To find the probability that the length of a randomly selected steel rod is between 119.9 cm and 122.9 cm, we can use the properties of the normal distribution.
Given:
- Mean (\(\mu\)) = 124.7 cm
- Standard deviation (\(\sigma\)) = 2.2 cm
We will first calculate the z-scores for the values 119.9 cm and 122.9 cm using the following formula for z-scores:
\[ z = \frac{X - \mu}{\sigma} \]
Calculating the z-score for 119.9 cm:
\[ z_{119.9} = \frac{119.9 - 124.7}{2.2} = \frac{-4.8}{2.2} \approx -2.1818 \]
Calculating the z-score for 122.9 cm:
\[ z_{122.9} = \frac{122.9 - 124.7}{2.2} = \frac{-1.8}{2.2} \approx -0.8182 \]
Next, we will find the probabilities corresponding to these z-scores using the standard normal distribution table (or a calculator).
- Find \(P(Z < -2.1818)\):
Using a z-table or calculator: \[ P(Z < -2.1818) \approx 0.0144 \]
- Find \(P(Z < -0.8182)\):
Using a z-table or calculator: \[ P(Z < -0.8182) \approx 0.2061 \]
Now, to find the probability that the length of a randomly selected steel rod is between 119.9 cm and 122.9 cm, we subtract the two probabilities:
\[ P(119.9 < X < 122.9) = P(Z < -0.8182) - P(Z < -2.1818) \] \[ P(119.9 < X < 122.9) = 0.2061 - 0.0144 = 0.1917 \]
Therefore, the probability that the length of a randomly selected steel rod is between 119.9 cm and 122.9 cm is approximately:
\[ \boxed{0.1917} \]
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