The lengths of rods made by a machine are normally distributed with mean 50 cm and standard deviation 4 cm. The probability that a rod chosen at random is longer than 56 cm is (Given \(P(0 \le Z \le 1.5) = 0.4332\), where Z is the standard normal variate)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0.0668
Let X be the length of a rod, so \(\mu = 50\) and \(\sigma = 4\). For X = 56,
\( \displaystyle Z = \frac{56 - 50}{4} = 1.5 \)
The area to the right of the mean is 0.5, and 0.4332 of it lies between Z = 0 and Z = 1.5:
\(P(X \gt 56) = P(Z \gt 1.5)\)
\(= 0.5 - 0.4332\)
\(= 0.0668\)
Why the other options are incorrect
Option (1) 0.9332 — Incorrect. This is 0.5 + 0.4332, the probability of a rod shorter than 56 cm.
Option (3) 0.4332 — Incorrect. This is the probability of a length between 50 cm and 56 cm.
Option (4) 0.5668 — Incorrect. This is 0.5 + 0.0668; the half of the curve below the mean is not longer than 56 cm.