2) on average, a house has 2 repairs per year. a) what is the probability that in the next 3 years, there will be less than 2 repairs? b) i had 3 repairs this year. what is the probability that i have to wait between 1 and 2 more years for my next repair?

Step-by-step explanation:

(a) Total repairs on an average in 3 years is given by:

2 repairs per year,

so , in 3 years it will be = 6 repairs

Now, from Poisson distribution,

P =          (1)

where,

= no. of times an event occurs

x = 0, 1, 2, .......

Now, using eqn (1):

P(X = x) =

To calculate the probability of less than 2 repairs in the 3 years:

for

Now,

= +

= 0.01735

(b)  To calculate the probability for 1 and 2 more years for next repair, if current repairs are 3:

The distribution function can be given by:

                (2)

Let Y denote the wait for next year and using eqn (2)

=

=

= 0.04731

(a) The probability for less than 2 repairs in next 3 years is 0.01735

(b) The probability to wait between 1 and 2 more years for next repair is 0.04731

answer: b.48 for

step-by-step explanation:

part 1 answer = 8

part 2 answer = 72

step-by-step explanation:

part 1 explanation = add all of the numbers, then divide by how many numbers you have. so, 200/25 = 8.

part 2 explanation = not gonna lie, this is hard to explain, so just google how to do it.