Most the questions have the final answers below them, I just need them to be wor
ID: 3156971 • Letter: M
Question
Most the questions have the final answers below them, I just need them to be worked out/ explained.
Assume that customers arrive at a small store at a rate of one customer per 5 minutes (on average). John works at the store from 8 AM till 1 PM and Bob works there from 12 noon till 6 PM. Let X be the number of customers coming to the store during John’s shift and Y be the number of customers coming to the store during Bob’s shift.
(a) Find the following: E(X) =
Answer: 60
Var(X) =
Answer: 60
(b) Use normal approximation to compute P(54 X < 63). Apply the histogram correction if necessary:
Solution: X is poisson(60), approximated by normal Y = N (60, 60), so P(54 X < 63) P(53.5 < Y < 62.5) = 62.5 60 60 53.5 60 60 = (0.32) (0.84) = 0.6255 (1 0.7995) = 0.4250.
(c) Find Cov(X, Y ) =
Answer: 12
(d) Find the correlation coefficient
Answer: X,Y = 12 66· 72 = 0.183
Are X and Y independent? Yes or No? Is the dependence strong or weak?
(e) Find E(X + Y ) =
Answer: E(X) + E(Y ) = 60 + 72 = 132
(f) Find Var(X + Y ) =
Answer: Var(X) + Var(Y ) + 2Cov(X, Y ) = 60 + 72 + 2 · 12 = 156
Explanation / Answer
a.) Number of hours Jonn works = 5 hours = 300 mins
Customer arriving per minute = 5
Expected number of customers during John's shift = 300 / 5 = 60
b.) Line models follow Poisson distribution. Hence, X ~ Poisson(60)
Now, we can use normal approximation to approximate Poisson distribution to Normal distribution.
Poisson(X) ~ N(X,X)
--> N( 60, 60) = N(60, 7.74)
P(54 X < 63) = P(X<63) - P(54)
= P(63-60/7.74 < Z) - P(54-60/7.74 Z)
= P(0.38<Z) - P(-0.77Z)
= 0.6517 - 0.2177
= 0.434
d.) Correlation Coefficient = 12 / (60*72)
= 0.183
No, they are not idependent but there dependence is weak as correlation coefficient is small.
e.) E(X) = 60
E(Y) = 6*60 / 5 = 72
f.) Var(X) = 60, Var(Y) = 72, Cov(X,Y) = 12
Var(X+Y) = Var(X) + Var(Y) + 2*Cov(X,Y) = 60 + 72 + 2*12 = 156
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