1 of 7 (7 Hw Score: 82.38%, 5.77 of 7 pts Section 8.2 Exercise 9 exam had a mean

Question

1 of 7 (7 Hw Score: 82.38%, 5.77 of 7 pts Section 8.2 Exercise 9 exam had a mean of 71 and a standard deviation of 16 points; the second had a mean of 82 and a standard deviation of 3 points. Anna scored a 81 The first statistics on both tests. Megan scored 88 on the first exam and 74 on the second. They both totaled 162 points on the two exams, but Anna claims that her total is bettr. Explain. The total of Anna's z-scores is © which is greater than Megan's total of (Round to one decimal place as needed.) and then click Check Answer All parts showing

Explanation / Answer

Z-score = (marks - mean)/standard deviation

Anna's Z-score in 1st test = (81-71)/16 = 5/8 = 0.625

Anna's Z-score in 2st test = (82-71)/3 = 11/3 = 3.67

Sum of Z-scores of Anna = 3.67+0.625 = 4.3

Megan's Z-score in 1st test = (88-71)/16 = 17/8 = 2.125

Megan's Z-score in 2st test = (74-71)/3 = 3/3 = 1

Sum of Z-scores of Megan= 2.125+1= 3.1

So Z-score of Anna is 4.3 which is greater than the z-score of Megan which is 3.1

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