Question

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 124.7-cm and a standard deviation of 2.2-cm.

Find the probability that the length of a randomly selected steel rod is between 119.9-cm and 122.9-cm.
P(119.9-cm < X < 122.9-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 \( P(119.9 < X < 122.9) \), where \( X \) is normally distributed with a mean \( \mu = 124.7 \text{ cm} \) and standard deviation \( \sigma = 2.2 \text{ cm} \), we will first convert the lengths 119.9 cm and 122.9 cm into z-scores.

The formula for calculating the z-score is:
\[
z = \frac{(X - \mu)}{\sigma}
\]

1. **Find the z-score for 119.9 cm**:
\[
z_1 = \frac{(119.9 - 124.7)}{2.2} = \frac{-4.8}{2.2} \approx -2.1818
\]

2. **Find the z-score for 122.9 cm**:
\[
z_2 = \frac{(122.9 - 124.7)}{2.2} = \frac{-1.8}{2.2} \approx -0.8182
\]

Next, we will find the probabilities corresponding to these z-scores using the standard normal distribution table or a calculator.

3. **Find the probability for \( z_1 \)**:
Using a standard normal distribution table or calculator:
\[
P(Z < -2.1818) \approx 0.0146
\]

4. **Find the probability for \( z_2 \)**:
Using a standard normal distribution table or calculator:
\[
P(Z < -0.8182) \approx 0.2061
\]

5. **Now, we can find the probability that \( X \) is between 119.9 cm and 122.9 cm**:
\[
P(119.9 < X < 122.9) = P(Z < -0.8182) - P(Z < -2.1818)
\]
\[
P(119.9 < X < 122.9) \approx 0.2061 - 0.0146 = 0.1915
\]

Thus, the final answer, represented as a number accurate to four decimal places, is:
\[
\boxed{0.1915}
\]