
