Question

Since we've determined that the Central Limit Theorem applies, let's calculate the probability: Assume that the standardized test scores of a certain group of high school students has an unknown distribution with a mean of 90 percent and a standard deviation of 15 percent. A sample size of n=31 is randomly drawn from a population. Use a calculator to find the probability that the sample mean is between 85 and 92 percent. If needed, round to the nearest hundredth.

Answer 1

To find the probability that the sample mean is between 85 and 92 percent, we calculate the standard error, convert 85 and 92 to z-scores, and then use the standard normal distribution to find the difference between these probabilities.

Step-by-step explanation:

You've asked to calculate the probability that the sample mean of standardized test scores is between 85 and 92 percent for a sample size of n=31, given that the population mean is 90 percent and the standard deviation is 15 percent. According to the Central Limit Theorem, for a sufficiently large sample size, the distribution of sample means will be approximately normal, or bell-shaped.

First, we need to calculate the standard error of the mean (SEM), which is equal to the population standard deviation (σ) divided by the square root of the sample size (n):
SEM = σ / √n = 15 / √31 ≈ 2.69.

Next, we convert the range of 85 to 92 into z-scores, which are defined as:
z = (X - μ) / SEM
For X = 85: z1 = (85 - 90) / 2.69 ≈ -1.86
For X = 92: z2 = (92 - 90) / 2.69 ≈ 0.74

Now, we use the standard normal distribution (z-table) or a calculator with statistical functions to find the probabilities corresponding to z1 and z2. The difference between these two probabilities will give us the probability that the sample mean lies between 85 and 92 percent.