IE 227 RECITATION - 9 29.11.2013 1. A service station has both self

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IE 227 RECITATION - 9
29.11.2013
1. A service station has both self-service and full-service islands. On each island, there is a
single regular unleaded pump with two hoses. Let X denote the number of hoses being used on
the self-service island at a particular time, and let Y denote the number of hoses on the fullservice island in use at that time. The joint PMF of X and Y appear in the accompanying
tabulation.
y
P(x,y)
x
0
1
2
0
0.10
0.04
0.02
1
0.08
0.20
0.06
2
0.06
0.14
0.30
a) What is 𝑃(𝑋 = 1 π‘Žπ‘›π‘‘ π‘Œ = 1)=?
b) Compute 𝑃(𝑋 ≤ 1 π‘Žπ‘›π‘‘ π‘Œ ≤ 1).
c) Give a word description of event {𝑋 ≠ 0 π‘Žπ‘›π‘‘ π‘Œ ≠ 0}and compute the probability of this
event.
d) Compute the marginal PMF of X and of Y. Using 𝑃π‘₯ (π‘₯) what is P(X ≤ 1)?
e) Are X and Y independent random variables? Explain.
Answer: a) 𝑃(𝑋 = 1, π‘Œ = 1) = 𝑃𝑋,π‘Œ (1, 1) = 0.20
b) 𝑃(𝑋 ≤ 1 π‘Žπ‘›π‘‘ π‘Œ ≤ 1) = 𝑃𝑋,π‘Œ (0,0) + 𝑃𝑋,π‘Œ (0, 1) + 𝑃𝑋,π‘Œ (1, 0) + 𝑃𝑋,π‘Œ (1, 1) = 0.42
c) At least one hose is in use at both islands. 𝑃(𝑋 ≠ 0 π‘Žπ‘›π‘‘ π‘Œ ≠ 0) = β‹―
2. Let X denotes the number of Canon digital cameras sold during a particular week by a certain
store. The PMF of X is
x
0
1
2
3
4
P(x) 0.1 0.2 0.3 0.25 0.15
Sixty percent of all customers who purchase these cameras also buy an extended warranty. Let
Y denote the number of purchasers during this week who buy an extended warranty.
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a) What is 𝑃(𝑋 = 4, π‘Œ = 2)?
b) Calculate 𝑃(𝑋 = π‘Œ).
c) Determine the joint PMF of X and Y and then the marginal PMF of Y.
Answer: a) P(𝑋 = 4, π‘Œ = 2) = 𝑃(π‘Œ = 2|𝑋 = 4). 𝑃(𝑋 = 4)
You have four purchases and look for the buying 2 extended warranty in 4 purchases.
So Y|X ~ Bin(n=4, p=0.6)
P(𝑋 = 4, π‘Œ = 2) = 𝑃(π‘Œ = 2|𝑋 = 4). 𝑃(𝑋 = 4) = (42)(0.6)2 (0.4)2 × (0.15) = 0.0518
b)…
3. Arrivals at the emergency rooms of two hospitals occur at average hourly rates of “α” and “µ”,
respectively. The joint PMF of the number of arrivals during any particular hour is
𝑃𝑋,π‘Œ (π‘₯, 𝑦) = 𝑒 −(𝛼+πœ‡) (
𝛼π‘₯πœ‡π‘¦
π‘₯!𝑦!
) π‘“π‘œπ‘Ÿ π‘₯, 𝑦 = 0,1,2 … and zero otherwise.
a) Determine and identify each conditional distribution of Y given X=x.
b) Determine and identify each conditional distribution of X given Y=y.
c) What is the relationship between each conditional distribution of Y given X=x and the
marginal distribution of X.
Answer: a) 𝑃(𝑋 = π‘₯) = ∑~𝑦=0 𝑃𝑋,π‘Œ (𝑋 = π‘₯, π‘Œ = 𝑦) = ∑~𝑦 𝑒 −(𝛼+πœ‡) (
𝑒 −𝛼 𝛼 π‘₯
π‘₯!
𝛼π‘₯πœ‡π‘¦
π‘₯!𝑦!
)=
𝑒 −𝛼 𝛼 π‘₯
π‘₯!
∑~𝑦=0 𝑒 −πœ‡
πœ‡π‘¦
𝑦!
=
π‘“π‘œπ‘Ÿ π‘₯ = 0,1,2 …
𝑃(π‘Œ = 𝑦|𝑋 = π‘₯) =
𝑃(𝑋 = π‘₯, π‘Œ = 𝑦) 𝑒 −πœ‡ πœ‡ 𝑦
=
π‘“π‘œπ‘Ÿ 𝑦 = 0,1,2 …
𝑃(𝑋 = π‘₯)
𝑦!
So Y|X~POI(µ)
b) 𝑃(π‘Œ = 𝑦) = ∑~𝑦=0 𝑃𝑋,π‘Œ (𝑋 = π‘₯, π‘Œ = 𝑦) =…
4. Suppose that 15 percent of the families in a certain community have no children, 20 percent
have one, 35 percent have two, and 30 percent have three; and suppose also that in each family,
each child is equally likely (independently) to be a boy or girl. If a family is chosen at random
from this community, then X, the number of boys, Y the number of girls, in this family.
a) Determine the joint PMF of X and Y.
b) Determine the marginal PMF of X and Y.
c) Find P(X < Y).
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5. Among ten students, three are freshmen, four are sophomores and three are seniors.
Suppose that three of these students are selected at random with replacement. Let X, Y and Z
be the number of freshmen, sophomores and seniors, respectively, in the sample.
a) Identify and obtain the Joint PMF of X, Y and Z.
b) Use the result of part (a) and the FPF to determine and interpret each of P(X>Y>Z),
P(X≤Y<Z) and P(X≤Z, Y≥Z).
c) Without doing any computations, identify and obtain the univariate marginal PMF’s of X,
Y and Z.
d) Without doing any computations, identify and obtain the PMF of each of the three
random variables X+Y, X+Z, and Y+Z.
e) Find P(1≤X+Z≤2).
Answer: a) X, Y, Z ~ Multinomial Distr. (n=3, px=0.3, py= 0.4, pz=0.3)
With joint PMF 𝑃𝑋,π‘Œ,𝑍 (π‘₯, 𝑦, 𝑧)
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= (π‘₯,𝑦,𝑧
) (0.3)π‘₯ (0.4)𝑦 (0.3) 𝑧 π‘ π‘’π‘β„Ž π‘‘β„Žπ‘Žπ‘‘ π‘₯ + 𝑦 + 𝑧 =
3 π‘Žπ‘›π‘‘ π‘₯, 𝑦, 𝑧 ≥ 0
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b) 𝑃(𝑋 > π‘Œ > 𝑍) = 𝑃𝑋,π‘Œ,𝑍 (2,1,0) = (2,1,0
) (0.3)2 (0.4)1 (0.3)0 = 0.108
c)…
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