The Normal Distribution

Size: px

Start display at page:

Download "The Normal Distribution"

Erik Cameron
6 years ago
Views:

1 Chapter 7 The Normal Distribution Section 7-1 Continuous Random Variables and Density Functions Recall the definitions Let S be the sample space of a probability experiment. A random variable X is a function from the set S into a target set T. X is a discrete random variable if the target set is (a subset of) the integers. In other words, discrete random variables have values that can be counted. X is a continuous random variable if the target set is (a subset of) the real numbers. Continuous random variables occur in probability experiments whose outcomes are measured rather than counted. Examples of continuous random variables: the weight of a cat, the amount of rainfall in Seattle, the scores (out of 100) on an exam, the length of a phone call at a switchboard, the value of the Dow Jones Industrial Average, the waiting time at a subway station, the amount of garbage per household. Remarks Note that since continuous random variables have an infinite range of outcomes, it does not make sense to assign probabilities to individual outcomes. E.g., if X is the weight of a cat (in pounds), then it does not make sense to assign a value to P (X = 14) (the probability that a randomly chosen cat will weigh 14 pounds), because that would mean that the cat would have to weigh exactly pounds which cannot be measured to that precision anyway. Instead, for a continuous random variable X, we only determine the probability for an interval in which X will lie. Density Functions random variable X, if The function f(x) is called a density function for the continuous P (a <X<b) = area under the graph of y = f(x) and between x = a and x = b. In words: the probability that X is between a and b is given by the area between the x-axis and the graph of y = f(x), starting at x = a, and ending at x = b. (If you know calculus: this means P (a <X<b)= b f(x) dx.) a If X is a continuous random variable with density function f(x), then 41

2 P (a <X<b)=P (a X b) =P (a <X b) =P (a X<b), and 0 P (a <X<b) 1. In particular, the total area under the graph of the density function f(x) is1. Example 1 Let X denote the waiting time (in minutes) at a subway station. Suppose the density function of X is given by the following graph. Find each of the following: (1) The probability of waiting between 2 and 4.5 minutes: 42

3 (2) The probability of waiting less than 2 minutes. (3) The probability of waiting more than 4.5 minutes. 43

4 Normal Distributions A special continuous distribution is the normal distribution. It is used e.g. to model certain traits in homogeneous populations, e.g., the amount of garbage per household, the weight of male army recruits, the scores on a test, the blood pressure of all Americans, the lifetime of batteries, the body temperature of rats. What is homogeneous? The fish population in the picture is not homogeneous - rather, there seem to be two species of fish present. However, each species constitutes a homogeneous population. If we take one species of fish, and take samples of, e.g., the weight (in grams) of the fish, we might get the following histogram: What is normal? It appears that there is a value for the weight of the fish, 13 grams, that can be considered normal, and most of the other values are clustered around this normal value. If X is the continuous random variable that describes the weight of this species of fish, this observation leads to the assumption that X is normally distributed. 44

5 Section 7-2 Properties of the Normal Distribution Definition A random variable X is normally distributed with mean µ and standard deviation σ if it has the following density function: f(x) = 1 e (x µ)2 2σ 2 2πσ For short, we write X N(µ, σ). Graphs of the density function are bell curves. The total area under these curves is always 1. For X N(0, 1), the graph of the density function looks like this: The first parameter, the mean µ, determines the center of the graph - if X N(3, 1), we have the density function: The second parameter, the standard deviation s, is related to the width of the graph - we compare the graphs of X N(0, 1) and X N(0, 2)

6 In the distribution of the weight X of the fish we see that the assumption X N(13, 4) approximates the histogram nicely:

7 Section 7-3 The Standard Normal Distribution We now consider random variables X N(0, 1). Their distribution is called the Standard Normal Distribution. Recall that if the continuous random variable X has the density function f(x), then the probability that X is between a and b is given by the area between the x-axis and the graph of y = f(x), starting at x = a, and ending at x = b. P (a <X<b) = area under the graph of y = f(x) and between x = a and x = b. Question: If X N(0, 1), and a and b are given, what is P (a <X<b)? We note the following rough result: This means the following: About 68% of all values lie within 1 standard deviation from the mean. About 95% of all values lie within 2 standard deviations from the mean. About 99.7% of all values lie within 3 standard deviations from the mean. Note: these rules are true not only for X N(0, 1), but also for X N(µ, σ), for any choice of the parameters µ and σ. Example 1 If X N(0, 1), P ( 2 <X<2) is about 95% = If X N(0, 1), P ( 1 <X<3) is about 34.1% % % + 2.1% = 83.9% =

8 Question: but what about other probabilities? E.g., what is P ( 0.48 < X < 2.63), for X N(0, 1)? z-values and Probability Tables Suppose X N(0, 1) andz > 0. Then the probability P (0 < X < z) corresponds to the following area: z WecanlookupthevaluesofP(0<X<z) in a probability table. This is Table E in your book. Example 1 Suppose X N(0, 1); find the probability P (0 <X<2.34). P (0 <X<2.34) corresponds to the area between 0 and 2.34, under the graph of the normal distribution curve From the table, we find that P (0 <X<2.34) = Example 2 Suppose X N(0, 1); find the probability P ( 1.75 <X<0). P ( 1.75 <X<0) corresponds to the area between and 0, under the graph of the normal distribution curve

9 We note that P ( 1.75 <X<0) = P (0 <X<1.75): 1.75 Example 3 Suppose X N(0, 1); find the probability P (X >1.11). P (X >1.11) corresponds to the area to the right of We note that P (X >1.11) and P (0 <X<1.11) add up to 0.5 = 50% (the total area under the curve is 100%). So, P (X >1.11) = 0.5 P (0 <X<1.11). Now, P (0 <X<1.11) is given by the following area:

10 Example 4 Suppose X N(0, 1); find the probability P (X < 0.93). P (X < 0.93) corresponds to the area to the left of We note that P (X < 0.93) = P (X >0.93) and P (X >0.93) = 0.5 P (0 <X<0.93). We have to find P (0 <X<0.93) using Table E: Example 5 Suppose X N(0, 1); find the probability P (1 <X<2.47). Example 6 Suppose X N(0, 1); find the probability P ( 1.37 <X<1.68). 50

11 Example 7 Suppose X N(0, 1); find the probability P (X > 1.16). Practice Problems for , 7-7, 7-11, 7-17, 7-19, 7-25, 7-27, 7-29, 7-31, 7-33, 7-35, 7-37, 7-41, 7-43, 7-45,

12 Section 7-4 Applications of the Normal Distribution So far, for a normally distributed random variable X with mean µ = 0 and standard deviation σ = 1 (i.e., X N(0, 1)), we know how to find the probabilities P (0 <X<b), P (a <X< b), P (X >b), P (X <a). Question: What do we do for a normally distributed random variable X with arbitrary mean µ and standard deviation σ? Answer: If X N(µ, σ), then for the random variable Z = X µ, Z N(0, 1). So for σ example, if X N(µ, σ), and we want to find P (a <X<b), we compute the z-values and then z 1 = a µ σ and z 2 = b µ σ, P (a <X<b)=P (z 1 <Z<z 2 ). Example 1 Each month, an American household generates an average of 28 pounds of newspaper for garbage or recycling. The standard deviation is 2 pounds. Assume the random variable X describing the monthly amount of newspaper is normally distributed. Find the probability that a randomly selected household will generate (a) between 27 and 31 pounds of newspaper per month: (b) more than 30.2 pounds of newspaper per month: 52

13 Sometimes, we can use percentages to find the expected number of times a random variable will fall into a certain interval. These questions are usually of the form: If n samples are taken, how many of the samples will lie between a and b? Example 2 The AAA reports that the average time it takes to respond to an emergency call is 25 minutes. Assume the random variable describing the response time is normally distributed and has a standard deviation of 4.5 minutes. If 80 calls are randomly selected, how many would you expect to have a response time (a) between 20.5 and 29.5 minutes: Let X be the random variable describing the response time (in minutes). Then X N(25, 4.5) and (b) less than 15 minutes? Another type of problem is of the form that the probability (or percentage) is given, and cutoff values corresponding to the probability should be found. Example 3 In order to qualify for a police academy, candidates must score in the top 10% on a general abilities test. The test has a mean of 200 and a standard deviation of 20. Find the lowest possible score to qualify (i.e., the cutoff score). Assume the test scores are normally distributed. The graph of the density function of the normal distribution with µ = 200 and σ =20is given below. We have to find the value a such that the shaded area is 0.1 = 10%. 53

14 a First, we need to find the z-value z 1 such that the shaded area in the graph of the standard normal distribution is 0.1 = 10%: z_1 Example 4 For a medical study, a researcher wishes to select people in the middle 60% of the population based on systolic blood pressure. If the mean systolic blood pressure is 120 and the standard deviation is 8, find the upper and lower readings that would qualify people to participate in the study. The graph of the density function for X N(120, 8) is given below. We have to find the values a and b such that the shaded area is 0.6 = 60%. a 120 b 140 First, we need to find the z-values z 1 and z 2 such that the shaded area in the graph of the standard normal distribution is 0.6 = 60%: 54

15 z_1 z_2 Practice Problems for , 7-61, 7-71,

16 Section 7-5 The Central Limit Theorem As we have seen in section 3-2, there are two different types of means: one is the population mean µ = X 1 + X X N, N where X 1,X 2,...,X N represent data values from every member of the population, i.e. N is the population size. It may not be feasible to collect data values from the entire population. Then we take samples and obtain sample data X 1,X 2,...,X n,wheren isthesamplesize. Thesample mean is then X = X 1 + X X n. n For large n N, we expect that X µ. We call the error the sampling error for the mean. err µ (X) =X µ We can state the result of this section, the Central Limit Theorem, in two ways: Theorem 1 (Central Limit Theorem) (1) Suppose the random variable X N(µ, σ). (1a) The sampling error err µ (X) is N(0,σ/ n) distributed. (1b) The sample means X are N(µ, σ/ n) distributed. (2) Suppose the random variable has mean µ and standard deviation σ. (2a) If the sample size n is large, then the sampling error err µ (X) will be approximately N(0,σ/ n) distributed. (2b) If the sample size n is large, then the sample means X will be approximately N(µ, σ/ n) distributed. In practice, we use this theorem as follows: (1) If X N(µ, σ), and X is a sample mean, then X N ( ) σ µ, n 56

17 (2) If X is a sample mean, and the sample size is greater or equal to 30, then X N ( ) σ µ, n (approximately) Example 1 A (fair) die is cast 87 times. Find the probability that the sample mean lies between 3.5 and 3.8. We saw in Example 3 in Section 6.3 that µ =3.5 andσ =1.7. Now, σ = 1.7 =0.18 n 87 Since n =87 30, X is approximately N(3.5, 0.18)-distributed. For a =3.5, the z-value is =0,andfora =3.8, the z-value is =1.67, and P (0 <Z<1.67) = (using Table E). So, the probability that the sample mean lies between 3.5 and 3.8 is approximately 45%. Example 2 A.C. Nielsen reported that children between the ages of 2 and 5 watch an average of 25 hours of television per week. Assume the variable describing the weekly viewing time is normally distributed and the standard deviation is 3 hours. Find the following probabilities: (a) A randomly selected child between the age of 2 and 5 watches more than 26.3 hours of television per week. This is a problem just like the ones in the previous section: if X is the random variable describing the weekly viewing time, then X N(25, 3); the z-value of a =26.3 is =0.43, and P (Z >0.43) = 0.5 P (0 <Z<0.43) = = The probability that randomly selected child between the age of 2 and 5 watches more than 26.3 hours of television per week is 33.36%. (b) The sample mean of a random sample of 20 children between the age of 2 and 5 will be greater than 26.3 hours of television per week. Here, we use that since X N(25, 3), the sample mean X is normally distributed with mean 25 and standard deviation 3/ 20 = Now, the z-value of a =26.3 is =1.94, and P (Z >1.94) = 0.5 P (0 <Z<1.94) = = The probability that a sample mean of a random sample of 20 children between the age of 2 and 5 is greater than 26.3 hours of television per week is 2.62%. 57

18 Remarks In Example 2, we note the following: since we assume that the random variable X is normally distributed, the sample mean X will be again normally distributed, for any sample size. This means that we do not require the sample size to be large (i.e., n 30), as in Example 1; also, the probabilities we compute are exact probabilities, not just approximate ones as in Example 1. Example 3 The average age of a vehicle registered in the United States is 8 years, or 96 months. Suppose 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. Practice Problems for , 7-103, 7-107,

The Normal Distribution Review

The Normal Distribution Review A. Properties of the Normal Distribution Many continuous variables have distributions that are bell-shaped and are called approximately normally distributed variables. The