MA131 Lecture 9.1. = µ = 25 and σ X P ( 90 < X < 100 ) = = /// σ X

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1 The Central Limit Theorem (CLT): As the sample size n increases, the shape of the distribution of the sample means taken with replacement from the population with mean µ and standard deviation σ will approach a normal distribution. This distribution will have a mean µ and a standard deviation σ/ n. Take note of the following when using the CLT. 1. When the original variable is normally distributed, the distribution of the sample means will be normally distributed, for any sample size n. 2. When the distribution of the original variable departs from normality, a sample size of 30 or more is needed to use for the normal distribution to approximate the distribution of the sample means. Example 1 It is reported that children between 2 and 5 yrs old watch an average of 25 hrs of TV per week. Assume the variable is normally distributed and the standard deviation is 3 hrs. If 20 children between the ages of 2 and 5 are randomly selected, find the probability that the mean of the number of hrs they watch TV will be greater than 26.3 hrs. Since the variable is approximately normally distributed, µ X = µ = 25 and σ X = 3 20 = The z value is / 20 = The area between 0 and 1.94 is Therefore, P ( X > 26.3 ) = = or 2.62%. /// Example 2 The average age of a vehicle in Fiji is 8 yrs, or 96 months. Assume the standard deviation is 16 months. If a random sample of 36 vehicles is selected, find the probability that the mean of their age is between 90 and 100 months. The associated z values are / 36 = 2.25 and / 36 = By adding the appropriate areas, P ( 90 < X < 100 ) = = /// Remember and notice the difference. The formula X µ σ X = X µ σ/ n should be used to analyze and gain information about a sample mean, X. The formula X µ σ should be used to analyze and gain information about individual data value, X, from the population. c 2004, RSHavea, MaCS, USP 1 File updated: September 13, 2004

2 Example 3 The average number of pounds (lbs) of meat a person consumes a year is (or 98.9 kg). Assume that σ = 25 lbs and the distribution is approximately normal. (a) Find the probability that a person selected at random consumes less than 224 lbs/yr (101.5 kg/yr). This is a question about an individual person. The z value is The area between 0 and 0.22 is Therefore, = P (X < 224) = = /// (b) If a sample of 40 individuals is selected, find the probability that the mean of the sample will be less than 224 lbs/yr. This is a question about the mean of a sample. The z value is / 40 = The area between 0 and 1.42 is Therefore, P ( X < 224 ) = = /// In the preceding example, notice the large difference between the probabilities! This is due to the fact that the distribution of sample means is much less variable than the distribution of individual values. Normal Approximation to the Binomial Distribution Recall: Now: The binomial distribution is applied to a discrete variable. Each repetition trial of a binomial experiment results in one of two possible outcomes: a success or a failure. The probabilities of the two (possible) outcomes remain the same for each repetition of the experiment. The trials are independent. The binomial formula, which gives the probability of X successes in n trials, is ( ) n P (X) = p X q n X. X Binomial formula becomes tedious when n is large. In such cases the normal distribution can be used to approximate the binomial probability. c 2004, RSHavea, MaCS, USP 2 File updated: September 13, 2004

3 The two distributions are very close to each other when n is large and p is close to 0.5. If p = 0.5 it does not mean that the two distribution are equal because not every bell shaped curve is a normal distribution! Usually, the normal distribution is used as an approximation to the binomial distribution when np > 5 and nq > 5. A correction for continuity is correction employed when a continuous distribution is used to approximate a discrete distribution. The following table gives a summary of the normal approximation to the binomial distribution. Binomial For all cases, µ = np, σ = npq, np 5, and nq 5. Normal When finding: Use: 1. P (X = a) P (a 0.5 < X < a + 0.5) 2. P (X a) P (X > a 0.5) 3. P (X > a) P (X > a + 0.5) 4. P (X a) P (X < a + 0.5) 5. P (X < a) P (X < a 0.5) Procedure for the normal approximation to the binomial distribution. Step 1. Check: If np 5 and nq 5, then go to Step 2. Step 2. Find µ and σ. Step 3. Write the problem in probability notation using X. Step 4. Rewrite the problem using the continuity correction factor. Identify the corresponding area under the normal curve. Step 5. Find the corresponding z values. Step 6. Find the solution. Example 4 A magazine reported that 6% of American drivers read the newspaper while driving! If 300 drivers are selected at random, find the probability that exactly 25 say they read the newspaper while driving. Here p = 0.06, q = 0.94, and n = 300. Step 1. np = 18 5, nq = Step 2. µ = 18, σ = 4.11 Step 3. P (X = 25) Step 4. P ( < X < ) = P (24.5 < X < 25.5) c 2004, RSHavea, MaCS, USP 3 File updated: September 13, 2004

4 Step 5. The z values are z 1 = Step 6. From the normal distribution table, = 1.82 and z 2 = = P (X = 25) P (1.58 < z < 1.82) = = Hence, the probability that exactly 25 people read the newspaper while driving is 2.27%. /// Example 5 Of the members of a bowling league, 10% are widowed. If 200 bowling league members are selected at random, find the probability that 10 or more will be widowed. Here, p = 0.1, q = 0.9, and n = 200. Step 1. np = 20 5, nq = Step 2. µ = 20, σ = 4.24 Step 3. P (X 10) Step 4. P (X > ) = P (X > 9.5) Step 5. The z value is Step 6. Therefore, = 2.48 P (X 10) P (z 2.48) = = That is, the probability of 10 or more widowed people in a random sample of 200 bowling league members is 99.34%. /// Example 6 According to an estimate, 50% of the people in America have at least one credit card. In a random sample of 30 persons is selected, what is the probability that 19 of them will have at least one credit card? Since With n = 30, p = 0.5, X = 19, the exact probability is P (19) = 30! 19!11! (0.5)19 (0.5) 11 = np = 30(0.5) = 15 5 and nq = 30(0.5) = 15 5, we can use normal distribution as an approximation. Now, With the correction for continuity, µ = 30(0.5) = 15 and σ = 30(0.5)(0.5) = The z corresponding z values for the endpoints are Therefore, P (X = 19) P (18.5 X 19.5) = 1.28 and = P (X = 19) P (18.5 X 19.5) = P (1.28 z 1.64) = = /// c 2004, RSHavea, MaCS, USP 4 File updated: September 13, 2004

5 Confidence Intervals and Sample Size Consider the following statements. One out of 4 Fijians is currently dieting. The average good IT graduate makes $32,786 a year. 72% of Fijians have flown in commercial airlines. These above values are only estimates of the true parameters and are derived from data collected from samples. Question: How large should the sample be in order to make an accurate estimate? Confidence intervals for the mean The assignment of value(s) to a population parameter based on value of the corresponding sample statistic is called estimation. Estimation procedure involves the following steps. Step 1. Select a sample. Step 2. Collect the required information from the members of the sample. Step 3. Calculate the value of the sample statistic. Step 4. Assign value(s) to the corresponding population parameter. The value(s) assigned to a population parameter based on the value of a sample statistic is called a point estimate. The sample statistic used to estimate a population parameter is called an estimator. A Good estimator has the following three properties. 1. Unbias It should be unbiased. That is, the expected value or the mean of the estimates obtained from samples of a given size is equal to the parameter being estimated. 2. Consistent It should be consistent, as sample size increases, the value of the estimator approaches the value of the parameter estimated. 3. Relatively Efficient All the statistics that can be used to estimate a parameter, the relatively efficient estimator has the smallest variance. c 2004, RSHavea, MaCS, USP 5 File updated: September 13, 2004