Figure 1: 2πσ is said to have a normal distribution with mean µ and standard deviation σ. This is also denoted

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1 Figure 1: Math 223 Lecture Notes 4/1/04 Section 4.10 The normal distribution Recall that a continuous random variable X with probability distribution function f(x) = 1 µ)2 (x e 2σ 2πσ is said to have a normal distribution with mean µ and standard deviation σ. This is also denoted X N (µ, σ) Empirical rule This rule, also called the rule, says that if X N (µ, σ), then approximately 68% of the observed values of X lie between µ σ and µ + σ 95% of the observed values of X lie between µ 2σ and µ +2σ 99.7% of the observed values of X lie between µ 3σ and µ +3σ 1

2 Example1:IQtestscores. Scores on the Wechsler Adult Intelligence Scale (a standard "IQ test") for the 20 to 34 age group are normally distributed with µ = 110 and σ =25. (a) About what percent of people in this age group have scores above 110? (b) About what percent have scores above 160? (c) In what range do the middle 95% of all scores lie? Standard normal distribution If X N(0, 1), then X is said to have a standard normal distribution. If X N(µ, σ), then the random variable is N(0, 1).z is called the z score. z = X µ σ Example 2: If X N(0, 1), use the table on page 576 to find P (X >0.42). 2

3 Example 3: In fourteen-year-old boys, the level of blood cholesterol X is N(170, 30). Levles above 240 mg/dl may require medical attention. What percent of fourteen-year-old boys have more than 240 mg/dl of cholesterol? Example 4: How hard do locomotives pull? (From TheBasicPracticeof Statistics by Moore) An important measure of the performance of a locomotive is its "adhesion", which is the locomotive s pulling force as a multiple of its weight. The adhesion of one 4400-horsepower diesel locomotive model varies in actual use according to a normal distribution with mean µ = 0.37 and standard deviation σ =0.04. What proportion of the adhesions are between 0.40 and 0.50? Finding a value given a proportion Example 5: Scores on the SAT verbal test in recent years are approximately N(505, 110). How high must a student score in order to place in the top 10% of all students taking the SAT? 3

4 Linear combinations of random variables (not in textbook) Definition: Let X 1,X 2,...,X n be random variables (continuous or discrete). Let c 1,c 2,...,c n be constants. Then then random variable Y defined by Y = c 1 X 1 + c 2 X c n X n is a linear combination of X 1,X 2,...,X n. In this course we will only consider linear combinations of independent random variables. Example 6: Suppose that we flip a penny and a nickel. Let X 1 =0if the penny comes up tails and let X 1 =1if the penny comes up heads. Similarly, let X 2 =0if the nickel comes up tails and let X 2 =1if the nickel comes up heads. Let Y =2X 1 3X 2. (a) Find the probability distribution function for Y. (b) What is E(Y )? 4

5 (c) What is Var(Y )? (d) What is 2E(X 1 ) 3E(X 2 )? (e) What is 2 2 Var(X 1 )+( 3) 2 Var(X 2 )? In general, if X 1,X 2,...,X n are independent random variables and Y = c 1 X 1 + c 2 X c n X n, then E(Y ) = µ Y = c 1 E(X 1 )+c 2 E(X 2 )+...+ c n E(X n ) Var(Y ) = σ 2 Y = c 2 1Var(X 1 )+c 2 2Var(X 2 )+...+ c 2 nvar(x n ). In fact, if X 1,X 2,...,X n are independent and normally distributed, wecansay more: Reproductive property of the Normal Distribution: If X 1,X 2,...,X n are independent, normal random variables with E(X i )=µ i and Var(X i )=σ 2 i for i =1,...,n, and if c 1,c 2,...,c n are constants, then the random variable Y defined by Y = c 1 X 1 + c 2 X c n X n is also normally distributed with E(Y )=µ Y = c 1 µ 1 + c 2 µ c n µ n and Var(Y )=σ 2 Y = c2 1 σ2 1 + c2 2 σ c2 nσ 2 n. 5

6 Example 7: The width of a casing for a door is normally distributed with a mean of 24 inches and a standard deviation of inches. The width of a door is normally distributed with a mean of 23 and a standard deviation of inches. Assume independence. What is the probability that the width of the casing minus the width of the door exceeds 0.25 inches? 6

7 Section 4.12 Sampling distributions In this section we study the distributions of sample means. For example, we could 1. Select ten deceased Americans at random. 2. Compute the average age at death x 1 of this sample. 3. Repeat steps 1 and times, calling the average ages at death x 2, x 3,...,x 100. Now, it is known that the distribution of lifespans is skewed to the right. Does this mean that if we made a histogram of x 1, x 2,...,x n, then the histogram would be skewed right? No, in fact the histogram would be roughly bell-shaped. This is what the Central Limit Theorem tells us. Roughly speaking, the Central Limit Theorem tells us that under very reasonable assumptions, sample means have an approximately normal distribution. Example 8: Use Minitab to simulate the following experiment: 1. Roll a die 20 times. Plot a histogram of the numbers. Is it bell-shaped? 2. Repeat step times (but without the histograms) 3. For each sequence of 20 rolls, compute the average number. 4. Make a histogram of the averages from step 3. Describe the shape, center and spread. So the sample averages tend to cluster around expected number on a roll of the die. The sample averages are somewhat spread out. We can decrease the spread of the sample averages by increasing the sample size. Example 9: Repeat Example 8, but this time simulate 1000 sequences of a hundred rolls. Compare the spreads from Examples 8 and 9. 7