A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of $253.6 \mathrm{~cm}$ and a standard deviation of $2.1 \mathrm{~cm}$.
Find $P_{51}$, which is the length separating the shortest $51 \%$ rods from the longest $49 \%$.
Round your answer to at least one decimal place.
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Answer
Final Answer: The length separating the shortest 51% of rods from the longest 49% is approximately \(\boxed{253.7}\) cm.
Step 1 ：The problem is asking for the length of the steel rod that separates the shortest 51% of rods from the longest 49%. This is equivalent to finding the 51st percentile of the lengths of the steel rods.
Step 2 ：The lengths of the steel rods are normally distributed with a mean of \(253.6 \, \text{cm}\) and a standard deviation of \(2.1 \, \text{cm}\).
Step 3 ：In a normal distribution, the percentile of a value can be found using the z-score formula. The z-score is the number of standard deviations a particular value is from the mean.
Step 4 ：We need to find the length that corresponds to a z-score that cuts off the bottom 51% of the data. We can use the z-score formula to find this length. The z-score formula is: \(Z = \frac{X - \mu}{\sigma}\), where \(Z\) is the z-score, \(X\) is the value we're trying to find, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.
Step 5 ：We can rearrange the z-score formula to solve for \(X\): \(X = Z\sigma + \mu\).
Step 6 ：The z-score corresponding to the 51st percentile is approximately 0.02506890825871106.
Step 7 ：Substituting the values into the formula, we get: \(X = 0.02506890825871106 \times 2.1 + 253.6\).
Step 8 ：Solving for \(X\), we get approximately 253.65264470734328 cm.
Step 9 ：However, the problem asks for the answer to be rounded to at least one decimal place. So, we round 253.65264470734328 cm to 253.7 cm.
Step 10 ：Final Answer: The length separating the shortest 51% of rods from the longest 49% is approximately \(\boxed{253.7}\) cm.
