Molly earned a score of 940 on a national achievement test. the mean test score was 850 with a standard deviation of 100. what proportion of students had a higher score than molly? (assume that test scores are normally distributed.)

Solution: We are given:

μ=850,σ=100

We have to find P(x>940)

We need to find the z score:

z=(x-μ)/σ

=(940-850)/100

=0.9

Now we need to find P(z>0.9)

Using the standard normal table, we have:

P(z>0.9)=0.1841

Therefore 0.1841 or (18.41%) of students had a higher score than molly

1. 3.4/4

2. 0.8/7

3. 0.5 x 3

4. 0.9 x 6

5. 10.8

6. 2.35

7. 0.567

8. 9.35

9. 0.02

10. 0.06

11. 0.16

12. 0.42

13. 0.63

14. 5.6

15. 8.1

16. 9.9

17. 0.28

18. 0.48

19. 0.27

20. 0.49

21. 0.45

22. 0.72

23. 0.9

24. 0.88

25. 0.75

26. 0.16

27. 0.328

28. 0.495

29. 0.72

30. 1.068

sine

step-by-step explanation: