)Batches of chemical are manufactured by a production process. Samples of 20 batches from a production run are selected for testing. If the standard deviation of the percentage of impurity contents in the sample batches exceeds 2.5%, the production process is thoroughly checked. Assume that the population distribution of percentage impurity concentrations is normal. What is the approximate probability that the production process will be thoroughly checked if the population standard deviation of percentage impurity concentrations is 2%?

The probability that the sample standard deviation of the percentage impurity of batches manufactured that exceeds 2.5% is 0.0559

Step-by-step 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

u rich

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