Suppose the distribution of grades ought final exam is approximately normal. The

Question

Suppose the distribution of grades ought final exam is approximately normal. The class is 72 and the standard deviation is 10. show your work) What is the z-score of a student who obtained 52 on the exam: Find a range of values (i.e. a lower bound and an upper hound for a ranch of values) would contain the middle 95% of grades on the exam: You have been told that you scored at the 97.5 percentile. What is your approximate score? Having scored at the 97.5 percentile. what is you approximate grade?

Explanation / Answer

1) from empirical formual as 82 lies above 1 standard deviation away from mean;around 16% obtained higher then that.

2)zscore =(X-mean|)/std deviation =(52-72)/10=-2

3)for middle 95% lies at 2 standard deviation away: hence values are =mean -/+2*std deviation=72-/+2*10

=52 ,92

4)as 97.5 % is at 2 standard deviation away, hence score =2

5) corresponding grade =mean +z*std deviation =72+2*10=92

please reply as it seems all of your answers are right, did not know why u have asked this

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