ZABLE 23 Joint and Conditional Dbetibutions omputer Cree W and Joint and Conditi
ID: 3041321 • Letter: Z
Question
ZABLE 23 Joint and Conditional Dbetibutions omputer Cree W and Joint and Conditional Distributions of Computer Crashes (M) and Computer Age(A) A. Joint Distribution Total 0.50 0.00 D 0,50 M-1 M-3 0.05 0.01 Old computer (A 0) New computer (A 1) Total 0.025 0.35 B 0.065 0.45 0.035 0.01 AD201] 0.10-_006 _003 0.01 1.00 B. Conditional Distributions of M given A Total 1.00 1.00 M-2 0.05 0.02 0.13 C0.10 0.02 0.70 Pr(M A 0) Pr(MIA 1) 0.90 0.07 0.01 0.00 1. Using the terminology from the course and based on the scenario, provide precise interpretations of the values indicated by Boxes A through D. How many times do you expect your computer to crash given you are assigned an old computer? Given your expectation, compute var(MIA 2. 0). The conditional distributions of M (computer crashing) given A (computer age) are shown in Table 2.3 Create two.tables, one showing the conditional distribution of A given M-0 and another showing the conditional distribution of A given M-1. Hint: Using the information in Panel A, apply the equation fo computing conditional probabilities. 3. 4. Reproduce the value in Box C using the appropriate joint and marginal probabilities. Then, reproduce value in Box B using the appropriate conditional and marginal probabilities. Hint: You must rearrang equation for conditional probabilities to obtain an equation for computing the joint probability. Are the random variables M and A independent? Explain.
Explanation / Answer
1.
A. Pr[M = 0] = 0.80
Probabilty of no computer crashes is 0.80
B. Pr[A = 0, M = 1] = 0.065
The probability that number of computer crashes in old computer is 1 is equal to 0.065
C. Pr[M = 2 | A = 0] = 0.10
Given the old computer, the probability of number of computer crashes is 2 is equal to 0.10
D. Pr[A = 1] = 0.50
Probability of new computer is 0.5
2.
E[M | A = 0] = 0 * Pr[M = 0 | A = 0] + 1 * Pr[M = 1 | A = 0] + 2 * Pr[M = 2 | A = 0] + 3 * Pr[M = 3 | A = 0] + 4 * Pr[M = 4 | A = 0]
= 0 * 0.70 + 1 * 0.13 + 2 * 0.1 + 3 * 0.05 + 4 * 0.02
= 0.56
Given old computer, the computer crashes 0.56 times
Var[M | A = 0] = 02 * Pr[M = 0 | A = 0] + 12 * Pr[M = 1 | A = 0] + 22 * Pr[M = 2 | A = 0] + 32 * Pr[M = 3 | A = 0] + 42 * Pr[M = 4 | A = 0]
= 02 * 0.70 + 12 * 0.13 + 22 * 0.1 + 32 * 0.05 + 42 * 0.02
= 1.3
3.
4.
Pr[M = 2 | A = 0] = Pr[M = 2, A = 0] / Pr(A = 0)
= 0.05 / 0.5 = 0.10
Pr[A = 0, M = 1] = Pr[M = 1] Pr[A = 0 | M = 1]
= 0.10 * 0.65 = 0.065
5.
Pr[M = 0, A = 0] = 0.35
Pr[M = 0] * Pr[A = 0] = 0.80 * 0.50 = 0.40
As, Pr[M = 0, A = 0] is not equal to Pr[M = 0] * Pr[A = 0], M and A are not independent.
A=0 A=1 Total Pr(A | M=0) Pr(A =0 | M=0)/Pr(M=0) = 0.35/0.8 = 0.4375 Pr(A =1 | M=0)/Pr(M=0) = 0.45/0.8 = 0.5625 1.00 Pr(A | M=1) Pr(A =0 | M=1)/Pr(M=1) = 0.065/0.1 = 0.65 Pr(A =1 | M=1)/Pr(M=1) = 0.035/0.1 = 0.35 1.00Related Questions
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