Statistics - Define a random variable X to be the number of machines that will break down in a day.?

A manufacturing company has 5 identical machines that produce nails. The probability that a machine will break down on any given day is 0.1. Define a random variable X to be the number of machines that will break down in a day.


a) What is the appropriate probability distribution for X? Explain how X satisfies the properties of the distribution.


b) Compute the probability that 4 machines will break down.


c) Compute the probability that at least 4 machines will break down.


d) What is the expected number of machines that will break down in a day?


e) What is the variance of the number of machines that will break down in day?

Statistics - Define a random variable X to be the number of machines that will break down in a day.?
a)


Binomial is the appropriate distribution:


P(X = 0) = 5C0*.1^0*.9^5 = .59049


P(X = 1) = 5C1*.1^1*.9^4 = .32805


P(X = 2) = 5C2*.1^2*.9^3 = .07290


P(X = 3) = 5C3*.1^3*.9^2 = .00810


P(X = 4) = 5C4*.1^4*.9^1 = .00045


P(X = 5) = 5C5*.1^5*.9^0 = .00001





X satisfies a binomial distribution by satisfying these four conditions:


1) X has only two options (break down/not break down).


2) X is a count.


3) The probability of success/failure is constant.


4) Trials are independent; whether a machine breaks down or not one day does not affect if it will break down the next.





b)


P(X = 4) = 5C4*.1^4*.9^1 = .00045





c)


P(at least 4) = P(X = 4) + P(X = 5)


=.00045 + .00001


=.00046





d)


E(X) = np = 5*.1 = .5





e)


variance = np(1-p) = 5*.1*.9 = .45





The guy below me gave you a wrong answer for part b).
Reply:a) It is a Binomial Distribution. X satisfies the properties of this distribution because: it is Discrete (the sum of the probabilities of all the possibile number of machine breakdowns = 1). The experiments are yes no in nature (yes it is or no it is not bad). and there is a probability (p) that any given machine is bad.


b) The probability mass function (pmf) is the probability of exactly k successes in n trials = p^k*(1-p)^n-k = .1^4*(1-.1)^5-4 = .00009


c) Add the pmf of k = 4 and k = 5


d) Mean number of breakdowns n*p = 5*.1


e) Variance np(1-p) = 5*.1(1-.1) =.5(.9)