Question

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 164.4-cm and a standard deviation of 1.4-cm.

Find the probability that the length of a randomly selected steel rod is greater than 162-cm.
P(X > 162-cm) =


Enter your answer as a number accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

Answer 1

To find the probability that the length of a randomly selected steel rod is greater than 162 cm, we first need to calculate the z-score for 162 cm using the formula:

\[
z = \frac{(X - \mu)}{\sigma}
\]

where:
- \(X\) is the value we are interested in (162 cm),
- \(\mu\) is the mean (164.4 cm), and
- \(\sigma\) is the standard deviation (1.4 cm).

Substituting the values into the formula gives us:

\[
z = \frac{(162 - 164.4)}{1.4} = \frac{-2.4}{1.4} \approx -1.7143
\]

Next, we need to find the probability \(P(X > 162)\), which is equivalent to finding \(P(Z > -1.7143)\). Using the standard normal distribution table or a calculator, we first find \(P(Z \leq -1.7143)\).

Using a standard normal distribution calculator or table, we get:

\[
P(Z \leq -1.7143) \approx 0.0436
\]

Thus, the probability \(P(Z > -1.7143)\) is:

\[
P(Z > -1.7143) = 1 - P(Z \leq -1.7143) \approx 1 - 0.0436 = 0.9564
\]

Therefore, the probability that the length of a randomly selected steel rod is greater than 162 cm is:

\[
P(X > 162 \text{ cm}) \approx 0.9564
\]

Final answer:

\[
\boxed{0.9564}
\]