Question

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 199.7 cm and a standard deviation of 0.9 cm.

Find the probability that the length of a randomly selected steel rod is less than 199.8 cm. Enter the probability as a decimal rounded to 4 decimal places.

P(X < 199.8 cm) =

Answer 1

To find the probability that the length of a randomly selected steel rod is less than 199.8 cm, we will use the properties of the normal distribution.

Given:
- Mean (\(\mu\)) = 199.7 cm
- Standard deviation (\(\sigma\)) = 0.9 cm

We need to standardize the value using the Z-score formula:
\[
Z = \frac{X - \mu}{\sigma}
\]
where \(X\) is the value we want to standardize (199.8 cm in this case).

Calculating the Z-score:
\[
Z = \frac{199.8 - 199.7}{0.9} = \frac{0.1}{0.9} \approx 0.1111
\]

Next, we will look up the Z-score in the standard normal distribution table or use a calculator to find the probability.

Using a Z-table or calculator for \(Z = 0.1111\):
\[
P(Z < 0.1111) \approx 0.5440
\]

Hence, the probability that the length of a randomly selected steel rod is less than 199.8 cm is:
\[
P(X < 199.8 \text{ cm}) \approx 0.5440
\]

So, the final answer rounded to four decimal places is:
\[
\boxed{0.5440}
\]