Answer: The probability that the sample standard deviation of the percentage impurity of batches manufactured that exceeds 2.5% is 0.0559
Explanation:
Let us consider a random sample selected from variable X of the size n chosen from population N. Let (μ) be the population mean and Σ σ be the variance of X and x bar be the sample mean and S^2 be the population mean.
S^2= 1/n-1Σx1
The sample variance has a mean of E(S^2) = σ^2
The variance is:
Var(s^2)= 2σ^2/(n-1)
The sample distribution of variance is given by:
Σ(x - x bar) /σ^2
The x has chi-square distribution with (n-1) degree of freedoms.
The percentage of impurity concentration of the batches of chemical X are normally distributed with a standard deviation of 2%. Here a random sample of 20 batches is considered.
σ=2
n=20
The mean of sample standard deviation is E(S^2) = 2^2 and the variance is:
Var(s^2)= 2×2^2/(20-1)
Therefore s^2= 19s^2/2^2 has chi-square distribution with 19 degree of freedoms.
So,
We are interested in calculating the probability of the standard deviation of the percentage of impurity concentrations of the batches that exceeds 2.5%. Thus, we need to calculate the probability that
P(s^2>2.5^2)
Ps^2> 2.5^2= (PX^2 19 > 19×2.5^2/2^2)
= P(X^2 19 > 29.6875)
= 0.0559
The probability that the sample standard deviation of the percentage impurity of batches manufactured that exceeds 2.5% is 0.0559