The times taken to complete a statistics test by all students are normally distributed with a mean of 120 minutes and a standard deviation of 12 minutes. Find the probability that the mean time taken to complete this test by a random sample of 16 students would be between 122 and 125 minutes.

Z = (score-mean)/SEm(Standard Error of the mean)

SEm = SD/√n

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to the Z scores.

To find the probability, we can use the Central Limit Theorem, which states that the sampling distribution of the sample means will be approximately normally distributed with a mean equal to the population mean and a standard deviation equal to the population standard deviation divided by the square root of the sample size.

Given:
Population mean (μ) = 120 minutes
Population standard deviation (σ) = 12 minutes
Sample size (n) = 16 students
We want to find the probability that the mean time taken to complete the test falls between 122 and 125 minutes.

Step 1: Calculate the standard error of the sample mean.
Standard error (SE) = σ / √n
SE = 12 / √16
SE = 12 / 4 = 3 minutes

Step 2: Convert the given values to z-scores using the standard normal distribution formula.
z1 = (x1 - μ) / SE
z1 = (122 - 120) / 3 = 2/3
z2 = (x2 - μ) / SE
z2 = (125 - 120) / 3 = 5/3

Step 3: Look up the z-scores in the standard normal distribution table.
P(z1 < Z < z2) = P(2/3 < Z < 5/3)

Looking up the z-scores in the standard normal distribution table, we find that the corresponding probabilities are:
P(Z < 2/3) = 0.7486
P(Z < 5/3) = 0.9522

Step 4: Find the probability between the two z-scores.
P(2/3 < Z < 5/3) = P(Z < 5/3) - P(Z < 2/3)
P(2/3 < Z < 5/3) = 0.9522 - 0.7486
P(2/3 < Z < 5/3) = 0.2036

Therefore, the probability that the mean time taken to complete the test by a random sample of 16 students falls between 122 and 125 minutes is approximately 0.2036 or 20.36%.

To find the probability, we need to standardize the distribution of sample means using the central limit theorem.

The central limit theorem states that the distribution of sample means tends to be approximately normal, regardless of the shape of the population distribution, when the sample size is large enough. In this case, the sample size is 16, which is considered large enough for the central limit theorem to apply.

First, we need to calculate the standard error of the sampling distribution, which is the standard deviation of the sample means. This is calculated by dividing the population standard deviation by the square root of the sample size.

Standard error (SE) = population standard deviation / √sample size
SE = 12 / √16
SE = 12 / 4
SE = 3

Next, we need to standardize the values of interest using the formula for the z-score:

z = (x - μ) / σ

where:
x = the value we are interested in (122 and 125 minutes, in this case)
μ = the mean of the population (120 minutes)
σ = the standard deviation of the population (12 minutes)

For 122 minutes:
z1 = (122 - 120) / 3
z1 = 2 / 3
z1 ≈ 0.667

For 125 minutes:
z2 = (125 - 120) / 3
z2 = 5 / 3
z2 ≈ 1.667

Now that we have the z-scores, we can find the probability using the standard normal distribution (z-distribution) table or a statistical calculator. We want to find the probability that the z-score falls between 0.667 and 1.667.

P(0.667 < z < 1.667) = P(z < 1.667) - P(z < 0.667)

Using the z-distribution table, we find that the probability of z < 0.667 is approximately 0.7475, and the probability of z < 1.667 is approximately 0.9525.

P(0.667 < z < 1.667) = 0.9525 - 0.7475
P(0.667 < z < 1.667) = 0.205

Therefore, the probability that the mean time taken to complete the test by a random sample of 16 students falls between 122 and 125 minutes is approximately 0.205 or 20.5%.