Math 227 Elementary Statistics Bluman 5 th edition
CHAPTER 6 The Normal Distribution 2
Objectives Identify distributions as symmetrical or skewed. Identify the properties of the normal distribution. Find the area under the standard normal distribution, given various z values. Find the probabilities for a normally distributed variable by transforming it into a standard normal variable. 3
Objectives (cont.) Find specific data values for given percentages using the standard normal distribution. Use the central limit theorem to solve problems involving sample means for large samples. Use the normal approximation to compute probabilities for a binomial variable. 4
Introduction Many continuous variables have distributions that are bell-shaped and are called approximately normally distributed variables. A normal distribution is also known as the bell curve or the Gaussian distribution. 5
Normal and Skewed Distributions The normal distribution is a continuous, bellshaped distribution of a variable. If the data values are evenly distributed about the mean, the distribution is said to be symmetrical. If the majority of the data values fall to the left or right of the mean, the distribution is said to be skewed. 6
Left Skewed Distributions When the majority of the data values fall to the right of the mean, the distribution is said to be negatively or left skewed. The mean is to the left of the median, and the mean and the median are to the left of the mode. 7
Right Skewed Distributions When the majority of the data values fall to the left of the mean, the distribution is said to be positively or right skewed. The mean falls to the right of the median and both the mean and the median fall to the right of the mode. 8
6.1 Normal Distribution I. Continuous Probability Distributions A continuous random variable is one that can theoretically take on any value on some line interval. We use to represent a probability density function. Unfortunately, does not give us the probability that the value x will be observed. To understand how a probability density function for a continuous random variable enables us to find probabilities, it is important to understand the relationship between probability and area. For the following given histogram, what is the probability that x is in between 2.5 to 5.5? A. B. Frequency Histogram Relative Frequency Histogram 5 20 4 3 2 10 1 0 0 1 2 3 4 5 6 7 8 C1 0 0 1 2 3 4 5 6 7 8 C1 9
Use the given frequency histogram to calculate P(2.5 < x < 5.5) : A: P (2.5 < x < 5.5) = (4 + 5 + 4) / (1 + 2 + 3 + 4 + 5 + 4 + 3 + 2 + 1) = 13 / 25 = 52% Use the corresponding relative frequency histogram to calculate P(2.5 < x < 5.5) : B: P(2.5 < x < 5.5) = 16% + 20% + 16% = 52% which is the same as the area of the three middle bars of the relative frequency histogram. The width of each bar is one and the height is the given percentage. For a continuous probability distribution, 1) for all values x of the random variable; 2) the total area under the graph of is 1; 3) P (a < x < b) can be approximated by the area under the graph of for a < x < b. 10
Note : P (x = a) = 0 for continuous random variables. This implies P(a x b) = P(a < x < b); P(x a) = P(x > a); and P(x a) = P(x < a). 11
II. The Normal Distribution Continuous probability distributions can assume a variety of shapes. However, the most important distribution of continuous random variables in statistics is the normal distribution, that is approximately moundshaped. Many naturally occurring random variables such as IQs, height of humans, weights, times, etc. have nearly normal distributions. 12
The mathematical equation for a normal distribution is 1 0.5 0.5 mean where e 2.718 3.14 = population mean = population standard deviation The mean is located at the center of distribution. The distribution is symmetric about its mean. 13
There is a correspondence between area and probability. Since the total area under the normal probability distribution is equal to 1, the symmetry implies that the area to the right of is 0.5 and the area to the left of is also 0.5. Large values of reduce the height of the curve and increase the spread. Small values of increase the height of the curve and reduce the spread. Almost all values of a normal random variable lie in the interval 14
III. Properties of the Normal Distribution The shape and position of the normal distribution curve depend on two parameters, the mean and the standard deviation. Each normally distributed variable has its own normal distribution curve, which depends on the values of the variable s mean and standard deviation. 15
Normal Distribution Properties The normal distribution curve is bell-shaped. The mean, median, and mode are equal and located at the center of the distribution. The normal distribution curve is unimodal (i.e., it has only one mode). The curve is symmetrical about the mean, which is equivalent to saying that its shape is the same on both sides of a vertical line passing through the center. 16
Normal Distribution Properties (cont.) The curve is continuous i.e., there are no gaps or holes. For each value of X, here is a corresponding value of Y. The curve never touches the x axis. Theoretically, no matter how far in either direction the curve extends, it never meets the x axis but it gets increasingly closer. 17
IV. The Standard Normal Distribution Since each normally distributed variable has its own mean and standard deviation, the shape and location of these curves will vary. In practical applications, one would have to have a table of areas under the curve for each variable. To simplify this, statisticians use the standard normal distribution. The standard normal distribution is a normal distribution with a mean of 0 and a standard deviation of 1. 18
Recall: z Values The z value is the number of standard deviations that a particular X value is away from the mean. The formula for finding the z value is: z value mean standard deviation or z X 19
Finding Areas Under the Standard Normal Distribution Curve Area To the Left of Any z Value Look up the z value in the table and use the area given. 0 z z 0 20
To the Right of Any z Value Look up the z value in the table to get the area. Subtract the area from 1. - z 0 21
Between Any Two z Values Look up both z values to get the areas. Subtract the smaller area from the larger area. z 0 +z 22
Between Any Two z Values Look up both z values to get the areas. Subtract the smaller area from the larger area. 0 z2 z1 23
Area Under the Curve The area under the curve is more important than the frequencies because the area corresponds to the probability! Note: In a continuous distribution, the probability of any exact Z value is 0 since area would be represented by a vertical line above the value. But vertical lines in theory have no area. So 24
Example 1 : (a) Find P ( z < 1.63) From table E 0.9484 (b) Find P (-2.48 < z < 0) 0.4934 0 1.63 P (z < 1.63) = 0.9484 From table E 2.48 0 Area for z=0? 0.5 P (-2.48 < z < 0) =0.5-0.0066 =0.4934 25
(c) Find P (-2.02 < z < 1.74) 0.0217 0.9591 From Table E -2.02 0.0217 1.74 0.9591 2.02 0 1.74 P (-2.02 < z < 1.74) = 0. 9591-0.0217 = 0.9374
(d) Find the probability that z is larger than 1.76. 0.9608 1 From Table E 1.76 0.9608 + 1 0 1.76 P (z >1.76) = 1 0.9608 = 0.0392 27
Example 2 : Assume the standard normal distribution. Fill in the blanks. (a) P (0 < z < 1.46 ) = 0.4279 0.4279 Add 0.5 to the given area of 0.4279 to get the cumulative area of 0.9279. 0 z? z = 1.46 (b) P (0 < z < 3.09 ) = 0.4997 0.4997 Add 0.5 to the given area of 0.4997 to get the cumulative area of 0.9997. 0 z? z 3.09 28
(c) P ( -1.53 < z < 0) = 0.4370 05-.4370=0.063 0.4370 z? 0 z = -1.53 because the z-value is to the left of the mean. (d) P (z < 2.16 ) = 0.9846 0.5 0.9846 From Table E 0 z? z = 2.16 29
(e) Find the z value to the left of the mean so that 71.90% of the area under the distribution curve lies to the right of it. 71.90% = 0.7190 0.7190 1 0.7190 = 0.2190 From Table E z? 0 0.2190-0.58 z = -0.58 30
(f) Find two z values, one positive and one negative, so that the areas in the two tails total to 12% 0.12 2 = 0.06 (one tailed area) From Table E 0 z = ±1.555 31
6.2 Applications of the Normal I. Calculating Probabilities for a Non-Standard Normal Distribution Consider a normal variable x with mean and standard deviation. 1. Standardize from x to z. Distribution 2. Use Table E to find the central area corresponding to z. 3. Adjust the area to answer the question. 32
Example 1 : Let x be a normal random variable with mean 80 and standard deviation 12. What percentage of values are (a) larger than 56? P (x > 56) Standardize from x to z: P (x > 56) = P (z > -2) 0.0228 0.9772 From Table E -2 0.0228 + 01 P (z > -2) = 1-0.0228 z 2 0 = 0.9772 33
(b) less than 62? P (x < 62) Standardize from x to z: P (x < 62) = P (z < -1.5) 0.0668 From Table E -1.5 0.0668 z 1.5 0 34
(c) Between 85 and 98? P (85 x 98) P (0.42 z 1.5) 0.6628 0.9332 From Table E 1.5 0.9332 0.42 0.6628 0 0. 42 1.5 P (0.42 z 1.5) = 0.9332 0.6628 = 0.2704 35
(d) outside of 1.5 standard deviations of the mean What is outside of 1.5 standard deviation of the mean? 0.0688 1. 5 From Table E -1.5 0.0668 1.5 0 1. 5 P (-1.5 < z < 1.5) =2 0.0668 = 0.1336 36
Example 2 : (Ref: General Statistics by Chase/Bown, 4 th ed.) The length of times it takes for a ferry to reach a summer resort from the mainland is approximately normally distributed with mean 2 hours and standard deviation of 12 minutes. Over many past trips, what proportion of times has the ferry reached the island in (a) less than 1 hour 45 minutes? P (z < -1.25) From Table E 0.1056-1.25 0.1056 1.25 0 37
(b) more than 2 hours, 5 minutes? P (z > 0.42) 0.1628 0. 5 0 0. 42 From Table E 0.42 0.6628 + 1 P (z > 0.42) =1-0.6628 = 0.3372 38
(c) between 1 hour, 50 minutes and 2 hours, 20 minutes? P (110 x 140) P (-0.83 z 1.67) 0.2967 0.9525 From Table E -0.83 0.2967 1.67 0.9525 0.83 0 1.67 P (-0.83 z 1.67) = 0.9525-0.2967 = 0.7492 39
II. Calculating a Cutoff Value Backward steps for calculating probabilities of a non-standard normal distribution. 1. Adjust to the corresponding central area. 2. Use Table E to find the corresponding z cutoff value. 3. Non-standardize from z to x. 40
Example 1 : Employees of a company are given a test that is distributed normally with mean 100 and variance 25. The top 5% will be awarded top positions with the company. What score is necessary to get one of the top positions? 2 Normal distribution, 100, 25 5 1 0.05 = 0.95 0 0.95 0.05 z? cutoff From Table E 0.95 1.645 z = 1.645 Non-standardize 41
Example 2 : Quiz scores were normally distributed with = 14 and = 2.8, the lower 20% should receive tutorial service. Find the cutoff score. Normal distribution, 14, 2. 8 0.2 From Table E 0.2 0.-84 z? 0 z = -0.84 Non-standardize 42
Section 6 3 The Central Limit Theorem I. Sampling Distribution of Sample Mean Example 1 : Population Distribution Table (a) Find the population mean and population standard deviations of the population distribution table. 43
(b) Construct a probability histogram for x P(x) 1 4 2 4 6 8 x 44
Example 2 : From the population distribution of example 1, 2 random variables are randomly selected. (a) List out all possible combinations (sample place) and for each combination. 2 4 6 8 2 4 6 8 2 4 6 8 2 4 6 8 2 4 6 8 45
(b) Construct a probability distribution table for. (c) Construct a probability histogram for. P(x) 4/16 3/16 2/16 1/16 1 2 3 4 5 6 7 8 x 46
(d) Find the mean of sampling distribution of. 47
(e) Find the standard deviation of the sampling distribution of. 48
(f) Compare with From (a) of Example 1, From (d) of Example 2, This shows that (g) Compare with From (a) of Example 1, From (e) of Example 2, This shows that ; however 49
Population parameter Sample statistics Mean Standard deviation Population Distribution Sampling Distribution P(x) P(x) 4/10 3/10 1 4 2 4 6 8 x 2/10 1/10 1 2 3 4 5 6 7 8 x 50
II. Central Limit Theorem If the population distribution is normally distributed, the sampling distribution of will be normally distributed for any size of n. P(x) P(x) x x If the population distribution is not normally distributed, the sampling distribution x of will be normally distributed for any size of n 30. P(x) P(x) x x 51
Example 1 : Population distribution P(x) Given : x (a) Find and for n = 4 (b) Is the sampling distribution normally distributed? According to central limit theory, will NOT be normally distributed because the population distribution is NOT normally distributed and n is NOT greater than 30. (c) If n is changed from 4 to 36, is the sampling distribution Yes, because n is greater than or equal to 30. normal distributed? 52
Example 2 : (Ref: General Statistics by Chase/Bown 4 th ed.) A population has mean 325 and variance 144. Suppose the distribution of sample mean is generated by random samples of size 36. (a) Find and (b) Find Recall : Standardize Now use : 0.5 0.3413 1 0 53
(c) Find 0.0228 0.8413 2 0 1 54
Example 3 : The average number of days spent in a North Carolina hospital for a coronary bypass in 1992 was 9 days and the standard deviation was 4 days (North Carolina Medical Database Commission, Consumer s Guide to Hospitalization Charges in North Carolina Hospitals, August 1994). What is the probability that a random sample of 30 patients will have an average stay longer than 9.5 days? 9 4 0.7517 0.2483 0 0.68 55
Example 4 : Suppose the test scores for an exam are normally distributed with = 75, = 8 (a) What percentage of the students has a score greater than 85? 0.8944 0.1056 0 1.25 56
(b) What is the probability that 4 randomly selected students will have a mean score higher than 85? 0.9938 0.0062 0 2.5 57
Section 6-4 Normal Approximation to the Binomial Distribution I. When to use a Normal distribution to approximate a Binomial distribution? Recall that a binomial distribution is determined by n and p. When p is approximately 0.5, and as n increases, the shape of the binomial distribution becomes similar to the normal distribution. In order to use a normal distribution to approximate a binomial distribution, n must be sufficiently large. It is known n will be sufficiently large if np 5 and nq 5. When using a normal distribution to approximate a binomial distribution, the mean and standard deviation of the normal distribution is the same as the binomial distribution. Now recall the formulas for finding the mean and standard deviation. np, npq 58
II. Continuity Correction In addition to the condition np 5 and nq 5, a correction for continuity is used in employing a continuous distribution (Normal distribution) to approximate a discrete distribution (Binomial distribution). Warning : The continuity correction should be used only when approximating the Binomial probability with a normal probability. Don t use the continuity correction with other normal probability problems. Continuity correction x ± 0.5 59
Example 1 : Use the continuity correction to rewrite each expression : (a) Binomial Distribution Normal Distribution P (x > 6) P ( x > 6.5) (b) Binomial Distribution Normal Distribution P (x 3) P ( x 3.5) (c) Binomial Distribution Normal Distribution P (x 9) P ( x 9.5) 60
(d) Binomial Distribution Normal Distribution P (1< x < 7) P ( 1.5 < x < 6.5) (e) Binomial Distribution Normal Distribution P (5 x 10) P (4.5 x 10.5) (f) Binomial Distribution Normal Distribution P (4 < x 6) P (4.5 < x 6.5) 61
III. Using a Normal Distribution to approximate a Binomial Distribution Step 1 : Check whether the normal distribution can be used. ( np 5 and nq 5) Step 2 : Find the mean and standard deviation. Step 3 : Write the problem in probability notation, using x. Step 4 : Step 5 : Rewrite the problem by using the continuity correction factor. Continuity correction x ± 0.5 Find the corresponding z value(s) Step 6 : Use the z table to find the center area and adjust the center area to answer the question. 62
Example 1 : (Ref: General Statistics by Chase/Bown, 4 th ed.) Assume that the experiment is a binomial experiment. Find the probability of 10 or more successes, where n = 13 and p = 0.4. (a) Use the Binomial table P (x 10) = P (x = 10) + P (x = 11) + P (x = 12) + P (x = 13) = 0.006 + 0.001 + 0 + + 0 + = 0.007 (b) Use the normal approximate to the binomial Step 1 : Check : np 5 13 0.4 5 5.2 5 nq 5 13 0.6 5 7.8 5 Step 2 : Find and 63
Step 3 : Step 4 : P (x 10) Binomial Distribution P (x 10) Normal Distribution P (x > 9.5) Step 5 : Step 6 : 0.9925 0.0075 0 2.43 64
Example 2 : A dealer states that 90% of all automobiles sold have air conditioning. If the dealer sells 250 cars, find the probability that fewer than 5 of them will not have air conditioning. p = 0.10, q = 0.9 n = 250 Step 1 : Check : np 5 250 0.1 5 25 5 nq 5 250 0.9 5 225 5 Step 2 : Find and Step 3 : P (x < 5) Step 4 : Binomial Distribution Normal Distribution P (x < 5) P (x < 4.5) 65
Step 5 : Step 6 : 0.0001 4.32 0 66
Example 3 : In a corporation, 30% of the people elect to enroll in the financial investment program offered by the company. Find the probability that of 800 randomly selected people, between 260 and 300 inclusive have enrolled in the program. p = 0.3, q = 0.7 n = 800 Step 1 : Check : np 5 800 0.3 5 240 5 nq 5 800 0.7 5 560 5 Step 2 : Find and Step 3 : P (260 x 300) Step 4 : Binomial Distribution Normal Distribution P (260 x 300) P (259.5 x 300.5) 67
Step 5 : Step 6 : 0.9332 0.9999 0 1.5 4.67 68
Summary The normal distribution can be used to describe a variety of variables, such as heights, weights, and temperatures. The normal distribution is bell-shaped, unimodal, symmetric, and continuous; its mean, median, and mode are equal. Mathematicians use the standard normal distribution which has a mean of 0 and a standard deviation of 1. 69
Summary (cont.) The normal distribution can be used to describe a sampling distribution of sample means. These samples must be of the same size and randomly selected with replacement from the population. The central limit theorem states that as the size of the samples increases, the distribution of sample means will be approximately normal. 70
Summary (cont.) The normal distribution can be used to approximate other distributions, such as the binomial distribution. For the normal distribution to be used as an approximation to the binomial distribution, the conditions np 5 and nq 5 must be met. A correction for continuity may be used for more accurate results. 71
Conclusions The normal distribution can be used to approximate other distributions to simplify the data analysis for a variety of applications. 72