Central limit theorem application
- 1. Previous Lesson What are the properties of sampling distribution of the sample means?
- 2. Refresher: A. Revisiting the A National Achievement Test was given to a group of high school graduating students. The result shows that the mean score is 95 with a standard deviation of 15. Determine the standard scores corresponding to the following scores. Score Standard Score 110 100 85 70
- 3. B. Areas Under the Normal Curve Find the area under the normal curve given the following conditions. Conditions Area Between z=0.5 and z=1.5 To the left of z=0.27 To the right of z= 1.23
- 4. Answer: A. B. Score Standard Score 110 z=1 100 z=0.33 85 z=-0.67 70 z=-1.67 Conditions Area Between z=0.5 and z=1.5 0.2417 or 24.17% To the left of z=0.27 0.6064 or 60.64% To the right of z= 1.23 0.1093 or 10.93%
- 5. Central Limit Theorem PREPARED BY: AILEEN D. LOSITAÑO SHS TEACHER
- 6. True or false? 1. If the population is normal, the sample size should be greater than 30. _______ 2. The Central Limit Theorem tells us that as the sample size increases, the sampling distribution of the sample means approaches a normal distribution regardless of the shape of the population.__________ 3. The sample mean can be standardized or converted to z-score if the population is normal and any sample size can be use. _____
- 7. The Central Limit Theorem (for the sample mean x) If a random sample of n observations is selected from a population (any population), then when n is sufficiently large, the sampling distribution of x will be approximately normal. (The larger the sample size, the better will be the normal approximation to the sampling distribution of x.)
- 8. The Central Limit Theorem This theorem allows us to use a sample to make inferences about a population because it states that if n is sufficiently large, the sampling distribution will be approximately normal no matter what the population distribution looks like.
- 9. To apply the Central Limit Theorem -If the population distribution is normal, any sample size can be used. -If the population distribution is not normal, then 𝑛 ≥ 30. -The distribution can be converted to standard scores provide that sample size is large.
- 10. Using the Central Limit Theorem to Convert to the Standard Normal Distribution
- 11. Example 1: Cholesterol Level The average number of milligrams (mg) of cholesterol in a cup of a certain brand of ice cream is 660mg, and the standard deviation is 35mg. Assume the variable is normally distributed. A. If a cup of ice cream is selected, what is the probability that the cholesterol content will be more than 670mg? B. If a sample of 10 cups of ice cream is selected, what is the probability that the mean of the sample will be larger than 670mg?
- 12. Solution: Given: µ=660; σ=35; X=670 a. Find P(X>670) 𝑧 = 𝑋 − 𝜇 𝜎 = 670 − 660 35 = 0.29 Find the area under the normal curve using the z-table: The P(X>670) has the area of 0.3859 or 38.59% b. Given: µ=660; σ=35; 𝑥 = 670; n=10 𝑧 = 𝑥 − 𝜇 𝜎 𝑛 = 670 − 660 35 10 = 0.90 Using the z-table: The P( 𝑥>670) is 0.1841 or 18.41%
- 13. Example 2: Time to complete a Jingle The average time it takes a group of college students to complete a certain jingle presentation in MAPEH is 46.2 minutes. The standard deviation is 8 minutes. Assume that the variable is normally distributed. If 50 randomly selected college students create a jingle, what is the probability that the mean time it takes the group to complete the jingle will be less than 43 minutes?
- 14. Solution: Given: µ=46.2 ; σ=8 ; 𝑥=43 ;n=50 find P( 𝑥<43) 𝑧 = 𝑥−𝜇 𝜎 𝑛 = 𝑧 = 43−46.2 8 50 = −3.2 1.131 = −2.83 Using z table, the area of z=-2.83 to the left is 0.0023 or 0.23% So, the probability that 50 randomly selected college student will complete the jingle in less than 43 minutes is 0.23%.
- 15. Board Work: 1. The average cholesterol content of a certain canned goods is 215 milligrams and the standard deviation is 15 milligrams. Assume the variable is normally distributed. If a sample of 25 canned goods is selected, what is the probability that the mean of the sample will be larger than 220 milligrams? 2. The average public high school has 468 students with a standard deviation of 87. If a random sample of 38 public elementary schools is selected, what is the probability that the mean number of students enrolled is between 445 and 485?
- 16. Solution: 1. Given: µ=215; σ=15; 𝑥 =220; n=25 Find P( 𝑥 >220) 𝑧 = 𝑥 − 𝜇 𝜎 𝑛 = 220 − 215 15 25 = 1.67 2. Given: µ=468; σ=87; 𝑥 = 445 𝑎𝑛𝑑 485; n=38 Find P(445 ≤ 𝑥 ≤ 485) 𝑧1 = 𝑥 − 𝜇 𝜎 𝑛 = 445 − 468 87 38 = −23 14.113 = −1.63 𝑧2 = 𝑥 − 𝜇 𝜎 𝑛 = 485 − 468 87 38 = 17 14.113 = 1.20 Using the z-table: The PP(445 ≤ 𝑥 ≤ 485) is 0.8849-0.0516=0.8333 or 83.33% Find the area under the normal curve using the z-table: The P( 𝑥 >220) has the area of 0.0475 or 4.75%
- 17. What do you think is the importance of Central Limit Theorem?
- 19. Seatwork: Try this! 1. An electrical company claims that the average life of the bulbs it manufactures is 1200 hours with a standard deviation of 250 hours. If a random sample of 100 bulbs is chosen, what is the probability that the sample mean will be: a. Greater than 1,150 hours? b. Less than 1,250 hours? c. Between 1,150 and 1,250 hours?
- 20. Solution: Given: µ=1200; σ=250; n=100 a. Find P( 𝑥 >1,150) 𝑧 = 𝑥 − 𝜇 𝜎 𝑛 = 1,150 − 1,200 250 100 = −2 Find the area under the normal curve using the z-table: The P( 𝑥 >1,150) has the area of 0.9772 or 97.72% b. Find P( 𝑥 <1,250) 𝑧 = 𝑥 − 𝜇 𝜎 𝑛 = 1,250 − 1,200 250 100 = 2 Find the area under the normal curve using the z-table: The P( 𝑥 <1,250) has the area of 0.9772 or 97.72%
- 21. c. Find P(1,150 ≤ 𝑥 ≤1,250) (Answer in A and B) The P( 𝑥 >1,150) has the area of 0.9772 or 97.72% The P( 𝑥 <1,250) has the area of 0.9772 or 97.72% Solution: .9972-.9972= 0 P(1,150 ≤ 𝑥 ≤1,250)=0 %
- 22. Evaluation: 1. Manufacturer of ball bearings claims that this product has a mean weight 502g and standard deviation of 30g. What is the probability that a random sample of 100 ball bearings will have combined weight: a. Between 496g and 500g? b. More than 510g?
- 23. Answer: Given: µ=502; σ=30; n=100 a. Find P(496 ≤ 𝑥 ≤ 500) 𝑧 = 𝑥 − 𝜇 𝜎 𝑛 = 496 − 502 30 100 = −2 𝑧 = 𝑥 − 𝜇 𝜎 𝑛 = 500 − 502 30 100 = −0.67 Find the area under the normal curve using the z-table: The 0.2514 – 0.0228=0.2286 or 22.86%
- 24. Thank you and God bless!!!