Of the parts produced by a particular machine, 1.5% are defective. If a random sample of 11 parts produced by this machine contains 2 or more defective parts, the machine is shut down for repairs. Find the probability that the machine will be shut down for repairs based on this sampling plan.
1-p(x = 0) + p(x = 1)
1-[(11C0)(.015)^0(.985)^11 + 11C1 *0.015^1 *.985^10)]
1-[.84684 + .14185]
1-.9887 = 0.0113
To find the probability that the machine will be shut down for repairs based on this sampling plan, we can use the binomial distribution.
Let's set up the problem:
n = 11 (sample size)
p = 0.015 (probability of a part being defective)
We want to find the probability of having 2 or more defective parts out of the 11 sampled parts.
Now, let's calculate the probability using the binomial distribution formula:
P(X ≥ 2) = 1 - P(X = 0) - P(X = 1)
P(X = k) = (n choose k) * (p^k) * ((1-p)^(n-k)), where (n choose k) = n! / (k! * (n-k)!)
P(X = 0) = (11 choose 0) * (0.015^0) * ((1-0.015)^(11-0))
= 1 * 1 * (0.985^11)
≈ 0.927337
P(X = 1) = (11 choose 1) * (0.015^1) * ((1-0.015)^(11-1))
= 11 * 0.015 * (0.985^10)
≈ 0.075588
P(X ≥ 2) = 1 - P(X = 0) - P(X = 1)
= 1 - 0.927337 - 0.075588
≈ 0.997075
Therefore, the probability that the machine will be shut down for repairs based on this sampling plan is approximately 0.997075, or 99.71%.
To find the probability that the machine will be shut down for repairs based on this sampling plan, we can use binomial distribution.
First, let's define some variables:
- Let X be the number of defective parts in a sample of 11 parts.
- Let p be the probability of a part being defective, which is 1.5% or 0.015 in decimal form.
- Let q be the probability of a part being non-defective, which is 1 - p or 0.985 in decimal form.
- Let n be the sample size, which is 11.
The probability of having 2 or more defective parts in the sample can be calculated as the sum of probabilities of having 2, 3, 4, ..., up to 11 defective parts.
P(X ≥ 2) = P(X = 2) + P(X = 3) + ... + P(X = 11)
Using the binomial probability formula P(X = k) = C(n, k) * p^k * q^(n-k), where C(n, k) is the number of combinations of n items taken k at a time, we can calculate the individual probabilities.
P(X = 2) = C(11, 2) * (0.015)^2 * (0.985)^(11-2)
P(X = 3) = C(11, 3) * (0.015)^3 * (0.985)^(11-3)
...
P(X = 11) = C(11, 11) * (0.015)^11 * (0.985)^(11-11)
Once we have the individual probabilities, we can sum them up to get the final probability:
P(X ≥ 2) = P(X = 2) + P(X = 3) + ... + P(X = 11)
By calculating these probabilities and adding them together, you can find the probability that the machine will be shut down for repairs based on this sampling plan.