Could you please show your work
1) The mean on a test is 5K with a standard deviation of k-2. A students score on a test is represented by 8k-16. If the students z-scores is 2, then calculate the students actual score.
2)The results of an exam were normally distributed with a mean of 65 and a standard deviation of 12. If 330 students recieved a mark greater than 80 how many wrote the test?
3)On a test Neil scored 81 while the class mean was 72 and the standard deviation was 15.Find the number of students who Neil bettered if 380 students wrote the test.
1) The mean on a test is 5K with a standard deviation of k-2. A students score on a test is represented by 8k-16. If the students z-scores is 2, then calculate the students actual score.
2)The results of an exam were normally distributed with a mean of 65 and a standard deviation of 12. If 330 students recieved a mark greater than 80 how many wrote the test?
3)On a test Neil scored 81 while the class mean was 72 and the standard deviation was 15.Find the number of students who Neil bettered if 380 students wrote the test.
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1.
z = (x – mean)/sigma
2 = ((8k – 16) – 5k)/(k – 2)
2k – 4 = 3k – 16
12 = k
The student’s score, 8k – 16, is 80.
2.
z = (x – mean)/sigma
z = (80 – 65)/12
z = 15/12
z = 1.2
This z-score equals or exceeds the scores of 88.49% of the test takers. That leaves 11.51% to score higher, which corresponds to the 330 students. Dividing then number of students by the percentage, 330/.1151, shows 2867 students took the test.
3.
z = (x – mean)/sigma
z = (81- 72)/15
z = 0.6
This z-score equals or exceeds the scores of 72.57% of the test takers. 380*.7257 = 275.76, so Neil’s score was better than 275 of the students who also took the test.
z = (x – mean)/sigma
2 = ((8k – 16) – 5k)/(k – 2)
2k – 4 = 3k – 16
12 = k
The student’s score, 8k – 16, is 80.
2.
z = (x – mean)/sigma
z = (80 – 65)/12
z = 15/12
z = 1.2
This z-score equals or exceeds the scores of 88.49% of the test takers. That leaves 11.51% to score higher, which corresponds to the 330 students. Dividing then number of students by the percentage, 330/.1151, shows 2867 students took the test.
3.
z = (x – mean)/sigma
z = (81- 72)/15
z = 0.6
This z-score equals or exceeds the scores of 72.57% of the test takers. 380*.7257 = 275.76, so Neil’s score was better than 275 of the students who also took the test.