To find the lowest possible score to qualify for the police academy by being in the top 10% of the general abilities test, we’ll need to determine the score that corresponds to the 90th percentile of a normal distribution with a mean of 200 and a standard deviation of 20.
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Identify the z-score for the 90th percentile: The 90th percentile corresponds to a z-score which we can find using a z-table or standard normal distribution calculator. From the z-table, the z-score associated with the 90th percentile is approximately 1.2816.
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Calculate the qualifying score using the z-score formula: The formula to convert a z-score to a raw score (X) in a normal distribution is: \[ X = \mu + z\sigma \] where:
- \( \mu \) (mean) = 200
- \( z \) = 1.2816
- \( \sigma \) (standard deviation) = 20
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Substituting in the values: \[ X = 200 + (1.2816)(20) \]
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Calculating: \[ X = 200 + 25.632 = 225.632 \]
Since scores are typically whole numbers, we round this up to the nearest whole number:
The lowest possible score to qualify is 226.