Given a standardized normal distribution (with a mean of O and a standard deviation of 1), determine the following probabilities. a. P(Z > 1.09) b. P(Z < -0.22) c. P(-1.96 d. What is the value of Z if only 8.08% of all possible Z-values are larger?

a) P(Z > 1.09) = 0.1379

b) P(Z < -0.22) = 0.4129.

c) P(Z < -1.96) = 0.025, P(Z > -1.96) = 0.025

d) Z = 1.4

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean and standard deviation , the z-score of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

a. P(Z > 1.09)

This is 1 subtracted by the pvalue of Z = 1.09.

Z = 1.09 has a pvalue of 0.8621

1 - 0.8621 = 0.1379, so

P(Z > 1.09) = 0.1379

b. P(Z < -0.22)

This is the pvalue of Z = -0.22.

Z = -0.22 has a pvalue of 0.4129. So

P(Z < -0.22) = 0.4129.

c) P(Z < -1.96) or P(Z > -1.96)

Z = -1.96 has a pvalue of 0.025. So

1 - 0.025 = 0.975

P(Z < -1.96) = 0.025

P(Z > -1.96) = 0.025

d. What is the value of Z if only 8.08% of all possible Z-values are larger?

Z with a pvalue of 100 - 8.08 = 91.92 = 0.9192, so Z = 1.4

ab: 2 n83b8eb7ebi. 8e yen in

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