Mauricio scored 84 on an exam (with a mean of 77 and standard deviation of 8) in
ID: 3177761 • Letter: M
Question
Mauricio scored 84 on an exam (with a mean of 77 and standard deviation of 8) in his psychology class at CSULA. Olivia scored 80 on an exam (with mean of 75 and standard deviation of 4) in her psychology class at CSUF. Each class is graded "on a curve", so the grades are normally distributed. (# 1) What is the z - score for Mauricio ? What is his letter grade ? (# 2) What is the z - score for Olivia ? What is her letter grade ? (# 3) Using his exam score of 84, what z - score and letter grade would have been assigned to Mauricio had he taken the exam in the psychology class at CSUF with Olivia ? (# 4) Using her exam score of 80, what z - score and letter grade would have been assigned to Olivia had she taken the exam in the psychology class at CSULA with Mauricio ? Determine the z - scores (standard normal distribution) for the following middle intervals: (#1) 90% (#2) 95% (#3) 99% (# 4) U. S. adults consume annually an average of 119 pounds of beef, with a standard deviation of 16 pounds. APPLY the z - scores in # 5, # 6, and # 7 above. SHOW your work. INTERPRET the intervals.Explanation / Answer
as zscore =(X-mean)/std deviation
zscore for Mauricio =(84-77)/8=0.875
zscore for olivia =(80-75)/4=1.25
3) Z score corresponding to 84 in CSUF =(84-75)/4=2.25
=4)z score =(80-77)/8=0.375
z scores :
90%=1.28155
95% =1.64485
99% =2.3263
4)as intercval =mean +/- z*std deviation
hence 90% interval =98.495 ; 139.505
95% interval =92.682 ; 145.318
99% Interval =81.778 ; 156.222