Question

Suppose that when the pH of a certain chemical compound is 5.00, the pH measured by a randomly selected beginning chemistry student is a random variable with a mean of 5.00 and a standard deviation of 0.2. A large batch of the compound is subdivided, and a sample is given to each student in a morning lab and each student in an afternoon lab. Let X be the average pH as determined by the morning students, and Y be the average pH as determined by the afternoon students.

If pH is a normally distributed variable and there are 25 students in each lab, compute P(−0.1≤X−Y≤0.1). [Hint: X−Y is a linear combination of normal variables, so it is normally distributed. Compute E(X−Y) and SD(X−Y).]

Answer 1

To find P(-0.1≤X-Y≤0.1), we use the Central Limit Theorem and normal distribution properties to calculate the mean and standard deviation of the difference in average pH levels between morning and afternoon labs, allowing us to determine the probability using standard normal distribution tools.

Step-by-step explanation:

To calculate P(-0.1≤X-Y≤0.1) for the average pH determined by morning students (X) and by afternoon students (Y), when the pH is normally distributed and each lab having 25 students, we begin by assessing the distributions of X and Y.

Since the distribution of individual student measurements is normal with a mean (μ) of 5.00 and a standard deviation (σ) of 0.2, the distribution of the average measurements over a sample size (n) of 25 will also be normal, according to the Central Limit Theorem.

The expected mean of X and Y, E(X) and E(Y), remains the same as the individual measurements, which is 5.00. The standard deviation of the sample means, SD(X) and SD(Y), would be σ/√n = 0.2/√25 = 0.04. Since X and Y are independent, the standard deviation of the difference X - Y will be √(SD(X)^2 + SD(Y)^2) = √(0.04^2 + 0.04^2) = √(0.0016 + 0.0016) = √(0.0032) ≈ 0.0566.

Now, with E(X - Y) = E(X) - E(Y) = 0 and SD(X - Y) ≈ 0.0566, we can find P(-0.1≤X-Y≤0.1) using a normal distribution table or calculator, since X - Y follows a normal distribution with the calculated mean and standard deviation.