4 Random Variables and Distributions

Transcription

1 4 Random Variables and Distributions Random variables A random variable assigns each outcome in a sample space. e.g. called a realization of that variable to Note: We ll usually denote a random variable by a capital letter (as above). We ll usually denote an unspecified realization (number) by a lower-case letter. e.g. P (X = x) is the probability random variable X realizes to the value x. The probability distribution of a random variable X describes the possible values X can take along with the probability of each subset of these values. A random variable represents a, while a set of its realizations represents a. e.g. Discrete random variables The possible values of a discrete random variable can be arranged in a (finite or infinite). The probability distribution for a discrete random variable X is its probability mass function (pmf) p defined by p(x) = P ( ). e.g. Let X heads = the number of heads in two tosses of a fair coin. X heads is a discrete random variable whose possible values are. The pmf of X heads is (draw) x p(x)

2 Mean and variance for discrete random variables The (population) mean (or expected value or expectation) of X is the of the possible values X can take: µ X = E(X) = x x p(x) (It s the of the distribution of X. For an X that takes on n equally-likely values, the formula is familiar: E(X) =.) e.g. µ Xheads = E(X heads ) = (draw on pmf) The (population) variance of X, denoted VAR(X) or σx 2, is the expected value of its from its mean: σ 2 X = E [ (X µ X ) 2] = x (x µ X) 2 p(x) (For an X that takes on n equally-likely values, the formula is almost familiar: σ 2 X =.) e.g. VAR(X heads ) = σ 2 X heads = The (population) standard deviation of X is σ X = σ 2 X. e.g. σ Xheads = (draw on pmf)

3 Properties of mean µ X (or expected value E(X)) and variance σx 2 (or VAR(X)) Theorem: If X and Y are random variables and c is a constant, E(c) = Optional proof: E(cX) = E(X + c) Optional proof: E(X + Y ) = VAR(c) = VAR(cX) = Optional proof: VAR(X + c) = if X and Y are independent, VAR(X + Y ) =

4 The binomial distribution Bernouli trials Bernoulli trials are repeated identical experiments in which each trial s outcome is either or, and π = P (success) is for all trials (so P (failure)= ) e.g. Toss a fair coin, with success = {head}, failure = {tail}: π =, 1 π = e.g. Roll a die, with success = {1, 2}, failure = {3, 4, 5, 6}: π =, 1 π = e.g. Randomly select a student from among the 29 Packer fans and 25 non-fans in STAT , calling the selection a success if the student is a Packer fan: π =, 1 π =. e.g. Given a simple random sample ( 5), each response to a is a Bernoulli trial. { 1, for success We can find the mean and variance of a Bernoulli trial if we define Y =. Then 0, for failure Y s probability mass function is Y 0 1 p(y ) µ Y = σ 2 Y = Binomial distribution The binomial distribution, Bin(n, π), is the distribution of the independent Bernoulli trials, each having probability of success. e.g. Let X = number of Packer fans in a random sample of 10 STAT 371 students. Then X Bin(n =, π = ). Many survey questions lead to Bin(n, π). We ll figure out the distribution of X Bin(n, π), first for an example, and then in general. in 1. e.g. Let X = #successes in rolling a die n = 3 times, with success = {1, 2}. That is, X Bin(n =, π = ). The possible outcomes are:

5 outcome P (outcome) x FFF (1 p) 3 0 FFS p(1 p) 2 1 FSF p(1 p) 2 1 FSS p 2 (1 p) 2 SSF p 2 (1 p) 2 SSS p 3 3 Reorganize the data to find X s mass function: x #outcomes with x successes P (each outcome) p(x) More generally, for X Bin(n, p), how many outcomes have x successes? That is, in how many ways can we write x successes (S) (and n x failures (F)) into this n-letter word,...? (#places for 1st success) (#places for last (x th )) #ways to order x successes ( ) n! n = x!(n x)! = x = n(n 1) (n (x + 1)) x(x 1) What s the probability that X = x, that is, that n Bernoulli trials will have x successes? P (X = x) = (#ways to choose x trials to be successful) (factor of π for each success) (factor of 1 π for each failure) ( ) n = π x (1 π) n x, for x = 0, 1,, n x

6 To summarize, For X Bin(n, ( ) π), the probability of x successes, where x is one of {0, 1, 2,..., n}, is n P (X = x) = π x (1 π) n x. x e.g. Suppose 99.6% of a warehouse full of memory cards work. Four are chosen randomly and installed in a computer. What is the probability that exactly three of the computer s cards work. Check with R: dbinom(x=3, size=4, prob=.996) What is the probability that no more than three work? Check with R: pbinom(q=3, size=4, prob=.996) The Mean and Variance of a Binomial Random Variable Since X Bin(n, π) is the number of successes in independent Bernoulli trials, each having probability of success, we can write X = n i=1 Y i, where Y i Bernoulli(π). Then µ X (mean) = σ 2 X (variance) = σ X (standard deviation) = nπ(1 π) e.g. For X = #working cards, above, µ = and σ =. e.g. What is the expected number of heads in 100 tosses of a fair coin?

7 Continuous random variables The values of a continuous random variable : between any two is a third. Here is a density histogram of ant weights from 3. If we make the bins smaller (while increasing ), the skyline approaches an approximating curve: Ant Weights Ant Weights (smaller bins) Ant Weights (approximate) Density Density Density Weight (mg) Weight (mg) Weight (mg) The proportion of weights in an interval (like (2.75, 3.25)) is the area of the histogram bars over the interval, or, in the limit, the under the curve (draw). The probability distribution for a continuous random variable X is its probability density function (pdf) f defined by y = f(x) such that P (a X b) = under f between a and b (draw). The cumulative distribution function of a random variable The cumulative distribution function (CDF) F of a random variable X is F (x) = P ( ), which is the probability X realizes to a value less than or equal to x. e.g. Draw CDF for X heads and CDF for ant weights.

8 Normal distributions The ant density curve is approximated by a special form called a normal (or Gaussian) curve. If X follows the normal distribution with mean µ and variance σ 2, then its pdf is X ~ N(µ, σ 2 1 ) f(x) = σ 2π e (x µ) 2 2σ 2 Density f(x) σ µ σ x (Note three famous,, in this formula. Then.) Find µ and σ by eye: µ is the location of the σ is distance from the center to the point at which the getting steeper to getting flatter from The normal distribution, the most important in statistics, is (exactly or ) the distribution of the mean, X, of a large sample from (almost) distribution with finite µ and σ ( 5); this is important for procedures we ll learn soon many effects that are the sum of many small additive and independent effects repeated measurements of the same quantity, where each is understood as a plus a. velocities of molecules in an, long-duration amounts like monthly totals, and many other natural quantities (in some important cases)

9 Finding normal probabilities The ant farm data were randomly sampled from a normal distribution, N(3,.25 2 ), so it s a reasonable approximation to the ant weight distribution. Let s calculate some probabilities from N(3,.25 2 ). Three possible approaches: Find the area under its density curve, f(x), via calculus: Homework and : use R via pnorm(q, mean=3, sd=.25), which gives the area under f(x) left of q, that is, it gives F (x) = P (X q), where X N(3,.25 2 ) (draw) and life before 1970: standardize and look in a table (below) Standardize the scale on a N(µ, σ 2 ) distribution by measuring in units about the. Subtracting µ centers distribution at, and dividing by σ makes the standard deviation. In particular, if x is an observation from N(µ, σ 2 ), then the standardized value of x is z = ; it s also called the. The standard normal distribution is N(0, 1 2 ). Theorem: If X N(µ, σ 2 ), then Z = X µ σ N(0, 1 2 ) (draw). (Also Z N(0, 1 2 ) = X = Zσ + µ N(µ, σ 2 ).) Proof: Suppose X N(µ, σ 2 ). Then: E(Z) = VAR(Z) = Z N(...)

10 e.g. Let X be a random ant weight. Then X N(3,.25 2 ) ( ). Find these ( ) probabilities: P (random ant weighs < 2.75mg) (.2 is correct from the actual population) Note: to find P (Z < 1.00 = ( ), look in row and column of the Cumulative N(0, 1) Distribution table, which gives F (z) = P (Z z). P (random ant weighs > 2.75 mg) (.8 is correct from the actual population) P (random ant weighs between 2.75 and 3.25 mg) (.59 is correct from the actual population) e.g. Eleanor scored 680 on the SAT math test, which had N(518, ) scores. Gerald scored 27 on the ACT math test, which had N(20.7, ) scores. Find the standardized score for both. Assuming the tests measure the same ability (and are taken by the same population), who performed better? e.g. Eleanor scored higher than what proportion of students on SAT math?

11 Finding a value given a proportion e.g. IQ scores are N(100, 15 2 ). (a) What scores fall in lowest 25%? Check with R: qnorm(p=.25, mean=100, sd=15) gives x such that P (X < x) =.25, where X N(100, 15 2 ). (b) How high a score is needed to be in top 5%? e.g. ACT scores are N(20.9, ). SAT scores are N(1026, ). José scored 1287 on the SAT. Assuming that both tests measure the same thing, what is the ACT equivalent of José s SAT score? The rule (for when you re without your ) For a normal distribution N(µ, σ 2 ), µ. 68% 95% 99.7% of observations fall within of the mean,

12 Extra exercises (for use if time permits) e.g. The intake valve clearances on new engines of a certain type are normally distributed with mean 200 µm and standard deviation 10 µm. a. What is the probability that the clearance is greater than 215 µm? Check with R: b. What is the probability that the clearance is between 180 and 205 µm? Check with R:

13 c. Find the clearance, x, such that 99% of clearances are less than x. Check with R: d. An engine has six intake valves. What is the probability that exactly two of them have clearances greater than 215 µm? Check with R: