In a single-server queuing system, if 12 customers arrive per hour with Poisson distribution and the customer service time is 2 minutes each with a negative exponential distribution. The average time a customer can expect to wait in line is: 1) 1.33 minutes. 2) 4.5 minutes. 3) Undecided.

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A: 3.73 4.37 4.33 3.39 3.68 4.68 3.55 4.55 2.47 4.13 4.47 3.22

Q: P(−1.37<z<1.68)=

A: standard normal distribution Z~N(0,1) P(−1.37&lt;z&lt;1.68) = ?

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A: d¯=3.125sd=2.911n=8CL=0.95

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A: Q-2)normal distributionμ = 60σ = 6P(55&lt;x&lt;66) = ?Q-3)μ = 69σ = 12n = 36μx¯ ?σx¯ = ?

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A: Answer:- Given data is,  Overtime hours  0 1 2 3 4 5 6 Employees  5 11 28 54 44 27 15 Total…

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A:

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Q: 16 26 36 46 56 66 15 25 35 45 55 56 14 24 34 44 54 64 13 23 33 43 53 63 12 22 32 42 52 62 11 21 31…

A: Here given that 2 dice are rolled . So total possible outcomes = 6* 6 = 36

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A: 36 35 26 58 51 18 44 43 56 36 29 44 41 48 32 24 51 36 46 57 43 53 39 31 48 59 27 44 57 16 Sample…

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