Class Notes (834,991)
Canada (508,850)
Statistics (280)
STAB22H3 (222)
Ken Butler (34)
Lecture 16


19 Pages
Unlock Document

Ken Butler

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
More Less

Related notes for STAB22H3

Log In


Join OneClass

Access over 10 million pages of study
documents for 1.3 million courses.

Sign up

Join to view


By registering, I agree to the Terms and Privacy Policies
Already have an account?
Just a few more details

So we can recommend you notes for your school.

Reset Password

Please enter below the email address you registered with and we will send you a link to reset your password.

Add your courses

Get notes from the top students in your class.