A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 230.3 cm and a standard deviation of 0.9 cm.

Find the probability that the length of a randomly selected steel rod is greater than 230 cm. *Enter the probability as a decimal rounded to 4 decimal places.*

P(X>230cm)=  

To find \( P(X > 230) \), where \( X \) is the length of the steel rod, we will use the standard normal distribution.

Given:
- Mean (\( \mu \)) = 230.3 cm
- Standard deviation (\( \sigma \)) = 0.9 cm

We need to standardize the value \( X = 230 \) to find the corresponding z-score:

\[
z = \frac{X - \mu}{\sigma} = \frac{230 - 230.3}{0.9}
\]

\[
z = \frac{-0.3}{0.9} = -0.3333
\]

Now, using the standard normal distribution table or a calculator, we find the probability for \( z = -0.3333 \).

The probability that \( Z \leq -0.3333 \) is approximately 0.3707. Therefore, the probability that \( Z > -0.3333 \) is:

\[
P(Z > -0.3333) = 1 - 0.3707 = 0.6293
\]

Thus, the probability that the length of a randomly selected steel rod is greater than 230 cm is:

\[
\boxed{0.6293}
\]

Would you like additional details or have any questions?

---

1. How does standard deviation affect the probability of lengths falling within a range?
2. What if we need to find the probability of rods longer than 231 cm?
3. How would the calculation change if the mean length were slightly different?
4. What is the probability of a rod being shorter than 230 cm?
5. How could we calculate the probability for a range, like 229 cm to 231 cm?

**Tip:** In normal distribution, converting values to z-scores is helpful to standardize data for probability calculations.

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 230.3 cm and a standard deviation of 0.9 cm. Find the probability that the length of a randomly selected steel rod is greater than 230 cm. Enter the probability as a decimal rounded to 4 decimal places.

Learn how to calculate the probability of a steel rod length exceeding 230 cm using normal distribution. Given a mean of 230.3 cm and a standard deviation of 0.9 cm, discover how to use Z-scores to determine the probability. Perfect for students in Grades 10-12.

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