It has been reported that the average exam score in a class isnormally distribut

Question

It has been reported that the average exam score in a class isnormally distributed with a mean of 80 and a standard deviation of10. Note: For each of the following areas, draw the normal curveand highlight the corresponding areas. (a). What is the median exam score of the class? (b). What is the exam score in the first quartile (Q1)? Roundz-score to the SECOND decimal place. (c). What is the exam score in the 95th percentile? Roundz-score to the THIRD decimal place. (if you could help me solve this and describe how to draw thegraphs, it'd be greatly appreciated. will rate!) It has been reported that the average exam score in a class isnormally distributed with a mean of 80 and a standard deviation of10. Note: For each of the following areas, draw the normal curveand highlight the corresponding areas. (a). What is the median exam score of the class? (b). What is the exam score in the first quartile (Q1)? Roundz-score to the SECOND decimal place. (c). What is the exam score in the 95th percentile? Roundz-score to the THIRD decimal place. (if you could help me solve this and describe how to draw thegraphs, it'd be greatly appreciated. will rate!)

Explanation / Answer

Given X~Normal(=80, =10) (a) P(X P((X-)/ < (c-80)/10)=0.5 --> (c-80)/10 =0 (check normal table) --> c= 80 (i.e. median exam score) (b) P(Z (c-80)/10 = -0.674 (check normal table) --> c= 80-10*0.674= 73.26 (c) P(Z (c-80)/10 =1.645 (check normal table) --> c= 80 + 10*1.645 = 96.45

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