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

step-by-step explanation:

this is not possible to solve with the info given.   you need at least an one angle somewhere.