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A library subscribes to two different weekly news magazines, each of which is su

ID: 3247172 • Letter: A

Question

A library subscribes to two different weekly news magazines, each of which is supposed to arrive in Wednesday's mail. In actuality, each one could arrive on Wednesday (W), Thursday (T), Friday (F), or Saturday (S). Suppose that the two magazines arrive independently of one another and that for each magazine P(W) = .30, P(T) = .25, P(F) = .25, and P(S) = .20. Define a random variable y by y = the number of days beyond Wednesday that it takes for both magazines to arrive. For example, if the first magazine arrives on Friday and the second magazine arrives on Wednesday, then y = 2, whereas y = 1 if both magazines arrive on Thursday. Obtain the probability distribution of y. (Hint: Draw a tree diagram with two generations of branches, the first labeled with arrival days for Magazine 1 and the second for Magazine 2.)



You may need to use the appropriate table in Appendix A to answer this question.

Statistical Tables 860 Table 1 Random Numbers 860 Table 2 Standard Normal Probabilities (Cumulative z Curve Areas) 862 Table 3 t Critical Values 864 Table 4 Tail Areas for t Curves 865 Table 5 Curves of b 5 P(Type II Error) for t Tests 868 Table 6 Values That Capture Specifi ed Upper-Tail F Curve Areas 869 Table 7 Critical Values of q for the Studentized Range Distribution 873 Table 8 Upper-Tail Areas for Chi-Square Distributions 874 Table 9 Binomial Probabilities 876

Value of y Probability 0 1 2 3

Explanation / Answer

When y=0,

This means both magazine arrived on wednesday itself.

Probability= (0.30)(0.30)= 0.09

When y=1,

This means one magazine arrived on wednesday and anither on thursday.This can happen in 3 ways: 1st magazine arrives on wednesday and 2nd magazine arrives on thursday OR 2nd magazine arrives on wednesday and 1st magazine arrives on thursday OR both magazines arrive on thursday

Probability= (0.30)(0.25)+(0.25)(0.30)+(0.25)(0.25)= 0.2125

When y=2,

Possible cases

Probability= (0.30)(0.25)+(0.25)(0.25)+(0.25)(0.25)+(0.25)(0.30)+(0.25)(0.25)= 0.3375

When y=3

Possible cases

Probability= (0.30)(0.20)+(0.25)(0.20)+(0.25)(0.20)+(0.20)(0.20)+(0.20)(0.30)+(0.20)(0.25)+(0.20)(0.25)= 0.36

So ,Probability Distibution is,

Magazine 1 Magazine 2 W F T F F F F W F T
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