Statistics Presentation week 6
- 1. STATISTICS & PROBABILITY Normal Distribution
- 2. Normal Distribution X
- 3. µ=85 Mo =85 ~= x 85 sx =5 0.5 0.5 X 70 75 80 85 90 95 100 The Normal Distribution 1. The mean = median = mode. 2. It is symmetrical about the mean. 3. The tails or ends are asymptotic relative to the horizontal axis. 4. The total area under the normal curve is 1.0 or 100%. 5. The normal curve area maybe subdivided into standard deviations, at least 3 to the left and 3 to the right of the mean
- 4. µ = 85 and s = 5 x −µ Fig. I z= σ X 70 75 80 85 90 95 100 The Normal Distribution µ − 3σ µ − 2σ µ − 1σ µ + 1σ µ + 2σ µ + 3σ Z -3 -2 -1 0 1 2 3 The Standard Normal Distribution
- 5. µ = 80 and s = 10 x −µ Fig. II z= σ X 50 60 70 80 90 100 110 The Normal Distribution µ − 3σ µ − 2σ µ − 1σ µ + 1σ µ + 1σ µ + 1σ Z -3 -2 -1 0 1 2 3 The Standard Normal Distribution
- 6. x−µ z= σ µ − 3σ µ − 2σ µ − 1σ µ + 1σ µ + 2σ µ + 3σ Z -3 -2 -1 0 1 2 3 The Standard Normal Distribution Properties 1. The mean is equal to zero. 2. The standard deviation is equal to one.
- 7. ALGEB ENGLISH x − µ 85 − 80 5 80 75 Z of Juan in ALGEB = = = = 1.0 σ 5 5 5 5 x − µ 82 − 75 7 85 82 Z of Juan in ENGLISH = = = = 1 .4 σ 5 5 ALGEB ENGLISH Example 1: Z -3 -2 -1 0 1 2 3 Juan’s score in ALGEB is 1 standard deviation above the mean, while his score in ENGLISH is 1.4 standard deviation above the mean Therefore, Juan performed better in English test than in ALGEB.
- 8. Example 2: A classmate of Juan in Algebra has a Z-score of 2.1. What was his score? Solution: Using the formula for Z-score, we have: Given: Z = 2.1, µ = 80, σ = 5 x−µ x − 80 Z= ; 2.1 = σ 5 x − 80 = 2.1(5) x − 80 = 10.5 x = 90.5 ≈ 91
- 9. Finding the areas under the normal curve is the same as finding the Probability. Example 1: 1. Find the area from Z = 0 to Z = 1.35 Z -3 -2 -1 0 1 2 3 Step1. Shade the required region. Step2. Find the area using the table. A = 0.4115 Area is at Z = 1.3 at column 5. P = 41.15%
- 10. Example 2: Find P(-2.0 ≤ Z ≤ 2.5) 8 A=0.493 772 0.4 A= Z -3 -2 -1 0 1 2 3 Step1. Shade the required region. Step2. Find the area using the table. A = 0.9710 Area from Z = 0 to Z = -2.0 is 0.4772 Add! P = 97.10% Area from Z = 0 to Z = 2.5 is 0.4938 0.9710
- 11. Example 3: Find P(Z ≥ -2.5) 38 0 00 49 0.5 0. Z -3 -2 -1 0 1 2 3 Step1. Shade the required region. A = 0.9938 Step2. Find the area using the table. P = 99.38% Area from Z = 0 to Z =- 2.5 is 0.4938 Area of half of + 0.5000 the normal curve A = 0.9938
- 12. Example 4: Find P(0.5 ≤ Z≤ 2.5) 0.4938 0.1915 A= 0. 30 23 Z -3 -2 -1 0 1 2 3 Step1. Shade the required region. A = 0.3023 Step2. Find the area using the table. P = 30.23% Area from Z = 0 to Z = 2.5 is 0.4938 Subtract! Area from Z = 0 to Z = 0.5 is 0.1915 A = 0.3023
- 13. Example 5: Find P(Z ≥ 0.5) 0.5000 0.1915 A =0 .30 85 Z -3 -2 -1 0 1 2 3 Step1. Shade the required region. A = 0.3085 Step2. Find the area using the table. P = 30.85% 0.5000 Subtract Area from Z = 0 to Z = 0.5 is 0.1915 A = 0.3085
- 14. Finding the Z given the areas under the Normal Curve Example 6: Find Z if the area to the right of +Z is 0.3085. A A = 0.1915 = 0. 3 08 5 Z 0 Z=? Z = 0.5 Step1. Shade the given area. Since Z is “+” it is at the right of Z=0. Since half of the Normal curve has A = 0.5000 Subtract the given area A = 0.3085 A = 0.1915
- 15. Finding the Z given the areas under the Normal Curve Example 7: Find Z if the area to the right of Z is 0.7612. A = 0.2612 A = 0.5000 0.7612 Z Z=-? 0 Step1. Shade the given area. In this case, where is Z? Step2. Find the area from Z = 0 to the unknown -Z. And the given area is 0.7612 Subtract Since half of the Normal curve has A = 0.5000 A = 0.2612 Z = -0.71
- 16. APPLICATIONS: 1. In an examination in Statistics, the mean grade is 72 and the standard deviation is 6. Find the probability that a particular student will have a score: a. higher than or equal to 75 b. from 65 to 80 c. lower than or equal to 60.
- 17. 2. The scores of 400 applicants who took the examination in a certain company approximate a normal distribution, with a mean of 75 and a standard deviation of 5. a. How many applicants got a score of 80 and above? b. How many applicants got a score of 70 and below? c. How many applicants got a score from 70 to 80? d. How many applicants got a score of 88?
- 18. 3. The mean weight of 200 male students at a certain college is 68.5 kgs. and the standard deviation is 6.8. Assuming that the weight are normally distributed, find how many students weigh: a. more than or equal to 83.9 kgs.? b. less than or equal to 58.1 kgs.? c. between 54.2 and 70.3kgs.? d. exactly 60.5 kgs.
- 19. 4. A teacher decided to fail 20% of the class. Exam marks are approximately normally distributed with a mean of 72 and a standard deviation of 6. What mark must a student get to pass? 5. The actual amount of instant coffee that a filling machine deposits into a “6-ounce” jars varies from jar to jar, and may be looked upon as a random variable having a normal distribution with a standard deviation of 0.04 ounce. If only 2% of the jars are to contain less than 6 ounces of coffee, what must be the mean fill of these jars?
- 20. 6. The sardines processed by a cannery have a mean length of 4.54 inches with a standard deviation of 0.25 inch. If the distribution of the lengths of the sardines are normally distributed, what percentage of the sardines are : a. shorter than 4.00 inches; b. between 4.40 and 4.60 inches long? c. longer than 5.3 inches 7. In reference to problem 6, find x so that only 20% of the sardines processed by the cannery are longer than x.