Step 1 :The mean score for the California Peace Officer Standards and Training test is \( \mu = 50 \)
Step 2 :The standard deviation of the scores is \( \sigma = 9 \)
Step 3 :The agency will only hire applicants with scores in the top 5%, which corresponds to the \( 95^{th} \) percentile
Step 4 :To find the lowest score for the top 5%, we need to find the z-score that corresponds to the \( 95^{th} \) percentile
Step 5 :The z-score for the \( 95^{th} \) percentile is \( z = 1.6448536269514722 \)
Step 6 :We use the z-score formula to find the lowest score: \( X = \mu + z \cdot \sigma \)
Step 7 :Substitute the known values into the formula: \( X = 50 + 1.6448536269514722 \cdot 9 \)
Step 8 :Calculate the lowest score: \( X = 64.80368264256325 \)
Step 9 :Round the lowest score to the nearest whole number: \( X = 65 \)
Step 10 :The final answer is \(\boxed{65}\)