Question

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 175.6-cm and a standard deviation of 1.7-cm.

Find the probability that the length of a randomly selected steel rod is greater than 180.4-cm.
P(X > 180.4-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 180.4 cm, we need to calculate the z-score for the length of 180.4 cm using the given mean and standard deviation.

The formula for the z-score is:

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

Where:
- \(X\) is the value we are interested in (180.4 cm),
- \(\mu\) is the mean (175.6 cm),
- \(\sigma\) is the standard deviation (1.7 cm).

Now, substituting the values:

\[
z = \frac{180.4 - 175.6}{1.7} = \frac{4.8}{1.7} \approx 2.8235
\]

Next, we will find the probability \(P(X > 180.4)\). This is equivalent to finding \(P(Z > 2.8235)\) in the standard normal distribution.

Using a standard normal distribution table or a calculator, we find \(P(Z > 2.8235)\). This is equal to \(1 - P(Z < 2.8235)\).

From the standard normal distribution table or using a calculator, we find:

\[
P(Z < 2.8235) \approx 0.9975
\]

Thus,

\[
P(Z > 2.8235) = 1 - P(Z < 2.8235) \approx 1 - 0.9975 = 0.0025
\]

Finally, the probability that the length of a randomly selected steel rod is greater than 180.4 cm is approximately:

\[
P(X > 180.4) \approx 0.0025
\]

Therefore, the final answer, rounded to 4 decimal places, is:

\[
\boxed{0.0025}
\]