Measures of Variation, Study notes of Calculus

Download Measures of Variation and more Study notes Calculus in PDF only on Docsity! Measures of Variation Summary of Section 9.2 Range The difference Largest Data - Smallest Data in a Sample. Deviation from the Mean 1 Variance σ2 = s2 = ∑ x2i −nx 2 n−1 = ∑ (xi−x)2 n−1 2 Standard Deviation σ = s = √ s2 These are random variables called Sample Variance and Sample Standard Deviation. For a random variable X , µ = E (X ) is called the mean. The variance Var(X ) is σ2 = Var(X ) = E ((X − µ)2). Main Property/ Explanation for dividing by n − 1: If Xi are i.i.d with distribution X , then if you set S2 = ∑ (Xi−X )2 n−1 , its expected value is E (S2) = σ2. This is not true for the standard deviation, E (S) 6= σ. Grouped Data s = √∑ fix2M,i − nx2 n − 1 . Dan Barbasch Math 1105 Chapter 9 Week of October 2 1 / 1 Examples I Example (Range) Data 15,−3, 4, 7, 18. The smallest is −3, the largest 18 so Range = 18− (−3) = 21. Always a nonnegative number. Example (Deviation from the Mean) In the previous example, x = 15−3+4+7+18 5 = 8.2. So 15− 8.2 = 6.8, −3− 8.2 = −11.2, 4− 8.2 = −3.8, 7− 8.2 = −1.2, 18− 8.2 = 9.8. Example (Variance and Standard Deviation) s2 = 6.82+11.22+3.82+1.22+9.82 4 = 152+32+42+72+182−5·8.22 4 s = √ s2. Dan Barbasch Math 1105 Chapter 9 Week of October 2 2 / 1 Examples IV Example X 2 3 −1 1 X 2 4 9 1 1 (X − µ)2 (2− 1/4)2 (3− 1/4)2 (−1− 1/4)2 (1− 1/4)2 P(X ) 1/2 1/8 1/4 1/8 Computing the expected values is below. µ = E (X ) = (2)× (1/2) + (3)× (1/8) + (−1)× (1/4) + (1)× (1/8) = 1/4. Var(X ) =(2− 1/4)2 · (1/2) + (3− 1/4)2 · (1/8) + (−1− 1/4)2 · (1/4)+ +(1− 1/4)2 · (1/8) = 47/16. Dan Barbasch Math 1105 Chapter 9 Week of October 2 5 / 1 Normal Distribution I Definition Data are said to be normally distributed if the rate at which the frequencies fall off is proportional to the distance of the score from the mean, and to the frequencies themselves. This definition requires Calculus. We don’t assume or do Calculus in this course. We will however learn how to work with this distribution. It is very useful in that many phenomena can be modeled by this. We will see how the binomial distribution is related to the normal distribution later in the chapter. Suppose you have a random variable X , and you would like to know about its mean µ. So you perform many n independent trials, and draw a histogram. The larger the n, the closer the outcome will look like the curve f (x) = 1√ 2πσ e− (x−µ)2 2σ2 . The pictures in the text show what it looks like. The resulting probability is called N(µ, σ2), normal with mean µ and Dan Barbasch Math 1105 Chapter 9 Week of October 2 6 / 1 Normal Distribution II variance σ2. There is a precise statement called the Central Limit Theorem which says that for large n, √ n(Sn − µ) “looks” like a normal distribution N(0, σ2). it is used in practice to model large populations and “ errors”. There are many examples that can be approximated by normal distributions. Heights of people, and scores on tests are examples. This is not a finite distribution. For a random variable that is normally distributed, we write N(µ, σ2), P(X ≤ a) = the area under the normal curve from −∞ to a. This is tabulated for µ = 0 and σ = 1. The rest is computed by simple formulas involving arithmetic. Dan Barbasch Math 1105 Chapter 9 Week of October 2 7 / 1 z−value The principle is X normal N(µ, σ2) ⇐⇒ Z = X−µ σ normal N(0, 1). So P(X ≤ a) = P(Z ≤ a− µ σ ). z = a−µ σ is called the z−value. This is what you look up in the tables. Dan Barbasch Math 1105 Chapter 9 Week of October 2 10 / 1 Example with Grades Example A professor (not this one!) of a course wants to give grades so that A top 8% F bottom 8% B next 20% below A D next 20% above the F C the rest The mean is µ = 67 and the standard deviation is σ = 17. Find the cutoffs. Answer. P(≤ A) = 0.92 z = 1.41 a = µ+ zσ = 67 + 17 · 1.41 = 91 P(≤ B) = 0.72 z = 0.58 a = µ+ zσ = 67 + 17 · 0.58 = 77 P(≤ C ) = 0.28 z = −.59 a = µ+ zσ = 67 + 17 · (−.59) = 57 P(≤ D) = 0.08 z = −1.39 a = µ+ zσ = 67 + 17 · (−1.39) = 43 from the tables. In Excel or alike you can write norminv(0.92, 67, 17) ∼= 91. Dan Barbasch Math 1105 Chapter 9 Week of October 2 11 / 1 Approximate Binomial Distribution by the Normal Dstribution I Theorem Let B = X1 + · · ·+ Xn be the binomial Distribution, coming from adding up Xi = X i.i.d. with P(X = 1) = p,P(X = 0) = 1− p. Then E (B) = np, Var(B) = np(1− p). The normal approxinmation of the binomial distribution is P(B ≤ a) ' P(Z ≤ a− np√ np(1− p) ). where Z has the normal distribution N(0, 1). In other words, the binomial distribution is approximately the normal distribution with the same mean and variance, N(np, np(1− p). Dan Barbasch Math 1105 Chapter 9 Week of October 2 12 / 1 Drug Effectiveness I Example (Drug Effectiveness, Problem 24 in 9.4) A new drug cures 80% of the patients to whom it is administered. It is given to 25 patients. Find the probabilities that among these patients, the following results occur. a. Exactly 20 are cured. b. All are cured. c. No one is cured. d. Twelve or fewer are cured. Dan Barbasch Math 1105 Chapter 9 Week of October 2 15 / 1 Drug Effectiveness II Answer. a) C (25, 20)(0.8)20(0.2)5 b) C (25, 25)(0.8)25 c) C (25, 0)(0.2)25 d) Sum over C (25, k)(0.8)k(0.2)25−k for 0 ≤ k ≤ 12. Use the interval −0.5 ≤ B(24, 0.8) ≤ 12.5.5. The mean is 25 · 0.8 = 20. The standard deviation is √ 25 · 0.8 · 0.2 = 2 P ( Z ≤ 12.5− 20 2 ) − P ( Z ≤ −0.5− 20 2 ) = = P (Z ≤ −7.5)− P(Z ≤ −20.5) You can also write P ( Z ≤ 12.5− 20 2 ) = P (Z ≤ −3.75) < 0.0002. This is not in the tables. 12.5 is more than 4 standard deviations away from the mean. Dan Barbasch Math 1105 Chapter 9 Week of October 2 16 / 1 Continuity Correction I This is an example off the web from a statistics course (Yale). Note: Because the normal approximation is not accurate for small values of n, a good rule of thumb is to use the normal approximation only if np > 10 and np(1− p) > 10. Sometimes people use 5 instead of 10. Example Consider a population of voters in a given state. The true proportion of voters who favor candidate A is equal to 0.40. Given a sample of 200 voters, what is the probability that more than half of the voters support candidate A? Dan Barbasch Math 1105 Chapter 9 Week of October 2 17 / 1