A paper manufacturer claims that fewer than 1 in 100 of its reels (2 ton rolls) of paper is flawed. A customer has just received a large shipment of these reels and proceeds to check a random sample of 600 of them for flaws. Of this sample, 14 reels are found to be flawed. What is the probability of finding at least 14 flawed reels in this sample?
You can use the normal approximation to the binomial distribution.
Your values are the following:
p = 1/100 = .01, q = 1 - p = 99/100 = .99, x = 14, and n = 600
You need to find mean and standard deviation.
mean = np = (600)(.01) = 6
standard deviation = √npq = 2.44 (this is a rounded value)
Now use z-scores to find probability:
z = (x - mean)/sd -->sd = standard deviation
With your data:
z = (14 - 6)/(2.44) = 3.28
Using a z-table, you will find the probability to be very small.
I hope this will help.
To find the probability of finding at least 14 flawed reels in this sample, we can use the normal approximation to the binomial distribution.
Let's start by finding the mean and standard deviation.
The mean (μ) can be calculated by multiplying the probability of success (p) by the number of trials (n). In this case, p = 1/100 and n = 600. So, μ = 600 * (1/100) = 6.
The standard deviation (σ) can be calculated using the formula σ = √(npq), where q is the probability of failure (1 - p). In this case, q = 99/100. So, σ = √(600 * (1/100) * (99/100)) = 2.44 (rounded to two decimal places).
Now, we can calculate the z-score using the formula z = (x - μ) / σ, where x is the number of flawed reels we want to find. In this case, we want to find at least 14 flawed reels, so x = 14. So,
z = (14 - 6) / 2.44 = 3.28 (rounded to two decimal places).
Next, we can use a z-table or a calculator to find the probability associated with this z-score. Since z = 3.28 is quite large, the probability is expected to be very small.