The length of telephone conversation in booth has been an exponential distribution and found on an average to be 5 minutes. Find the probability that a random call made from this booth (i) ends less than 5 minutes (ii) between 5 and 10 minutes

0.632 ; 0.233

Step-by-step explanation:

Given that :

Conversation average λ = 1/5 = 0.2

Using the poisson distribution function :

P(x) = 1 - e^(-λ/x)

p(< 5) = 1 - e^-(1/5 * 5)

P(<5) = 1 - e^-1

P(< 5) = 1 - 0.3678794

= 0.6321205

= 0.632

B.) between 5 and 10 minutes

5 < x < 10

P(x< 5) - p(x< 10)

P(< 5) = 1 - 0.3678794 = 0.6321205

P(< 10) = 1 - (e^-1/5 * 10)

P(< 10) = 1 - e^-2 = 1 - 0.1353352 = 0.8646648

P(< 10) - P(<5)

0.8646648 - 0.6321205

=0.2325443

= 0.233

c.

step-by-step explanation:

because first you have to add the 4x, 9x, and the 7x together and you get 20x and then you add the last numbers up and you get 5and then you put the subtraction sign in there and that's how you get the answer.