STAT 111 Recitation 3
Xin Lu Tan
[email protected]
September 20, 2013
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Miscellaneous
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Please turn in homework 2.
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Please have your name written on your answers, with family
name written last and written in CAPS.
Please write the number of the class (e.g. 201 or 202) that
you are ACTUALLY SITTING IN on your answers (and if you
are registered for other section please specify this as well).
Please pick up homework 3 and the graded homework 1.
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Miscellaneous
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Your TA’s name is Xin Lu Tan, and her email address is
[email protected].
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Grade for homework 1 has been posted on Canvas. Please
check your grade and let me know during the next recitation if
there are any grade discrepancies (please show me your
graded homework as well).
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I will not be dealing with grade discrepancies for homework
that has been returned for more than a week —— you need
to inform me about grade discrepancies during the following
recitation once you get your homework back!
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Late homework should be turned in to Professor Ewens during
the lectures (no guarantee that it will be accepted!).
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Homework 1: Problem 2
A probability calculation shows that if a newborn baby is equally
likely to be a boy or a girl, the probability that a random sample of
2500 newborn babies will contain between 1185 and 1315 boys is
calculated to be about 0.99. Suppose that a random sample of
2500 newborns contained 1334 boys. What reasonable inductive
(i.e. statistical) statement do you think you can make?
Based on the probability calculation, if a newborn baby is equally
likely to be a boy or a girl, then there is a 0.01 probability that the
number of boys out of 2500 newborns will stay outside the range
of 1185-1315. So we have good evidence that a newborn baby is
not equally likely to be a boy or a girl.
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Homework 1: Problem 4, 5
The events are not independent, so we cannot use the formula
P(A ∩ B) = P(A)P(B). We can use
P(A ∩ B) = P(A) + P(B) − P(A ∪ B)
if P(A ∪ B) is first computed. To check independence of events A
and B, we then need to compute P(A), P(B), and P(A ∩ B) and
see if P(A ∩ B) = P(A)P(B).
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Homework 1: Problem 6
An unfair coin (probability 0.4 for head) is tossed twice. Given that
on at least one toss a head occurred, what is the probability that a
head occurred on both tosses?
Event A: Heads on both tosses.
Event B: At least one head in two tosses.
P(A|B) =
P(A ∩ B)
P(B)
We have
P(A ∩ B) = P(A) = 0.4 × 0.4 = 0.16,
and
P(B) = 1 − P(no heads) = 1 − P(both tails)
= 1 − 0.6 × 0.6 = 1 − 0.36 = 0.64.
So we get
P(A|B) =
0.16
= 0.25.
0.64
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Recall: Mean and Variance of a Discrete Random Variable
If X is a random variable that takes on possible values v1 , v2 . . . , vk
with respective probabilities P(v1 ), P(v2 ), . . . , P(vk ), then
mean of X = µ
= v1 P(v1 ) + v2 P(v2 ) + · · · + vk P(vk )
=
k
X
vi P(vi )
i=1
2
variance of X = σ
= (v1 − µ)2 P(v1 ) + · · · + (vk − µ)2 P(vk )
=
k
X
(vi − µ)2 P(vi )
i=1
Both µ and σ 2 are unknown constants, that is, they are parameters
of the probability distribution of X . We also call the square root of
the variance, denoted by σ, the standard deviation.
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Many random variables
Given n random variables X1 , X2 , . . . , Xn , we can derive two other
random variables
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The sum, denoted by Tn ,
Tn = X1 + X2 + . . . + Xn
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The average, denoted by X̄ ,
X̄ =
Tn
X1 + X2 + . . . + Xn
=
n
n
Both the sum and the average, being random variables, each have
a probability distribution, and thus each has a mean and variance.
An average is a totally different concept from a mean — a mean is
a parameter.
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Mean and Variance of Tn and X̄
When X1 , X2 , . . . , Xn are independent and identically distributed
(i.i.d.) random variables, each having a probability distribution
with mean µ and variance σ 2 ,
mean of Tn = nµ,
variance of Tn = nσ 2
mean of X̄ = µ,
variance of X̄ =
σ2
n
When X1 , X2 , . . . , Xn are independent random variables with
respective means µ1 , µ2 , . . . , µn and respective variances
σ12 , σ22 , . . . , σn2 ,
mean of Tn = µ1 + µ2 + · · · + µn , variance of Tn = σ12 + σ22 + · · · + σn2
mean of X̄ =
µ1 + µ2 + · · · + µn
,
n
variance of X̄ =
σ12 + σ22 + · · · + σn2
n2
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Mean and Variance of D = X1 − X2
When X1 , X2 are independent and identically distributed (i.i.d.)
random variables, each having a probability distribution with mean
µ and variance σ 2 , and
D = X1 − X2 ,
Then we have
mean of D = 0,
variance of D = 2σ 2
When X1 , X2 are independent random variables with respective
means µ1 , µ2 and respective variances σ12 , σ22 ,
mean of D = µ1 − µ2 ,
variance of D = σ12 + σ22
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Summary: Three concepts
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Parameter: µ (mean)
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Random variable: X̄ (average, before experiment)
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Observed value (or data): x̄ (average, after experiment)
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Continuous random variables
A random variable X is continuous if it can take any value in some
continuous range of values. Associated with any continuous
random variable is its (usually unknown) density function f(x).
I The probability that the continuous random variable takes a
value in the range (a,b) is
Z
P(a < X < b) =
b
f (x) dx.
a
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The mean and variance of X are, respectively,
Z H
mean of X = µ =
xf (x) dx,
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Z H
2
variance of X = σ =
(x − µ)2 f (x) dx.
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Normal distribution
A continuous random variable X is said to follow the normal (or
Gaussian) distribution with mean µ and variance σ 2 , written
X ∼ N(µ, σ 2 ), if it has density
f (x) = √
1
2
2
e −(x−µ) /2σ
2πσ
for −∞ < x < ∞.
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When µ = 0, σ 2 = 1, X is said to follow the standard normal
distribution.
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If X ∼ N(µ, σ 2 ), then Z =
standardized version of X .
X −µ
σ
∼ N(0, 1). Z is the
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The “Z ” chart
The “Z ” chart is designed for a standard normal random variable,
i.e. Z ∼ N(0, 1). It gives “less than” probabilities for positive
values of z, i.e. P(Z ≤ z) (Note: z is denoted as x in the table).
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