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

8/k

step-by-step explanation:

55.44 inches

step-by-step explanation:

part a. assuming a perfectly linear relationship, we can find the slope from the first two data points.

slope = m = (change in rainfall)/(change in years)

= (61.80 -62.33)/(2005 -2004) = -0.53/1 = -0.53

then the point-slope form of the equation of the line can be written as

y = m(x -h) +k . . . . . m = -0.53, (h, k) = (0, 62.33)

y = -0.53x +62.33 . . . x = years after 2004

part b. in 2017, x = 2017 -2004 = 13. then the predicted rainfall is

y = -0.53·13 +62.33 = 55.44 . . inches

the predicted rainfall in 2017 is 55.44 in.