3. [9 points] Let X denote the lag time in a printing queue at a particular comp

Question

3. [9 points] Let X denote the lag time in a printing queue at a particular computer center That is, X denotes the difference between the time that a program is placed in the queue and the time at which printing begins. Assume that X is normally distributed with mean 16 minutes and standard deviation 4 minutes (a) Find the probability that a program will reach the printer within 3 minutes of arriving in the queue. (b) Would it be umusual for a program to stay in the queue (c) Would you be surprised if it took longer than 30 minutes for the program to reach the printer? (d) Find the first and third quartiles of the lag time in the queue. (e) Find the lag time where only 3% of the jobs will wait until they start printing. (f) If a job has been in the queue for 10 minutes, what is the probability that it will begin printing in less than 3 minutes?

Explanation / Answer

Question 3

Here, mean time lag = 16 minutes

standard deviation of time lag = 4 minutes

(a) Here is x is the time in queue

Pr(x < 3) = NORM (x < 3 ; 16 ; 4)

Z = (3 - 16)/4 = -3.25

Pr(x < 3) = NORM (x < 3 ; 16 ; 4) = Pr(Z < -3.25) = 0.0006

(b) Here,

Pr(12 mins < x < 20 mins) = Pr( - < x < + ) = Pr (Z < 1 ) - Pr(Z < -1) = 0.8413 - 0.1587 = 0.6826

so not it is not unusual for a program to stay in between 12 and 20 minutes

(c) Here,

Pr(x > 30 minuteS) = 1 - Pr (x < 30 minutes) = 1 - NORM (x < 30 minutes ; 16 mins; 4 mins)

Z = (30 - 16)/4 = 3.5

Pr(x > 30 minuteS) = 1 - Pr (x < 30 minutes) = 1 - NORM (x < 30 minutes ; 16 mins; 4 mins) = 1 - Pr(Z < 3.5)

= 1 - 0.9998 = 0.0002

(d) Here first quartiles means 25 percentile

Z value for 25 percentile is = -0.6745

so, Z = (X - 16)/4 = -0.6745

x = 16 - 4 * 0.6745 = 13.30 mins

Z value for 75 percentile is = 0.6745

so, Z = (X - 16)/4 = 0.6745

x = 16 + 4 * 0.6745 = 18.70 mins

(e) Here if we take that lag time = c where only 3% of jobs have to wait for print

so,

Pr(x > c) = 0.03

NORM (x < c; 16 mins ; 4 mins)= 1- 0.03 = 0.97

relative z value from z table is

Z = 1.8808

Z = (c - 16)/4 = 1.8808

c= 16 + 4 * 1.8808 = 23.52 mins

(f) Pr(x < 13 mins l x > 10 mins) = Pr(10 mins < x < 13 mins) / Pr(x > 10 mins)

Pr(x > 10 mins) = 1 - NORM (x > 10 mins ; 16 mins ; 4 mins)

Z= (10 - 16)/4 = -1.5

Pr(x > 10 mins) = 1 - NORM (x > 10 mins ; 16 mins ; 4 mins) = 1 - Pr(Z < -1.5) = 1 -0.0668 = 0.9332

Now,

Pr(10 mins < x < 13 mins) = Pr(x < 13 mins) - Pr(x < 10 mins)

Z (x = 13) = (13 - 16)/4 = -0.75

Pr(Z < -0.75) = 0.2266

Pr(10 mins < x < 13 mins) = Pr(x < 13 mins) - Pr(x < 10 mins) = Pr(Z < -0.75) - Pr(Z < -1.5) = 0.2266 - 0.0668 = 0.1598

so,

Pr(x < 13 mins l x > 10 mins) = Pr(10 mins < x < 13 mins) / Pr(x > 10 mins) = 0.1598/0.9332 = 0.1712

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