Question

A certain standardized test has scores which range from 0 to 500 , with decimal scores possible Scors an the eam deen nemuly distributed with a mean of 321 and a standard deviation of 42 . Find the percentile P₈₂ for the scores of students taking the exam.

Answer 1

The 82nd percentile for the standardized test with a mean of 321 and a standard deviation of 42 is found at approximately 359.23, indicating that 82% of test scores are the same or lower than this score.

Step-by-step explanation:

To find the percentile P82 for a standardized test score using the given information (mean of 321 and standard deviation of 42), one can utilize the properties of the normal distribution. Here are the steps to determine the percentile:

1. Find the z-score corresponding to the 82nd percentile using a standard normal distribution table or calculator.
2. Convert the z-score to the actual test score using the formula: score = mean + (z-score × standard deviation).

Firstly, the z-score for P82 is typically around 0.915 as per standard normal distribution tables. Then, calculate the corresponding score using the formula,

Score = 321 + (0.915 × 42) = approximately 359.23.

Hence, the 82nd percentile score is approximately 359.23, which means that 82% of the scores are the same as or lower than 359.23.