Question

An IT user support service has monitored the daily number of phone calls requesting for IT services,

No of calls 0 1 2 3 4 5 6
Probability K 0.15 0.3 0.2 K K 0.05

i) determine the value of K.

ii) over 3 days, what is the probability of IT services being requested more than twice on each day?

Answer 1

k = .1

Answer 2

how you get that?

Answer 3

P = 1

Answer 4

the sum of that must be 1.

Answer 5

(0xK)+(1x0.15)+(2x0.3) + (3x0.2)+(4xK)+(5xK)+(6x0.05) = 1

K = 0.18

Answer 6

is this correct?

Answer 7

ii)
(1-p(0)-p(1))³

Answer 8

no just the sum of prob must be one(K + 0.15 +0.3 +0.2 +K + K +0.05)

Answer 9

so its K + 0.15 +0.3 +0.2 +K + K +0.05 = 1 ?

Answer 10

yes

Answer 11

so k = 0.1 .

ii) (1-p(0)-p(1))³

Answer 12

@RaphaelFilgueiras That's clever =)

Answer 13

i dont know how to get p(0) and p(1)

Answer 14

= [1 - ( .1 + .15) ] = .44

Answer 15

thank you so much!