To find the probability that the length of a randomly selected steel rod is between 92.3 cm and 93 cm, we can use the properties of the normal distribution. We first need to convert the lengths into z-scores.
The formula for a z-score is given by:
\[ z = \frac{(X - \mu)}{\sigma} \]
where \( X \) is the value, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.
Given:
- Mean \( \mu = 93 , \text{cm} \)
- Standard deviation \( \sigma = 0.8 , \text{cm} \)
Now, we calculate the z-scores for 92.3 cm and 93 cm.
- For \( X = 92.3 , \text{cm} \):
\[ z_{92.3} = \frac{(92.3 - 93)}{0.8} = \frac{-0.7}{0.8} = -0.875 \]
- For \( X = 93 , \text{cm} \):
\[ z_{93} = \frac{(93 - 93)}{0.8} = \frac{0}{0.8} = 0 \]
Next, we will find the probabilities corresponding to these z-scores using the standard normal distribution table or a calculator with normal distribution functions.
- The probability for \( z = 0 \):
\[ P(Z < 0) = 0.5 \]
- The probability for \( z = -0.875 \):
Using a z-table or a normal distribution calculator:
\[ P(Z < -0.875) \approx 0.1894 \]
Now, we can find the probability that the length of a randomly selected steel rod is between 92.3 cm and 93 cm:
\[ P(92.3 < X < 93) = P(Z < 0) - P(Z < -0.875) = 0.5 - 0.1894 = 0.3106 \]
Thus, the final probability is:
\[ \boxed{0.3106} \]
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