Notes4-RandomVariables_S

Statistics University of Wisconsin Madison – Chelsey Green chelseygreen@wisc.edu N OTES 4: D EFINING R ANDOM V ARIABLES AND T WO C OMMON ONES Q UANTIFYING THE POSSIBLE OUTCOMES OF A STUDY Before a study is performed, we do not know with certainty what outcome will be observed. Sometimes outcomes have qualitative descriptions (E.g.: moderate, severe, or no nausea in response to a medicine) and sometimes numerical values (E.g.: length of bolt produced). By focusing on the numeric summaries of qualitative data, we can associate a numerical value with each outcome of an experiment. In these notes we will learn some theoretical tools to describe characteristics of the population we are drawing from and also the numeric outcomes of some studies. We will also apply the probability rules learned in Notes 3 and continue to develop our simulation and coding skills. R ANDOM V ARIABLE T ERMINOLOGY A random variable (RV) associates a numerical value with each outcome of a random process. It is customary to denote random variables with uppercase letters when considering all values it could take. It is called “random” because we don’t know the value observed until the experiment is completed. *You can think of the RV as the population of values that could be observed. E.g. Many measurements are taken on babies moments after birth. Examples of a few random variables of interest for newborn babies include: W=birth weight, X=Apgar score, Y=length of baby at birth, N=number of medical staff in room, G=number of hemlock seeds that germinate out of 4 A realization of the RV is the value that is observed when the experiment is performed/recording is made. Realizations are usually denoted by lower-case letters. *You can think of a sample as a collection of realizations of a RV. E.g.: w=3,321 grams is the weight of a recently born baby; x=9 is the Apgar score for a recently born baby, y1=20.0, y2=19.3, y3=19.4 are the lengths of the last 3 babies born, g=2 if only 2 of the 4 hemlock seeds germinate Weld Failures Example: According to past study of weld failures in a certain assembly, 85% of them occur in the weld metal itself, 10% occur in the base metal, and the cause is unknown in 5% of failures. Consider a scenario where 3 weld failures in this type of assembly are observed. 1 Weld Metal Failure 0.85 Base Metal Failure 0.10 Unknown Failure 0.05

Statistics University of Wisconsin Madison – Chelsey Green chelseygreen@wisc.edu Weld Failures Example a: Define a random variable X: the number of weld failures caused by the weld metal in the 3 failures observed . Discuss what we know about this random variable before and after the experiment is conducted. X: the number of weld metal cause failures in the observed. Before: After: Types of Random Variables A RV is called discrete if it has a countable number of values. If the values are arranged in order, there is a gap between each value and the next. The set of possible values may be infinite. E.g.: X: The Apgar score is on a scale from 0-10 based on skin condition, heart rate, muscle tone, breathing, and response when stimulated. Also, N: the number of nurses in the room for a procedure is a discrete random variable. A RV is called continuous if it is capable of taking an uncountable number of values in an interval. It represents some measurement on a continuous scale. E.g.: T: the daily maximum temperature in Madison, WI can be measured to any precision. Similarly for birthweight, W. D ESCRIBING THE V ALUES O F A D ISCRETE R ANDOM V ARIABLE A probability distribution of a random variable consists of the RV’s possible values along with the probabilities of realizations occurring. The descriptions of the possible values and probabilities can take the form of a probability histogram, table (discrete RV only), or formula. *probability distributions are often approximated from empirical studies Probability Mass Function (pmf) is the probability distribution for a discrete random variable and is a list of values that can be obtained, together with the probabilities of each value. *Values are mutually exclusive *Each value has a probability between 0 and 1 *Sum of the probabilities is 1 E.g.1: We can write out the probability mass function for the random variable F: number of dots on the face that lands up when rolling a fair 6-sided die. There are 6 possible outcomes: [1,2,3,4,5,6] which we assume are equally likely so each outcome has a probability of 1/6. E.g.2: Researchers recorded the Apgar scores of over 2 million newborn in a single year. The approximate probability mass function for Apgar score (X) 2 F 1 2 3 4 5 6 P(F= f) 1/6 1/6 1/6 1/6 1/6 1/6

Statistics University of Wisconsin Madison – Chelsey Green chelseygreen@wisc.edu Weld Failures c: An approximate probability histogram for X is given at right. Use it to compute the probability that in 3 weld failures, at least one of the failures will be in the weld metal, that is P ( X ≥ 1 ) . C ENTER AND S PREAD O F A D ISCRETE R ANDOM V ARIABLE The expected value or mean value of a RV X is denoted E(X) or μ X . It represents the mean of an infinite number of realizations of X (infinite number of replications of random process). Note: the mean doesn’t need to equal an observable value. *For discrete RV, we find the mean by finding the sum of the products of each value by its probability: μ X = E ( X ) = ∑ x x i ∗ P ( X = x i ) *This will be the balancing point of a histogram showing the distribution of a random variable. E.g.1: To compute the mean of F: number of dots on the face that lands up when rolling a fair 6-sided die: μ F = 1 6 ∗ 1 + 1 6 ∗ 2 + 1 6 ∗ 3 + 1 6 ∗ 4 + 1 6 ∗ 5 + 1 6 ∗ 6 = 1 6 ∗ ( 1 + 2 + 3 + 4 + 5 + 6 ) = 1 6 ∗ ( 21 ) = 21 6 = 3.5 E.g.2: To compute the mean Apgar score in this population of 2 million: μ X = 0.001 ∗ 0 + 0.006 ∗ 1 + 0.007 ∗ 2 + 0.008 ∗ 3 + 0.012 ∗ 4 + 0.02 ∗ 5 + 0.038 ∗ 6 + 0.099 ∗ 7 + 0.319 ∗ 8 + 0.437 ∗ 9 . The variance of a RV X, denoted VAR(X) or σ X 2 gives a measurement of the variability of an infinite number of realizations of X. *For discrete RV, we multiply the squared deviations - from the values of the random variable to the mean by the probability of each respective value- and then find the sum. (This is equivalent to how we computed variance for a population of values) σ X 2 = VAR ( X ) = ∑ i ( x i − μ x ) 2 ∗ P ( X = x i ) E.g. To compute the variance of F: number of dots on the face that lands up when rolling a fair 6-sided die: σ F 2 = 1 6 ∗ ( 1 − 3.5 ) 2 + 1 6 ( 2 − 3.5 ) 2 + 1 6 ( 3 − 3.5 ) 2 + 1 6 ∗ ( 4 − 3.5 ) 2 + 1 6 ∗ ( 5 − 3.5 ) 2 + 1 6 ∗ ( 6 − 3.5 ) 2 = 2.916667 4

Statistics University of Wisconsin Madison – Chelsey Green chelseygreen@wisc.edu The standard deviation of a RV X, denoted SD(X) or σ X is the standard deviation of an infinite number of realizations of X. It is the square root of the variance computed above. E.g. To compute the standard deviation of F: number of dots on the face that lands up when rolling a fair 6-sided die: σ F = √ 2.916667 = 1.707825 Weld Failures d: Calculate the mean and standard deviation of X: [the count of weld metal failures in 3 failures] using the approximate probability distribution given at right in the histogram. Check your computations in R. Interpret the value in context. (note: the probability values are rounded to 3 decimal places) Mean of X: μ X : Meaning of μ X : Variance of X: σ X 2 Standard Deviation of X: σ X : Meaning of σ X : 5

Statistics University of Wisconsin Madison – Chelsey Green chelseygreen@wisc.edu *Mean of X: μ X = nπ and Variance of X: σ X 2 = nπ ( 1 − π ) Weld Failures Example e: Consider X: the number of weld metal failures observed in 3 weld failures. Check the assumptions that need to be made so that X is reasonably approximated by a binomial random variable . Is X counting the number of successes in a fixed number of trials? Are they Bernoulli Trials? 1. Success? 2. P(Success) the same for each trial? 3. Are trials independent? Weld Failures Example f: Use the binomial formula and R to calculate P(X=2) and explain what that value means. Check your computations with R (Revisit parts b & c) Weld Failures Example g: Use R to calculate P ( X ≥ 1 ) and explain what that value means. Check your computations for parts (b and c) with R. Weld Failures Example h: Calculate the mean, variance, and standard deviation of X using the binomial shortcuts and compare these values to what we computed in Weld Failures Example d. Mean of X: μ X = ¿ Variance of X: σ X 2 = ¿ and Standard Deviation of X: σ X : 7 Weld Metal Failure 0.85 Base Metal Failure 0.10 Unknown Failure 0.05

Statistics University of Wisconsin Madison – Chelsey Green chelseygreen@wisc.edu Weld Failures Example i: Consider the random variable Y: The number of base metal failures in 20 weld failures. Is Y a binomial random variable? If so, calculate P(Y=2) using the formula and also using the R function. If not, explain why not. Weld Failures Example j: Consider the random variable U: The number of weld failures that occur before a base metal failure is observed. Is U a binomial random variable? If so, calculate P(U=2) using the formula and also using the R function. If not, explain why not. Type 0 Blood Example b: Suppose a set of parents has 4 children. Let X be the number of children born with type O blood. Calculate the probability of each possible value of X assuming a binomial model for X~Bin(4, 0.25) is reasonable and create a histogram of its distribution. Describe the shape, center, and spread of the distribution of X. *The probability that no children get type O blood: P(X=0)= ( 4 0 ) 0.25 0 ∗ 0.75 4 = 0.3164062 *The probability that one child gets type O blood: P(X=1)= ( 4 1 ) 0.25 1 ∗ 0.75 3 = 0.421875 *The probability that two children get type O blood: P(X=2)= ( 4 2 ) 0.25 2 ∗ 0.75 2 = 0.2109375 *The probability that three children get type O blood: P(X=3)= ( 4 3 ) 0.25 3 ∗ 0.75 1 = 0.046875 *The probability that all four children get type O blood: P(X=4)= ( 4 4 ) 0.25 4 ∗ 0.75 0 = 0.00390625 8

Statistics University of Wisconsin Madison – Chelsey Green chelseygreen@wisc.edu birthweights on any interval. We can use the area under the density curve between W=3 and W=4 to approximate the probability of a baby being born with birth weight between 3 and 4 kg. The expected value or mean value of a continuous RV X is denoted E(X) or μ X . It represents the mean of an infinite number of realizations of X (infinite number of replications of random process). *For generic continuous RVs, we need to use integration which we will not be doing in this class *This will be the balancing point of a histogram showing the distribution of a random variable. The variance and standard deviation of a continuous RV X, denoted VAR(X)= σ X 2 and SD(X) = σ X , respectively, gives a measurement of the variability of an infinite number of realizations of X. *For generic continuous RVs, we again need to use integration T RANSFORMING R ANDOM V ARIABLES Linear Transformations of RV: Suppose we have constants a, b, c and random variable X with E ( X ) = μ X and Var ( X ) = σ X 2 . We can define new random variables by transforming X by adding/subtracting and multiplying/dividing by a constant [a linear transformation] . The properties of the new random variables are described below: *Let Y = X + c , E ( Y ) = E ( X + c ) = E ( X ) + c = μ X + c and Var ( Y ) = Var ( X + c ) = Var ( X ) = σ X 2 ; SD ( Y ) = SD ( X ) = σ X *Let P = aX , E ( P ) = E ( aX ) = a ∗ E ( X ) = aμ X and Var ( P ) = Var ( aX ) = a 2 Var ( X ) = a 2 σ X 2 ; SD ( P ) = ¿ a ∨ SD ( X ) = ¿ a ∨ σ X *Let L = aX + c , E ( L ) = E ( aX + c ) = aE ( X ) + c = aμ X + c and Var ( L ) = Var ( aX + c ) = a 2 Var ( X ) = a 2 σ X 2 ; SD ( L ) = ¿ a ∨ SD ( X ) = ¿ a ∨ σ X Discrete E.g. A [hypothetical] college considers a student to be full-time if they are taking between 12 and 18 units in a semester. The number of credits C taken by a randomly selected full-time student at this college is given by the following distribution and is based on current enrollment . Confirm that the number of credits taken is on average: μ C = 14.73 with standard deviation σ C = 1.933 : C: 12 13 14 15 16 17 18 P(C=c) 0.21 0.08 0.07 0.37 0.10 0.02 0.15 10

Statistics University of Wisconsin Madison – Chelsey Green chelseygreen@wisc.edu If each credit is $75 and each full time student is assessed a student fee of $100 per semester, a probability distribution for T=75C+100 were T is the semester cost would be as given below: T: 1000 1075 1150 1225 1300 1375 1450 P(T=t) 0.21 0.08 0.07 0.37 0.10 0.02 0.15 We can compute that the average total cost μ T = 75 ∗ μ C + 100 = 75 ∗ 14.73 + 100 = 1204.75 with standard deviation: σ T = 75 ∗ σ C = 75 ∗ 1.933 = 144.975 . These values match what we would find calculating mean and standard deviation from the T probability distribution. Ferry Example: A [hypothetical] small car ferry runs every hour from one side of a river to the other. Based on historical records, the number of vehicles V on a randomly chosen ferry trip has the probability distribution shown below in the table and histogram. You can confirm that expected value is μ V = 3.87 and standard deviation is σ V = 1.286 . (rounded to 3 decimal places) Ferry Example a: The ferry charges $5 for each vehicle that makes the trip. Let C=5*V be the random variable for the amount of money collected on a randomly selected trip. Fill in the probability distribution for C and compute the expected value ( μ C ), standard deviation ( σ C ), and variance ( σ C 2 ), for C. μ C = ¿ σ C 2 = ¿ σ C = ¿ Ferry Example b: The ferry’s expenses are $20 per trip and they charge each car $5 for the trip. Let the random variable P=5*V-20 be the profit made by the ferry company on a randomly selected trip. Fill in the probability distribution for P and compute the expected value ( μ P ) and standard deviation ( σ P ) for P. 11 # of Vehicles, V: 0 1 2 3 4 5 Probability: 0.0 2 0.0 5 0.0 8 0.1 6 0.2 7 0.4 2 Collected, C: 1*5= 5 3*5=1 5 4*5=2 0 5*5=2 5 Probability: 0.02 0.05 0.16 0.27 0.42 Profit, P: -15 -5 5*4- 20= 0 5*5- 20= 5 Probability: 0.02 0.05 0.16 0.27 0.42

Statistics University of Wisconsin Madison – Chelsey Green chelseygreen@wisc.edu A RV X is called a Normal (Gaussian) RV if it has probability density function: f ( x ) = 1 √ 2 π σ 2 e − ( x − μ ) 2 2 σ 2 . *A Normal RV X has a bell-shaped distribution symmetric about mean μ X with st. dev σ X , and these properties are specified in the notation: X N ¿ ) *Within 1, 2, and 3 standard deviations of the mean, there is ~68%, ~95%, and ~99.7% of data, respectively. P ( μ − σ < X < μ + σ ) = .683 P ( μ − 2 σ < X < μ + 2 σ ) = .954 P ( μ − 3 σ < X < μ + 3 σ ) = .997 *Total area under the curve of f(x) from − ∞ < X < ∞ is 1. * If a RV X is distributed X N ( μ,σ 2 ) , the standardized variable Z = X − μ σ has the standard normal distribution Z N ( 0,1 ) . *If a RVs X is normally distributed, then any linear transformation of X is also normally distributed. *The area under the density function between two possible realizations gives the probability the RV will realize to a value in that range and is computed in R. We need advanced computation methods to compute this area. *Some common R functions for normally distributed random variable, X: pnorm(): compute the area below [ or above] a specified value x qnorm(): compute the specific value x with a desired amount of area below [or above] rnorm(): generate n random observations from a normal distribution AA Battery Voltage Example: All of the AA batteries of a certain model produced by a company had their voltage tested at production and the voltages were well approximated by a normal random variable with mean 1.6 and standard deviation 0.05. V N ( μ V = 1.6 ,σ V 2 = 0.05 2 ) . Each battery is marketed as being 1.5V . AA Battery Voltage a: Label a density histogram of V assuming a normal distribution and identify some of its characteristics . 13

Statistics University of Wisconsin Madison – Chelsey Green chelseygreen@wisc.edu AA Battery Voltage b: Draw where a 1.5V battery falls on the V distribution. Also, convert 1.5V to its standardized value (z-score) and draw it on the standard normal distribution. AA Battery Voltage c: What is the probability that a randomly chosen battery from this population has lower voltage than the advertised amount (1.5V)? AA Battery Voltage d: If you examine one AA battery off of the production line, what is the probability that it has a voltage between 1.58V and 1.64V? AA Battery Voltage e: Above what weight is the highest 1% of voltages? That is what voltage is at the 99 th percentile? What is the value’s z score? 14

Statistics University of Wisconsin Madison – Chelsey Green chelseygreen@wisc.edu Boys’ Heights Example: In the children’s growth chart that medical professionals reference, the heights of two-year old boys are nearly normally distributed with a mean of 34.5 inches and a standard deviation of 1.4 inches. We approximate the height of two-year old boys with a normally distributed random variable H N ( μ H = 34.5 ,σ H 2 = 1.4 2 ) . Boys’ Heights a: Show that the distribution Z = H − 34.5 1.4 has a Normal distribution with a mean of 0 and standard deviation of 1. Before we go through any mechanics, think of what the equation is doing- it is subtracting 34.5 from each value of H (which shifts H left by 34.5 units so the new mean is 0) and dividing all values of H by 1.4 (reducing spread). Let’s confirm with our transforming RV rules: E ( Z ) = E ( H − 34.5 1.4 ) = 1 1.4 ∗ [ E ( H ) − 34.5 ] = 1 1.4 ∗ ( 34.5 − 34.5 ) = 0 Var ( Z ) = Var ( H − 34.5 1.4 ) = 1 1.4 2 ∗ Var ( H ) = 1 1.4 2 ∗ 1.4 2 = 1 Boys’ Heights b: What is the probability that a randomly chosen two-year old boy has a height less than 33 in? We can convert the height to standard units P ( H < 33 ) = P ( Z < 33 − 34.5 1.4 ) = P ( Z ← 1.071429 ) . This height is more than 1 standard deviation below the mean height. R by default assumes we are plugging in the standardized score of the value of interest: pnorm(-1.071429)= 0.1419883 or we can specify the original value, mean, and sd: pnorm(33, mean=34.5, sd=1.4)= 0.1419884. There is a probability of ≈ .142 since ≈ 14.2% of heights (values of H) are below 33 inches. We say the height of 33 is around the 14th percentile of H. Boys’ Heights c: Approximately what percent of two-year old boys have heights between 33.1 and 35.9? We can convert to standard units: P ( 33.1 < H < 35.9 ) = P ( 33.1 − 34.5 1.4 < Z < 35.9 − 34.5 1.4 ) = P ( − 1 < Z < 1 ) to see we are looking for the percent of values of H within one standard deviation of the mean. Since H is approximately normally distributed, our rule of thumb tells us ≈ 68% of heights.We can also use R to check: pnorm(1)-pnorm(-1)= 0.6826895 16

Statistics University of Wisconsin Madison – Chelsey Green chelseygreen@wisc.edu Boys’ Heights d: What height of two-year-old boys is at the 95 th percentile? (ie, what value of H has ≈ 95% below and ≈ 5% above?) We can look for what standardized (z) score has .95 area below it with qnorm(.95, mean=0, sd=1)= 1.644854 and then convert that value to h by undoing the standardization (z score) transformation: 1.644854 = h − 34.5 1.4 , so h= 1.644854 ∗ 1.4 + 34.5 = 36.8028 inches. R will also do the conversion for us if we specify the mean and sd of H: qnorm(.95, mean=34.5, sd=1.4)= 36.8028 . 17