Download STAT 111 Recitation 4

STAT 111 Recitation 4
Xin Lu Tan
[email protected]
September 27, 2013
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Miscellaneous
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Please turn in homework 3.
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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).
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Please pick up homework 4 and the graded homework 2.
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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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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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Practice Problems
In the following, X ∼ N(µ, σ 2 ) means that X is a random variable
having a normal distribution with mean µ and variance σ 2 .
Facts:
X −µ
σ ,
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If X ∼ N(µ, σ 2 ), let Z =
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P(Z ≤ −z) = P(Z ≥ z) = 1 − P(Z ≤ z).
then Z ∼ N(0, 1).
Problems:
1. X ∼ N(2, 16), what is the probability that P(3 < X < 10)?
2. X ∼ N(5, 9), what is the probability that P(2 < X < 7)?
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95% Confidence Intervals of Normal distribution
If Z ∼ N(0, 1), then
P(−1.96 < Z < 1.96) = 0.95.
Since Z =
X −µ
σ ,
we have
P(µ − 1.96σ < X < µ + 1.96σ) = 0.95.
Often, we approximate 1.96 by 2 and use
P(µ − 2σ < X < µ + 2σ) ≈ 0.95
instead.
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The Central Limit Theorem (CLT)
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If the the random variables X1 , X2 , . . . , Xn are independently
and identically distributed (i.i.d.), then no matter what the
probability distribution of these random variables might be,
the average X̄ = (X1 + X2 + · · · + Xn )/n and the sum
Tn = X1 + X2 + · · · + Xn both have approximately a normal
distribution. The approximation becomes more accurate as n
increases.
Specifically, if X1 , X2 , . . . , Xn are independently and identically
distributed (i.i.d.) with mean µ and variance σ 2 , then when n
is sufficiently large, we have
σ2
),
n
·
Tn ∼ N(nµ, nσ 2 ),
·
X̄ ∼ N(µ,
·
where ∼ means approximately distributed.
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The Central Limit Theorem (CLT)
Graphically, the central limit theorem tells us that
Shown above are the resulting frequency distributions each based on 500
averages. For N = 4, 4 scores were sampled from a uniform distribution 500
times and the average computed each time. The same method was followed
with averages of 7 scores for N = 7 and 10 scores for N = 10.
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The Central Limit Theorem (CLT) on Binomial Example
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When Y ∼ Binomial(n, θ), we have mean of Y = nθ, variance
of Y = nθ(1 − θ), the proportion P = Yn of successes has
mean θ and variance θ(1−θ)
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We can think of Y = Tn = X1 + X2 + · · · + Xn , where Xi is a
random variable that takes value 1 with probability θ, and
takes value 0 with probability 1 − θ. Then P = X̄ , and the
CLT tells us that
·
Y ∼ N nθ, nθ(1 − θ)
θ(1 − θ)
·
P ∼ N θ,
n
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Why is the Central Limit Theorem (CLT) useful?
The Central Limit Theorem tells us that when X1 , X2 , . . . , Xn are
independently and identically distributed (i.i.d.) with mean µ and
variance σ 2 , then
σ2
·
X̄ ∼ N(µ, ),
n
and so we have
σ
σ
P µ − 1.96 √ < X̄ < µ + 1.96 √
≈ 0.95.
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The Central Limit Theorem enables us to make a probability
statement involving the normal distribution that we are already
familiar with!
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Practice Problem:
A fair coin is flipped 5000 times. Find two numbers between which
you are approximately 95% certain that the (i) number of heads,
(ii) proportion of heads, will lie. Hint: Use
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If Y ∼ Binomial(n, θ), P = Yn , then
·
Y ∼ N nθ, nθ(1 − θ)
θ(1 − θ)
·
P ∼ N θ,
n
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If X ∼ N(µ, σ 2 ), then
·
P(µ − 2σ < X < µ + 2σ) ≈ 0.95
When should you doubt if the coin is fair or not?
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