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Find a student's test score where the class has a mean of 83.2, a standard deviation of 4.65, and a z-score of 2.43.

a) 100.8
b) 91.38
c) 93.5
d) 87.21

Answer :

Final Answer:

The student's test score is 94.5, which corresponds to the 90th percentile of the test scores.

None of the given options is answer.

Explanation:

To find the student's test score given the mean, standard deviation, and z-score, we can use the formula for calculating a raw score from a z-score: X = Mean + (Z-score * Standard Deviation )

Given:

Mean = 83.2

Standard Deviation = 4.65

Z-score = 2.43

Substitute the values into the formula:

X = 83.2 + (2.43 * 4.65 )

X = 83.2 + 11.2695

X = 94.4995

Rounded to one decimal place, the student's test score is 94.5.

None of the given options is answer.

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