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Chapter 12 Elementary Statistics Applications of Normal Distribution Chapter 12 Elementary Statistics Applications of Normal Distribution ◮ Blood pressure of adults at certain age ◮ Length of time of a football game ◮ Speed of cars on a certain road ◮ Price of similar vehicles ◮ Monthly income ◮ And · · · Chapter 11 Elementary Statistics Chapter 12 Elementary Statistics Example: The filling process dispenses cookies with weight that follows the normal distribution with mean 510 grams and standard deviation 5 grams. A bag of cookies is underweight if it weighs less than 500 grams and is overweight if it weighs more than 520 grams. ◮ What is the probability that a randomly selected bag is underweight? ◮ What is the probability that a randomly selected bag is overweight? ◮ What is the probability that a randomly selected bag weighs between 505 and 520 grams? Solution: We have a normal probability distribution with µ = 510, and σ = 5. Chapter 12 Elementary Statistics Solution Continued: ◮ What is the probability that a randomly selected bag is underweight? ⇒ P(x < 500) P(x < 500) 500 | µ = 510 σ=5 Now we can use the normal distribution chart or different technology to compute the area which is the probability that we are looking for. In this case, P(x < 500) = 0.023. Chapter 12 Elementary Statistics Solution Continued: ◮ What is the probability that a randomly selected bag is overweight? ⇒ P(x > 520) P(x > 520) | µ = 510 520 σ=5 Now we can use the normal distribution chart or different technology to compute the area which is the probability that we are looking for. In this case, P(x > 520) = 0.023. Chapter 12 Elementary Statistics Solution Continued: ◮ What is the probability that a randomly selected bag weighs between 505 and 520 grams? ⇒ P(505 < x < 520) P(505 < x < 520) 505 | µ = 510 520 σ=5 Now we can use the normal distribution chart or different technology to compute the area which is the probability that we are looking for. In this case, P(505 < x < 520) = 0.8186. Chapter 11 Elementary Statistics Chapter 12 Elementary Statistics Example: The data from traffic analysts INRIX suggests that at the northbound 405, between the 10 and the 101 Freeways, over a two-week period, between 4 and 7 pm, a total distance of 10 miles, they found that traversing that stretch of the highway took 35 minutes, which suggests average speeds of 16 miles per hour. Let’s suppose that speed of cars on that segment of the freeway during those time follow a normal distribution with standard deviation 2 miles per hour. ◮ What is the probability that a randomly selected car has a speed between 15 and 20 miles per hour? ◮ Find the speed of cars that separates the top 5% from the rest of cars. Solution: We have a normal probability distribution with µ = 16, and σ = 2. Chapter 12 Elementary Statistics Solution Continued: ◮ What is the probability that a randomly selected car has a speed between 15 and 20 miles per hour?⇒ P(15 < x < 20) P(15 < x < 20) 15 | µ = 16 20 σ=2 Now we can use the normal distribution chart or different technology to compute the area which is the probability that we are looking for. In this case, P(15 < x < 20) = 0.6687. Chapter 12 Elementary Statistics Solution Continued: Find the speed of cars that separates the top 5% from the rest of cars. ⇒ x = P95 0.05 0.95 | µ = 16 P95 σ=2 Using any method, we get x = P95 ≈ 19.3, that is P(x < 19.3) = 0.95 and P(x > 19.3) = 0.05. So 5% of the cars have speed above 19.3 miles per hour. Chapter 12 Elementary Statistics The average teacher salary in California seems to always be higher than the national average. In 2009, the average teacher salary in California was $60,583, considerably higher than the national average of $49,720. Chapter 12 Elementary Statistics Example: Suppose the monthly salary of teachers with no more than 5 years teaching experience in California follows a normal distribution with the mean of $6250 and standard deviation of $320. ◮ What is the probability that a randomly selected teacher with no more than 5 years teaching experience in California makes less than $6000 or more than $6500 per month? ◮ Find a range of salaries, rounded to to the nearest ten, that separates the middle 80% from the rest. Solution: We have a normal probability distribution with µ = 6250, and σ = 320. Chapter 12 Elementary Statistics Solution Continued: ◮ What is the probability that a randomly selected teacher with no more than 5 years teaching experience in California makes less than $6000 or more than $6500 per month? ⇒ P(x < 6000 or x > 6500) P(x > 6500) P(x < 6000) 6000 | µ = 6250 6500 σ = 320 Since the total area under the normal curve is equal to 1, then P(x < 6000 or x > 6500) = 1 − P(6000 < x < 6500). Chapter 12 Elementary Statistics Solution Continued: Using any method, we get P(6000 < x < 6500) = 0.5653, therefore P(x < 6000 or x > 6500) = 1 − 0.5653 = 0.4347 Find a range of salaries, rounded to to the nearest ten, that separates the middle 80% from the rest. ⇒ P(x1 < x < x2 ) = .8 10% 10% Middle 80% x1 | µ = 6250 x2 σ = 320 Using any method, we get x1 = P10 ≈ 5840 , and x2 = P90 ≈ 6660 . Chapter 12 Elementary Statistics Estimating Binomial Distribution by Using Normal Distribution Whenever n · p ≥ 5 & n · q ≥ 5 Binomial Distribution Normal Distribution P(x = a) P(a − 0.5 < x < a + 0.5) P(x ≤ a) P(x < a + 0.5) P(x ≥ a) P(x > a − 0.5) P(a ≤ x ≤ b) P(a − 0.5 < x < b + 0.5) Chapter 12 Elementary Statistics Example: Consider a binomial probability distribution with n = 100, p = 0.5, µ = 50, σ = 5, and x is the number of successes. ◮ Show that n · p ≥ 5 and n · q ≥ 5. ◮ Find P(x = 48) ◮ Estimate P(x = 48) by using normal distribution. Solution: Since p = 0.5, then q = 1 − p = 1 − 0.5 = 0.5, ◮ Show that n · p ≥ 5 and n · q ≥ 5. ⇒ n · p = 100 · 0.5 = 50 ≥ 5 and n · q = 100 · 0.5 = 50 ≥ 5 ◮ Find P(x = 48) ⇒ P(x = 48) = binompdf(100, 0.5, 48) = 0.0735 Chapter 12 Elementary Statistics Solution Continued: ◮ Estimate P(x = 48) by using normal distribution. ⇒ P(x = 48) ≈ P(47.5 < x < 48.5) P(47.5 < x < 48.5) = normaldf(47.5, 48.5, 50, 5) = 0.0736 Example: Consider a binomial probability distribution with n = 180, p = .75, and x is the number of successes. ◮ ◮ Show that n · p ≥ 5 and n · q ≥ 5. Find P(x ≤ 145) ◮ Find µ ◮ Find σ ◮ Estimate P(x ≤ 145) by using normal distribution. Chapter 12 Elementary Statistics Solution: Since p = 0.75, then q = 1 − p = 1 − 0.75 = 0.25, ◮ Show that n · p ≥ 5 and n · q ≥ 5. ⇒ n · p = 180 · 0.75 = 135 ≥ 5 and n · q = 180 · 0.25 = 45 ≥ 5 ◮ Find P(x ≤ 145) ⇒ P(x ≤ 145) = binomcdf(180, 0.75, 145) = 0.9675 ◮ Find µ ⇒ µ = n · p = 180 · 0.75 = 135 √ √ Find σ ⇒ σ = n · p · q = 180 · 0.75 · 0.25 = 5.809 ◮ ◮ Estimate P(x ≤ 145) by using normal distribution. ⇒ P(x ≤ 145) ≈ P(x < 145.5) P(x < 145.5) = normaldf(−E 99, 145.5, 135, 5.809) = 0.965 Chapter 12 Elementary Statistics Example: A multiple-choice exam has several hundred questions, each with 4 possible answers of which only 1 is the correct answer. Suppose we randomly select 80 questions, and x is the number of correct answers by a student whose selections of answers is by sheer guesswork with no knowledge of the subject that is being tested on. ◮ Identify n, p and q. ◮ Show that n · p ≥ 5 and n · q ≥ 5. ◮ Find µ. ◮ Find σ. ◮ Use normal distribution to estimate the probability that sheer guesswork yields from 10 to 15 correct answers. Chapter 12 Elementary Statistics Solution: ◮ ◮ ◮ ◮ ◮ 1 Identify n, p and q. ⇒ n = 80, p = = 0.25, and 4 3 q = = 0.75 4 Show that n · p ≥ 5 and n · q ≥ 5. ⇒ n · p = 80 · 0.25 = 20 ≥ 5 and n · q = 80 · 0.75 = 60 ≥ 5 Find Find µ. ⇒ µ = n · p = 80 · 0.25 = 20 √ √ Find σ ⇒ σ = n · p · q = 80 · 0.25 · 0.75 = 3.873 Use normal distribution to estimate the probability that sheer guesswork yields from 10 to 15 correct answers. ⇒ P(10 ≤ x ≤ 15) ≈ P(9.5 < x < 15.5) P(9.5 < x < 15.5) = normaldf(9.5, 15.5, 20, 3.873) = 0.1193 Chapter 12 Elementary Statistics What is a Sampling Distribution? It is a probability distribution of a statistic such as sample mean x̄. What is the Central Limit Theorem? It is the conclusion of the sampling distribution of x̄ from any population with mean µ and variance σ 2 when random samples of size n are drawn from. The sampling distribution of x̄ ◮ is approximately normally distributed ◮ with mean µx̄ = µ, σ2 σ variance σx̄2 = , and standard deviation σx̄ = √ . n n ◮ Chapter 12 Elementary Statistics Example: Consider a discrete population consisting of values 2, 4, 6, 8 and 10. ◮ Find µ and σ 2 . ◮ List all possible samples of size 2 with replacement. ◮ Find the mean of each samples. ◮ Construct a table that contains the mean of each samples and the probability of each mean. ◮ Use the discrete probability distribution to compute the mean of x̄, that is µx̄ . ◮ Use the discrete probability distribution to compute the variance of x̄, that is σx̄2 . ◮ Use these results to verify the conclusion of the Central Limit Theorem. Chapter 12 Elementary Statistics Solution: ◮ ◮ Find µ and σ 2 . ⇒ µ = 6, and σ 2 = 8 List all possible samples of size 2 with replacement. 2,2 2,4 2,6 2,8 2,10 ◮ 4,2 4,4 4,6 4,8 4,10 6,2 6,4 6,6 6,8 6,10 8,2 8,4 8,6 8,8 8,10 Find the mean of each samples. 2 3 4 5 6 3 4 5 6 7 4 5 6 7 8 5 6 7 8 9 6 7 8 9 10 10,2 10,4 10,6 10,8 10,10 Chapter 12 Elementary Statistics Solution Continued: ◮ ◮ ◮ Construct a table that contains the mean of each samples and the probability of each mean. x̄ 2 3 4 5 6 7 8 9 10 P(x̄) 1 25 2 25 3 25 4 25 5 25 4 25 3 25 2 25 1 25 Use the discrete probability distribution to compute the mean of x̄, that is µx̄ . ⇒ µx̄ = 6 Use the discrete probability distribution to compute the variance of x̄, that is σx̄2 . ⇒ σx̄2 = 4 Chapter 12 Elementary Statistics Solution Continued: ◮ Use these results to verify the conclusion of the Central Limit Theorem. 8 σ2 = = 4. We can verify that µx̄ = µ = 8, and σx̄2 = n 2 Example: Use sampling distribution of x̄ when samples of size 16 are selected at random from a normally distributed population with mean 375 and variance 100. ◮ Find µx̄ . ◮ Find σx̄2 . ◮ Find σx̄ . Chapter 12 Elementary Statistics Solution: Using the Central Limit Theorem, ◮ ◮ ◮ Find µx̄ . ⇒ µx̄ = µ = 375 σ2 100 = = 6.25 n 16 q √ Find σx̄ . ⇒ σx̄ = σx̄2 = 6.25 = 2.5 Find σx̄2 . ⇒ σx̄2 = Example: Use sampling distribution of x̄ when samples of size 10 are selected at random from a normally distributed population with mean 82 and standard deviation 7.5. ◮ Find µx̄ . ◮ Find σx̄ . Chapter 12 Elementary Statistics Solution: Using the Central Limit Theorem, ◮ ◮ Find µx̄ . ⇒ µx̄ = µ = 82 σ 7.5 Find σx̄ . ⇒ σx̄ = √ = √ ≈ 2.372 . n 10 Z score & x̄ Sampling Distribution x −µ , now we can replace x with x̄, µ with µx̄ , σ x̄ − µ x̄ − µx̄ to get z = σ . σ with σx̄ , and reduce z = σx̄ √ n we know that z = Chapter 12 Elementary Statistics Example: Use sampling distribution of x̄ when samples of size 36 are selected at random from a normally distributed population with mean 6250 and standard deviation 275. ◮ Find the z score for x̄ = 6450. ◮ Find the z score for x̄ = 6200. Solution: Using the formula z = ◮ ◮ x̄ − µ σ , √ n 6450 − 6250 ≈ 4.364 275 √ 36 6200 − 6250 ≈ −1.091 Find the z score for x̄ = 5820. ⇒ z = 275 √ 36 Find the z score for x̄ = 6450. ⇒ z = Chapter 12 Elementary Statistics Example: The average life of a certain blender is 5.1 years with a standard deviation of 1.2 years. Assume the lives of these blenders are normally distributed. ◮ Find the probability that a mean life of a random sample of 9 such blenders fall between 4.5 and 5.5 years. ◮ Find the value of x̄ that separates the top 10% from the rest of the means computed from random samples of size 9. Solution: We have a normal probability distribution with µ = 5.1, σ = 1.2, and random sample of size 9. We can use the central limit theorem σ 1.2 to compute µx̄ = µ = 5.1 and σx̄ = √ = √ = 0.4 n 9 Chapter 12 Elementary Statistics Solution Continued: ◮ Find the probability that that a mean life of a random sample of 9 such blenders fall between 4.5 and 5.5 years. ⇒ P(4.5 < x̄ < 5.5) 4.5 | µx̄ = 5.1 σx̄ = 0.4 5.5 P(4.5 < x̄ < 5.5) = normaldf(4.5, 5.5, 5.1, 0.4) = 0.7745 Chapter 12 Elementary Statistics Solution Continued: ◮ Find the value of x̄ that separates the top 10% from the rest of the means computed from random samples of size 9. ⇒ P(x̄ > k) = 0.1 90% 10% | µx̄ = 5.1 σx̄ = 0.4 k x̄ = k = P90 = invNorm(0.9, 5.1, 0.4) ≈ 5.6 Chapter 12 Elementary Statistics Example: Suppose the hourly wages of all workers in a manufacturer company have a normal distribution with a mean of $15.50 and a standard deviation of $2.75. If we randomly select a sample of 10 workers from this company, find the probability that their mean hourly wages is ◮ less than $14.25. ◮ more than $16.50. Solution: We have a normal probability distribution with µ = 15.50, σ = 2.75, and random sample of size 10. We can use the central limit theorem to compute µx̄ = µ = 15.50 and σ 2.75 σx̄ = √ = √ ≈ 0.87 n 10 Chapter 12 Elementary Statistics Solution Continued: ◮ less than $14.25. ⇒ P(x̄ < 14.25) 14.25 | µx̄ = 15.50 σx̄ = 0.87 P(x̄ < 14.25) = normaldf(−E 99, 14.25, 15.5, 0.87) = 0.0754 Chapter 12 Elementary Statistics Solution Continued: ◮ more than $16.50. ⇒ P(x̄ > 16.50) P(x̄ > 16.50) | µx̄ = 15.50 σx̄ = 0.87 16.50 P(x̄ = 16.50) = normalcdf(16.50, E 99, 15.50, 0.87) ≈ 0.125 Chapter 11 Elementary Statistics