1. The quality-control inspector of a production plant will reject a batch of syringes if two or more defective syringes are found in a random sample of eight syringes taken from the batch. Suppse the batch contains 2% defective syringes. Find sigma.
Please show work.
Thanks
sigma = sqrt [np(1-p)]
http://stattrek.com/Lesson2/Binomial.aspx
.099
To find sigma, we need to calculate the standard deviation of the defective syringes in the batch.
Step 1: Let's calculate the probability of finding a defective syringe in a random sample of size 8.
The probability of finding a defective syringe in one trial is 2% or 0.02.
The probability of not finding a defective syringe in one trial is 1 - 0.02 = 0.98.
Since we have a sample of size 8, we can use the binomial distribution to calculate the probability of finding 0, 1, 2, or more defective syringes in the sample.
Let's consider the probability of finding 0 defective syringes:
P(X = 0) = (0.98)^8 = 0.851
Now, let's consider the probability of finding 1 defective syringe:
P(X = 1) = 8 * (0.02) * (0.98)^7 = 0.278
Next, let's consider the probability of finding 2 or more defective syringes:
P(X >= 2) = 1 - [P(X = 0) + P(X = 1)]
= 1 - (0.851 + 0.278)
= 0.128
Step 2: Now, let's calculate the variance of the binomial distribution.
The variance (sigma^2) of a binomial distribution with parameters n and p is given by the formula:
Variance = n * p * (1 - p)
In this case, n = 8 and p = 0.02.
Variance = 8 * 0.02 * (1 - 0.02)
= 0.1264
Step 3: Finally, to find sigma (the standard deviation), we take the square root of the variance.
Sigma = sqrt(Variance)
= sqrt(0.1264)
= 0.3555
Therefore, the value of sigma is approximately 0.3555.