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Answer :
Final answer:
The question asks for the 42nd percentile or P(58) of a normal distribution with mean 33.4 and standard deviation 37.6. Use the Z-score for the 42nd percentile, approximately -0.2, in the formula X = μ + Zσ to find X = 33.4 - 0.2*37.6, which is about 25.8. This answer is not among the provided options.
Explanation:
The question is asking for the value that separates the lower 42% from the upper 58% in a normal distribution. This is also known as the 42nd percentile or P(58). Given the normal distribution with mean μ = 33.4 and standard deviation σ = 37.6, you can use a standard normal distribution table or a calculator to find the Z-score associated with the 42nd percentile.
Typically, Z-scores are associated with a percentage or probability 'below' them. Therefore, the Z-score for the 42nd percentile can also be interpreted as the Z-score that separates the lower 42% from the upper 58% of the distribution.
In a standard distribution table, the Z-score for the 42nd percentile is approximately -0.2. This Z-score corresponds to a value slightly below the mean of the standard normal distribution, which is 0.
Now that you have the Z-score, you can calculate the corresponding value in the original distribution using the following formula:
X = μ + Zσ
Substitute the given values into the formula:
X = 33.4 - 0.2*37.6.
After calculating, you should find that X is approximately 25.8, which is not present among the options provided. Therefore, the correct answer appears to be missing.
Learn more about Normal Distribution here:
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