Question

The time required for an individual to be served in a cafeteria is a random variable that has an exponential distribution with an average of 8 minutes. What is the probability that a person is treated in less than 4.4 minutes in exactly 4 of the following 7 days?

Answer using 4 decimals.

Answer 1

We 60 for an random variable 9 q 90 have given the time required individual be served jo cafeteria is that has an exponential distribution with average of 8 minutes. i.e. xw esp ( -8) .(mean =0= The probability density function of exponential distribution is, fins = 8 o elsewhere First, lets Calculate the probability that a person is treated in less than 4.4 minutes, P(X < 444) = fons dan .4.4 O 4.4 x18 e dre 8 4.4 001 - dx -X48 -8 e - 4.418 *]** - (-888 éº] [1860-5769498104) + 3(1)] (-8 (0.5769498104) + 8 ] -0.55 8 + 8e

[8 (-0.57694981024 + +1) 8 -0.5769498104 + 1 I-0.5769498104 0.423050 1896 P (x 24.4) - 6. 42305 7 Now, we can determine the probability that a person is kreaked in less than 4.4 montes in exactly 4 of the following days Using Binomial distribution, we get the probability.. The parameters wo na 7 & p=0.42305 . The P.M.F. of Binomial distribution is, P(x=x) $(1) 60.42305) C0.576955=0,1,–,~ jelsewhere z that in . The probability person is treated less than 4,4 minutes in coactly 4 of the following 9 days is, P(X=4) = (1 Co.42305) (6.57695) 4107-4)! (0.423.5)*(0.57695) 7-4 7!

FX 6 X 5*4! 41 31 (0.42305)4 3 #X6x5 3) (0.57695 (0.4230534 0.57695)3 (0.42305) (0.57695} (0.423054 (0.57695) - 210 3x2x1 210 6 35 x 0.032036 72712 X 0.19205009798 0.215302649862 | P(x=4) 0.253