The number of customers arriving at customer service of a department store follo

Question

The number of customers arriving at customer service of a department store follows a Poisson distribution with a mean of 4.0 per time unit. A time unit has 10 minutes. Use X to denote the time between two customer arrivals.

30. The standard deviation of X in minutes is (2)

(A) 2.5000 (B) 1.2500 (C) 0.4000 (D) 0.2500 (E) none of the above
  
31. If one customer just arrived, the probability that the next customer will not arrive within the next 2.00 minutes is closest to (2)

(A) 0.2000 (B) 0.4493 (C) 0.5507 (D) 0.8000 (E) none of the above

32. If one customer just arrived, the probability that the next customer will arrive within the next 3.50 minutes is closest to (2)

(A) 0.2000 (B) 0. 2466 (C) 0.7534 (D) 1.4000 (E) none of the above


33. If one customer just arrived, the probability that the next customer will not arrive within the next 3.00 minutes is closest to (2)

(A) 1.2000 (B) 0.6988 (C) 0.3120 (D) 0.3012 (E) none of the above

34. If one customer just arrived, the probability that the next customer will arrive between the next 1.5 and 3.50 minutes is closest to (2)

(A) 0.3022 (B) 0.4512 (C) 0.5488 (D) 0.24660 (E) none of the above

Explanation / Answer

30.  standard deviation of X in minutes =10/4 =2.5 option (A) 2.5000

31)probability that the next customer will not arrive within the next 2.00 minutes =e-2/2.5 =0.4493

B) 0.4493

32)

probability that the next customer will arrive within the next 3.50 minutes is =1-e-3.5/2.5 =0.7534

(C) 0.7534

33)

probability that the next customer will not arrive within the next 3.00 minutes =e-3/2.5 =0.3012

(D) 0.3012

34)

probability that the next customer will arrive between the next 1.5 and 3.50 minutes =P(1.5<X<3.5)

=(1-e-3.5/2.5)-(1-e-1.5/2.5)=0.3022

(A) 0.3022

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