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Lecture 16

STAB22-LEC16-(16,17).docx

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Department
Statistics
Course
STAB22H3
Professor
Ken Butler
Semester
Fall

Description
STAB22, LEC16 CHAPTER 16: RANDOM VARIABLES (COVERS 16,17) [197] SD OF RANDOM VARIABLE (see last lec) Recall - random var's - var's that are random - ex. tossing coin and seeing how many heads you get - ex. rolling a red die and counting how many spots you get - list of val's, and probability of each one = probability model [198] LINEAR CHANGES TO RANDOM VARIABLE - Just like how in data we had rescaling (multiplying) and shifting (add/subtract) of data val's, we also have similar thing done with random var's -ex. multiply by 2, or add 3, so what will happen to mean, SD of data when this is done? => same type of q's are asked about random var's Adding/subtracting by constant - mean changes by the amt of addition/subtraction - SD stays the same Multiplying by constant - mean and SD both change; multiply them by constant VALUE OF Y 3 4 5 PROBABILITY 0.1 0.8 0.1 Y VALUE OF Y 6 8 10 PROBABILITY 0.1 0.8 0.1 2Y (ex) so what is 2Y? - do get 2Y, take all the val's of Y and multiply them by 2, and LEAVE PROBABILITIES AS THE SAME. - mean = 8, SD = 0.9, which is double of Y VALUE OF Y 6 7 8 PROBABILITY 0.1 0.8 0.1 Y + 3 (ex) so what is Y+3? - take all possible val's of Y, and then you add 3 to them - mean = 7, SD = 0.45 (stays the same) - took original Y and added 3 onto it - what's hapepend to mean and SD? The mean went up from 4 to 7, but SD did NOT change - same like in data; if take bunch of data and multiply by 2, eth becomes twice as big, but if add 3 to it, the whole distribution moves over by that much, so you do not get change in spread. => linear change to a random variable = Adding/subtracting/multiplying sth to random var, or doing a combination of these operations (another ex ex) converting Celsius to Farenheit [199] SUMMARY ADDING/SUBTRACTING MEAN: constant to random var., what happens to mean, Mean of (X ± a) = SD? mean of X ± A SD: SD of (X+-a) = SD of X - when add constant to random var, you add same constant to all val's, leave P's as they were - mean goes up by that constant, - SD stays same MULTIPLYING constant MEAN: to random var, what happens to mean, SD? Mean of (bX) = b*(mean of X) SD: SD of (bX) = b*(SD of X) - if multiply random var by constant, both mean and SD go up by what u multiplied by [200] TWO/MORE RANDOM VARIABLES (EX) - Let: >- X = #heads when tossing coin 5 times. >- Y = #spots when rolling a die 1 time. - think about tossing coins and rolling dice (ex) imagine tossing coin five times and counting how many heads we get, where we toss each coin once - also, we will take ordinary 6-sided die and roll it once Q: What can we say about total "score", given by X + Y our X and Y are indep. - we are given SD's for X and Y: SD(X) = 1.12 SD(Y) = 1.71 - now, to get variance for a particular random var, we square SD - but to get Var(X + Y), we do: ( ) ( ) ( ) (we are squaring the standard deviations of each random var. here and adding them up) ...which is equivalent to the previous expression: ( ) ( ) ( ) [!] X AND Y MUST BE INDEPENDENT, OTHERWISE THIS WAY OF GETTING SD DOESN'T WORK Problems with X and Y being dependent - if X and Y are positively correlated (if knew X was big, Y was big), then total's SD would be v.big - if X and Y are -vely correlated (X is big, Y is small), then you will get total's SD being too small NOTE - YOU CANNOT ADD UP SD'S of RANDOM VAR'S TO GET SD OF SUM OF RANDOM VAR. - ie. this is not true: ( ) ( ) ( ) - instead, √ ( ) ( ) and, as mentioned before: ( ) ( ) ( ) => have to get variance to get SD [202] ODD FACT: SD OF X-Y IS SAME AS SD OF X+Y (ex) coin toss and die roll - suppose instead of wanting to get total score (X + Y), we wanted to get difference of scores from coin and die (ie. X - Y) - then, >- E(X) = 2.5 - 3.5 = -1 >- Var(X) = (1.12)^2 + (1.71)^2 = 4.18 >- SD(X) = \sqrt{4.18} = 2.04 - notice that even though our new random var was a difference of random var's, we still used + sign for variance => varability of X + Y is same as variability of X - Y - suggests: we will score more on die than on coins on avg (b/c mean for die is higher than for coins), but large SD => at least sometimes score more on coins. - b/c relatively great variability (SD = 2.04) in what score is coming from How do we find difference in scores? - assume that it is normal distribution - ie. X - Y can be described using Normal model - then, we want to know when score from coin higher than it is for die - our y in the z-score is 0, and we want the probability for the interval y > 0 on our normal distribution. - the area under normal curve for y > 0 is the probability that the score for coins is higher than it is for die. - b/c then we will get a positive difference, implying that the term we subtracted from (val. of coin toss) is larger than what it was being subtracted from (val. of dice roll) - turn 0 into z-score - find corresponding probability of z-score from Z-table (that will give you y < 0 (less)) - to find y > 0, do 1 - the probability you got for y < 0 (retrieved from Z Table) [202] DISCRETE RANDOM VARIABLES - set of val's that random var. can take on are finite (ex) dice roll - can have 6 sep. things that you can get on dice - 1,2,3,4,5,6 - set of possible val's (often whole #'s), and probability for each one Recall: Normal distribution - can have ANY val possible for it, whether decimal or whole # - any val. at all is possible - ex. 6.78, 1.3035 - doesn't make sense to get probability of ONE value, but rather you get probability of interval on normal curve - no sep. set of val's or P's going w/ each single value, rather we have P's for intervals: - ex. less than 10 - ex. b/ween 10 & 15 - ex. greater than 15 CONTINUOUS RANDOM VARIABLES - continuous are those that have an infinite # of val's that the random var. can take on. (ex) normal random var. is continuous => any val. in some range is possible - to find mean and SD of cont random var's involves calculus (will not do this, will be given this) [204] HANDLING TWO NORMAL DISTIRBUTIONS "Betty & Clara go for a swim every morning. Time it takes each of them to complete their swim have INDEPENDENT normal distributions. " Given info: Betty mean = 10 min SD = 2 min Clara mean = 11 min SD = 1 min - betty is more quicker, but more variable in her time than Clara Question: - how likely is it that Clara will complete her swim first
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