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Psychology 406
Assignment \#4
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1) Hays --- Chapter 5, \#24, \#26, \#38; Chapter 6, \#2, \#8, \#12
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2) Let X be a random variable with the following values and
probabilities:
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\begin{tabular}{cc}
X & Probability \\
-1 & 1/4 \\
0 & 1/4 \\
+1 & 1/2 \\
\end{tabular}
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Suppose I make three independent observations on X, i.e.,
$\mathrm{X}_{1}$, $\mathrm{X}_{2}$, and $\mathrm{X}_{3}$.
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a) What is the sample space for the three observations? How many
members? List each possible outcome and its probability of
occurrence.
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b) let Y = $(1/3)(\sum_{i=1}^{3}\mathrm{X}_{i})$. What is the
probability distribution of Y? (i.e., the sampling distribution for
the mean of the three observations) (Hint: Use your listing in (a))
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c) E(X) = \rule{.5in}{.01in}? Var(X) = \rule{.5in}{.01in}?
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d) What is E(Y)? What is its relation to E(X)? Do you have a term
for this relationship?
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e) What is Var(Y)? What is its relation to Var(X)?
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f) Suppose I wish to estimate Var(X) on the basis of
$\mathrm{X}_{1}$, $\mathrm{X}_{2}$, and $\mathrm{X}_{3}$, and do
not have the exact probabilities associated with X. Assume
$\mathrm{X}_{1}$ = 1, $\mathrm{X}_{2}$ = 0, and $\mathrm{X}_{3}$ =
1. What is the unbiased estimator for Var(X)? Using this unbiased
estimate of Var(X), what is the estimate for the standard deviation
of Y?
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3) Suppose X is Normal with mean 2 and variance 9 (denoted by X
$\sim$ N(2,9)).
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a) P(1 $\le$ X $\le$ 3) = \rule{.5in}{.01in}?
P(X $\ge$ 4) = \rule{.5in}{.01in}?
P(X $\le$ -1) = \rule{.5in}{.01in}?
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b) Find the value b such that P(2 - b $\le$ X $\le$ 2 + b) is as
close to .9 as possible.
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c) Suppose I have two independent observations on X, i.e.,
$\mathrm{X}_{1}$ and $\mathrm{X}_{2}$. Let Y = 4$\mathrm{X}_{1}$ -
$\mathrm{X}_{2}$.
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E(Y) = \rule{.5in}{.01in}?
Var(Y) = \rule{.5in}{.01in}?
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What is the distribution of Y?
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Suppose I have 10 independent observations on X, i.e.,
$\mathrm{X}_{1}$, $\ldots$, $\mathrm{X}_{10}$.
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Let M = (1/10)$\sum_{i=1}^{10}$$\mathrm{X}_{i}$.
What is E(M)? Var(M)?
What is the distribution of M?
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4) Assume X is binomial with N = 100, p = 1/4.
Using the normal approximation, calculate the following quantities:
P(20 $\le$ X $\le$ 40) = \rule{.5in}{.01in}?
P(X $\le$ 80) = \rule{.5in}{.01in}?
P(X $\ge$ 20) = \rule{.5in}{.01in}?
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