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STAB22H3
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Ken Butler
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Lecture 16

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Statistics

STAB22H3

Ken Butler

Fall

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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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