P4_COMPUTING PROBABILITIES USING THE STANDARD NORMAL TABLE.pptx
- 2. PROBABILITY NOTATION 𝑃 (𝑎<𝑧 <𝑏) 01 denotes the probability that the z- score is between a and b 𝑃 ( 𝑧 >𝑎) denotes the probability that the z- score is greater than a 𝑃 ( 𝑧 <𝑎) denotes the probability that the z- score is less than a
- 3. To denote the area between z = 1 and z = 2, we use the notation: = 0.1359 “the probability that the z = 1 and z = 2 is 0.1359”
- 4. CASE “greater than z” “at least z” “more than z” “to the right of z” “above z” 0 1 Example 1: Find the probability of the area above z = -1. P (z-1) = 0.5 + 0.3413 P (z-1) = 0.8413 The area that corresponds to z = -1 is 0.3413
- 5. CASE “greater than z” “at least z” “more than z” “to the right of z” “above z” 0 1 Example 1: Find the probability of the area above z = -1. P (z-1) = 0.5 + 0.3413 P (z-1) = 0.8413 Thus, the probability is 0.8413 or 84.13%.
- 6. CASE “greater than z” “at least z” “more than z” “to the right of z” “above z” 0 1 Example 2: Find the probability of the area greater than z = 1. P (z1) = 0.5 - 0.3413 P (z1) = 0. 1587 The area that corresponds to z = -1 is 0.3413
- 7. CASE “greater than z” “at least z” “more than z” “to the right of z” “above z” 0 1 Example 2: Find the probability of the area greater than z = 1. P (z1) = 0.5 - 0.3413 P (z1) = 0. 1587 Thus, the probability is 0.1587 or 15.87%.
- 8. CASE “less than z” “at most z” “no more than z” “to the left of z” “below z” 0 2 Example 3: Find the probability of the area to the left of z = -1.5. P (z-1.5) = 0.5 – 0.4332 P (z-1.5) = 0.0668 The area that corresponds to z = -1.5 is 0.4332.
- 9. CASE “less than z” “at most z” “no more than z” “to the left of z” “below z” 0 2 Example 3: Find the probability of the area to the left of z = -1.5. P (z-1.5) = 0.5 – 0.4332 P (z-1.5) = 0.0668 Thus, the probability is 0.0668 or 6.68%.
- 10. CASE “less than z” “at most z” “no more than z” “to the left of z” “below z” 0 2 Example 4: Find the probability of the area below z = 1.5. P (z-1.5) = 0.5 + 0.4332 P (z-1.5) = 0.9332 The area that corresponds to z = -1.5 is 0.4332.
- 11. CASE “less than z” “at most z” “no more than z” “to the left of z” “below z” 0 2 Example 4: Find the probability of the area below z = -1.5. P (z-1.5) = 0.5 + 0.4332 P (z-1.5) = 0.9332 Thus, the probability is 0.9332 or 93.32%.
- 12. CASE “between and ” “between and ” 0 3 Example 5: Find the probability between z = -2 and z = -1.5. P (-2 z-1.5) = 0.4772 - 0.4332 P (-2 z-1.5) = 0.0440 The area that corresponds to z = -2 is 0.4772 The area that corresponds to z = -1.5 is 0.4332
- 13. CASE “between and ” “between and ” 0 3 Example 5: Find the probability between z = -2 and z = -1.5. P (-2 z-1.5) = 0.4772 - 0.4332 P (-2 z-1.5) = 0.0440 Thus, the probability is 0.0440 or 4.40%.
- 14. CASE “between and ” “between and ” 0 3 Example 6: Find the probability between z = 0.98 and z = 2.59. P (0.98 z) = 0.4951 – 0.3365 P (0.98 z) = 0.1586 The area that corresponds to z = 0.98 is 0.3365 The area that corresponds to z = 2.59 is 0.4951
- 15. CASE “between and ” “between and ” 0 3 Example 6: Find the probability between z = 0.98 and z = 2.59. P (0.98 z) = 0.4952 – 0.3365 P (0.98 z) = 0.1587 Thus, the probability is 0.1587 or 15.87%.
- 16. CASE “between and ” “between and ” 0 4 Example 7: Find the probability between z = -1.32 and z = 2.37. P (-1.32 z) = 0.4066 + 0.4911 P (-1.32 z) = 0.8977 The area that corresponds to z = -1.32 is 0.4066 The area that corresponds to z = 2.37 is 0.4911
- 17. CASE “between and ” “between and ” 0 4 Example 7: Find the probability between z = -1.32 and z = 2.37. P (-1.32 z) = 0.4066 + 0.4911 P (-1.32 z) = 0.8977 Thus, the probability is 0.8977 or 89.77%.
- 18. CASE “between and ” “between and ” 0 4 Example 8: Find the probability between z = 0.92 and z = -1.75. P (0.92 z) = 0.3212 + 0.4599 P (0.92 z) = 0.7811 The area that corresponds to z = 0.92 is 0.3212 The area that corresponds to z = -1.75 is 0.4599
- 19. CASE “between and ” “between and ” 0 4 Example 8: Find the probability between z = 0.92 and z = -1.75. P (0.92 z) = 0.3212 + 0.4599 P (0.92 z) = 0.7811 Thus, the probability is 0.7811 or 78.11%.
- 20. Example 9: The weights of the adults in Barangay X approaches a normal distribution with a mean of 52.9 kg and standard deviation of 7.2 kg. If an adult is chosen at random from Barangay X, what is the probability that his weight is above 48.5 kg. The area corresponding to z=-0.61 is 0.2291 Thus, the probability that the randomly selected adults weighs above 48.5 kg is 0.7291 or 72.91%
- 21. Example 10: A brisk walk at 4 miles per hour burns an average of 300 calories per hour. If the standard deviation of the distribution is 8 calories, find the probability that a person walks one hour at the rate of 4 miles per hour will burn the following calories. Assume the variable to be randomly distributed. a. more than 280 calories 0.4938 Thus, the probability is 0.9938 or 99.38%. The area corresponding to z=-2.5 is 0.4938
- 22. Example 10: A brisk walk at 4 miles per hour burns an average of 300 calories per hour. If the standard deviation of the distribution is 8 calories, find the probability that a person walks one hour at the rate of 4 miles per hour will burn the following calories. Assume the variable to be randomly distributed. b. less than 294 calories 0.2734 Thus, the probability is 0.2266 or 22.66%. The area corresponding to z=-0.75 is 0.2734
- 23. Example 10: c. between 278 and 318 calories 0.4878 Thus, the probability is 0.9848 or 98.48%. The area corresponding to z=-2.75 is 0.4970 The area corresponds to z=2.25 is 0.4878
- 24. GRACIAS!