A manufacturing company has 5 identical machines that produce nails. The probability that a machine will break down on any given day is .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 saisfies 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 umber of machines that will break down in a day?
e. What is the variance of the number of machines that will break down in a day?
a. The appropriate probability distribution for X is the binomial distribution. This distribution satisfies the properties of a random variable X because it is a discrete probability distribution that models the number of successes in a sequence of n independent yes/no experiments, each of which yields success with probability p. In this case, the number of experiments is 5, and the probability of success is .1.
b. The probability that 4 machines will break down is 0.0010.
c. The probability that at least 4 machines will break down is 0.0062.
d. The expected number of machines that will break down in a day is 0.5.
e. The variance of the number of machines that will break down in a day is 0.4.
a. The appropriate probability distribution for X is the binomial distribution because it satisfies the properties of a discrete probability distribution. The properties are:
1) There are a fixed number of trials (in this case, 5 machines).
2) Each trial has two possible outcomes - the machine breaks down or it doesn't.
3) The probability of success (machine breaking down) is the same for each trial.
4) The trials are independent, meaning the breakdown of one machine does not affect the breakdown of another.
b. To compute the probability that 4 machines will break down, we can use the binomial probability formula:
P(X = k) = (n C k) * p^k * (1-p)^(n-k)
where n is the number of trials, k is the number of successes, p is the probability of success.
P(4 machines break down) = (5 C 4) * (0.1)^4 * (0.9)^(5-4)
= 5 * 0.0001 * 0.9
= 0.00045
c. To compute the probability that at least 4 machines will break down, we need to calculate the probabilities of 4, 5 machines breaking down and add them together.
P(at least 4 machines break down) = P(4 machines break down) + P(5 machines break down)
P(at least 4 machines break down) = 0.00045 + (5 C 5) * (0.1)^5 * (0.9)^(5-5)
= 0.00045 + 1 * 0.00001 * 1
= 0.00046
d. The expected number of machines that will break down in a day can be calculated using the formula:
E(X) = n * p
where n is the number of trials and p is the probability of success.
E(X) = 5 * 0.1
= 0.5
Therefore, the expected number of machines that will break down in a day is 0.5.
e. The variance of the number of machines that will break down in a day can be calculated using the formula:
Var(X) = n * p * (1-p)
where n is the number of trials and p is the probability of success.
Var(X) = 5 * 0.1 * (1-0.1)
= 5 * 0.1 * 0.9
= 0.45
Therefore, the variance of the number of machines that will break down in a day is 0.45.
a. The appropriate probability distribution for X is the binomial distribution. This is because we have a fixed number of independent trials (5 machines) and each trial can have one of two outcomes (the machine breaks down or it doesn't). The properties of the binomial distribution that X satisfies are:
1. Fixed number of trials: In this case, the number of trials is fixed at 5, as we have 5 identical machines.
2. Independent trials: The breakdown of one machine does not affect the breakdown of any other machine. Each machine is independent of the others.
3. Only two possible outcomes: A machine can either break down or not break down on any given day.
4. Constant probability of success: The probability of a machine breaking down is constant for each trial, at 0.1.
b. To compute the probability that 4 machines will break down, we can use the formula for the binomial distribution: P(X = k) = C(n, k) * p^k * (1-p)^(n-k), where n is the number of trials, k is the number of successful outcomes, p is the probability of success, and C(n, k) is the number of combinations of n objects taken k at a time.
In this case, n = 5, k = 4, and p = 0.1. Plugging these values into the formula, we get:
P(X = 4) = C(5, 4) * 0.1^4 * (1 - 0.1)^(5 - 4)
c. To compute the probability that at least 4 machines will break down, we need to calculate the probabilities for X = 4, X = 5 and add them together. This is because "at least 4" includes the cases where exactly 4 or all 5 machines break down.
P(X ≥ 4) = P(X = 4) + P(X = 5)
d. The expected number of machines that will break down in a day can be found using the formula: E(X) = n * p, where n is the number of trials and p is the probability of success.
In this case, n = 5 and p = 0.1. Plugging these values into the formula, we get:
E(X) = 5 * 0.1
e. The variance of the number of machines that will break down in a day can be found using the formula: Var(X) = n * p * (1 - p), where n is the number of trials and p is the probability of success.
In this case, n = 5 and p = 0.1. Plugging these values into the formula, we get:
Var(X) = 5 * 0.1 * (1 - 0.1)