A machine Shop has 100 drill presses and other machines in constant use. The probablility that a machine will become inoperative during a given day is .002. During some days no machines are inoperative, but during some days one, two three or more are broken down. a. what type of distribution is this?b. What is the probability that fewer than 2 machines will be inoperative during a particular day?

a) Poisson Distribution

b) P(X<2) = 0.981

Step-by-step explanation:

a) This is a Poisson Distribution because a Poisson distribution is the one in which there is a time duration involved and here we are dealing with the probability that the machine will become inoperative during a given day. Hence, this is a Poisson Distribution.

b) Let X denote the number of inoperative machines. We need to find P(X<2) using the Poisson distribution formula

P(X=x) = (λˣ e^-λ)/x!

Where λ = mean no. of times the event occurs

           x = no. of successes

We don't have a value for λ in the question so we will make use of λ = np, where n=100 and p=0.002. So,

λ = (100)(0.002) = 0.2

So, for P(X<2) we have only two cases i.e. when X=0 and when X=1. So,

P(X<2) = P(X=0) + P(X=1)

           = ((0.2)⁰e^-0.2)/0! + ((0.2)¹e^-0.2)/1!

           = 0.818 + 0.163

P(X<2) = 0.981

step-by-step explanation:

5×5×5+2(8)÷2(2)+3(3)

15+16÷4+9

31÷13

do 31÷13 and u will get ur answer