a manufacturer of a cotter pins know that 5% of his product is defective if he sells cotter pins in boxes of 100 pins and guarantee that no more than 10 pins will be defective find the probability that a box will fail to meet the guaranteed quality
m=np
n=100
p=0.05
P(failing)= getting more than 10 defective pin= m¹¹e^(-11)/11!
To find the probability that a box will fail to meet the guaranteed quality, we can use the binomial distribution.
Let's define the variables:
p = Probability of a pin being defective = 0.05
n = Number of trials (number of cotter pins) = 100
x = Number of successes (defective pins) allowed = 10
The probability mass function of the binomial distribution is given by:
P(x) = (nCx) * (p^x) * ((1-p)^(n-x))
We want to find P(x ≥ 11), as we want no more than 10 pins to be defective.
P(x < 11) = P(x = 0) + P(x = 1) + P(x = 2) + ... + P(x = 10)
P(x < 11) = Σ [(nCx) * (p^x) * ((1-p)^(n-x))], from x = 0 to 10
Using this formula, we can calculate the probability:
P(x < 11) = (100C0 * 0.05^0 * (1-0.05)^(100-0)) + (100C1 * 0.05^1 * (1-0.05)^(100-1)) + ... + (100C10 * 0.05^10 * (1-0.05)^(100-10))
P(x < 11) = (1 * 1 * 0.95^100) + (100 * 0.05 * 0.95^99) + ... + (17310309456440 * 0.05^10 * 0.95^90)
Calculating this sum would require several calculations, but you can use a calculator or software to find the final probability.
Please note that this is an estimation based on assumptions. The exact probability may vary based on other factors and assumptions, such as the independence of each pin's defectiveness.
To find the probability that a box will fail to meet the guaranteed quality, we need to determine the probability of having more than 10 defective cotter pins in a box of 100.
Given that 5% of the cotter pins are defective, we can calculate the average number of defective cotter pins in a box as follows:
Average number of defective cotter pins = (Percentage of defectiveness / 100) x Total number of cotter pins
= (5 / 100) x 100
= 5
Now, we can use the Poisson distribution to calculate the probability of having more than 10 defective cotter pins in a box. The Poisson distribution is typically used when the average number of outcomes occurs per unit of time or space.
Let's define:
λ (lambda) = average number of defective cotter pins in a box = 5
k = number of defective cotter pins we want to calculate the probability for, in this case, k > 10.
The probability mass function (PMF) for the Poisson distribution is given by:
P(k; λ) = (e^(-λ) * λ^k) / k!
Where e is a mathematical constant approximately equal to 2.71828.
To find the probability that more than 10 defective cotter pins occur, we sum up the probabilities for k > 10:
P(k > 10) = 1 - P(k ≤ 10)
= 1 - [P(0) + P(1) + P(2) + ... + P(10)]
Now, we can calculate the probabilities:
P(0) = (e^(-5) * 5^0) / 0! = 0.0067 (using k = 0)
P(1) = (e^(-5) * 5^1) / 1! = 0.0337 (using k = 1)
P(2) = (e^(-5) * 5^2) / 2! = 0.0842 (using k = 2)
Since the probabilities decrease rapidly, we can use a calculator or a spreadsheet program to calculate the remaining probabilities for k = 3 to 10.
Now, we can sum up the probabilities for k ≤ 10:
P(k ≤ 10) = P(0) + P(1) + P(2) + ... + P(10)
Finally, we can calculate the probability that a box will fail to meet the guaranteed quality:
P(box fails quality) = 1 - P(k ≤ 10)
I encourage you to use a calculator, spreadsheet, or programming language (such as Python) to perform the calculations for the remaining probabilities and get the final result.