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ExamsGATEGeneral Aptitude

The lengths of a large stock of titanium rods follow a normal distribution with mean $\mu = 440$ mm and standard deviation $\sigma = 1$ mm. What is the percentage of rods whose lengths are between 438 mm and 441 mm?

  1. 81.85%
  2. 68.4%
  3. 99.75%
  4. 86.64%

Correct answer: 81.85%

Solution

Standardizing gives $z = (438-440)/1 = -2$ and $z = (441-440)/1 = 1$. The required probability is $P(-2 \le Z \le 1) = P(Z \le 1) - P(Z \le -2) = 0.8413 - 0.0228 = 0.8185$. Thus the percentage is 81.85%.

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