Question 581439: A national reading test has a mean of 60.3 and a standard deviation of 7.82. The school district in smithville gave the exam to all students in the designated grade. The district average for all students was 69.2. In Miss Browns class , the average was 75.9. A)Based on the national scores what is the z-score for smithville district average score?
B) Based on the national scores, what is the z-score for the average score earned in Miss Browns class?
C)Assume that the distribution of all scores was a normal distribution. Approximately what percent of students in the country scored lower than miss browns class average?
Im not sure if I got A and B correct but my answers were 1.138 and 1.994?! I did not know C... Please help me with this!
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A national reading test has a mean of 60.3 and a standard deviation of 7.82.
The school district in smithville gave the exam to all students in the designated grade. The district average for all students was 69.2.
In Miss Browns class , the average was 75.9.
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A)Based on the national scores what is the z-score for smithville district average score?
z(69.2) = (69.2-60.3)/7.82 = 1.1381
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B) Based on the national scores, what is the z-score for the average score earned in Miss Browns class?
z(75.9) = (75.9-60.3)/7.82 = 1.9949
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C)Assume that the distribution of all scores was a normal distribution. Approximately what percent of students in the country scored lower than miss browns class average?
P(z < 1.9949) = normalcdf(-100,1.9949) = 0.9770
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Cheers,
Stan H.
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