Question

The mean GPA for students in School A is \(3.0 ;\) the mean GPA for students in School B is 2.8 . The standard deviation in both schools is \(0.25 .\) The GPAs of both schools are normally distributed. If 9 students are randomly sampled from each school, what is the probability that: a. the sample mean for School A will exceed that of School B by 0.5 or more? b. the sample mean for School B will be greater than the sample mean for School A?

Short answer

a. Probability ≈ 0.0055 b. Probability ≈ 0.0455

Step-by-step solution

Identify the relevant distributions

Both schools' GPAs are normally distributed, with given means and standard deviations. School A: mean \( \mu_A = 3.0 \); School B: mean \( \mu_B = 2.8 \). Both schools have a standard deviation \( \sigma = 0.25 \).

Determining the sampling distribution parameters

For samples of size 9 from each school, the sampling distribution of the sample means will be normally distributed with the standard deviation \( \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{0.25}{3} \approx 0.0833 \).

Calculate the difference in mean GPAs

The distribution for the difference in mean GPAs \( \bar{X}_A - \bar{X}_B \) is normal with mean \( \mu_D = \mu_A - \mu_B = 3.0 - 2.8 = 0.2 \) and standard deviation \( \sigma_D = \sqrt{2 \times (0.0833^2)} \approx 0.118 \).

Compute the probability for part (a)

We want \( P(\bar{X}_A - \bar{X}_B > 0.5) \). Standardize this by finding the z-score: \( z = \frac{0.5 - 0.2}{0.118} \approx 2.54 \). Use a z-table to find \( P(Z > 2.54) \). This corresponds to \( 1 - P(Z \leq 2.54) \approx 1 - 0.9945 = 0.0055 \).

Compute the probability for part (b)

We want \( P(\bar{X}_B > \bar{X}_A) \) which is \( P(\bar{X}_A - \bar{X}_B < 0) \). Standardize with z-score: \( z = \frac{0 - 0.2}{0.118} \approx -1.69 \). Use a z-table for \( P(Z < -1.69) \), which is \( \approx 0.0455 \).

Key concepts

Sampling Distribution

Imagine you are picking a small group from a larger group. A sampling distribution is all about understanding the behavior of something, like the mean, in these smaller groups. When you take multiple samples from a population, the average of these samples forms its own distribution.

• This distribution is called the sampling distribution of the sample mean.
• It's an important concept because it helps us make predictions about the population.

When dealing with GPAs from both schools, the exercise illustrates how we find the sampling distribution using samples of 9 students. Since the original GPA scores are normally distributed, the mean of the sampling distribution is equal to the mean of the population. Moreover, the standard deviation of the sampling distribution (often referred to as the standard error) is smaller than that of the population. It's calculated using the formula:\[\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}\]Here, \(\sigma\) is the standard deviation of the population, and \(n\) is the sample size. For School A and School B, this helps in calculating probabilities about their sample means.

Normal Distribution

A normal distribution is like a magical bell-shaped curve where most numbers cluster around the average, and fewer numbers lie farther from it. It's a crucial concept in statistics because many real-life phenomena follow this pattern.

• In a normal distribution, about 68% of the data falls within one standard deviation from the mean, 95% within two, and 99.7% within three.
• This predictable pattern helps in making statistical inferences.

In our GPA problem, both schools have GPA scores that form a normal distribution, which is why we can use tools like z-scores and z-tables to find probabilities. The exercise successfully uses the idea that any sample mean will also follow a normal distribution when the original data is normal, known as the Central Limit Theorem. This makes it possible to compare things like means between two schools for any randomly chosen samples.

Standard Deviation

Standard deviation measures how "spread out" the numbers in a data set are. If you have a smaller standard deviation, your data points are closer to the mean. When the standard deviation is large, data points are spread out all over the place.

• It's a key descriptor of a data set's "variability" or "dispersion."
• In equations, it's denoted by the Greek letter \(\sigma\).

In the exercise with School A and School B, both have a standard deviation of 0.25. This means that within each school's GPA scores, most students have grades that are pretty close to the school’s average GPA. By calculating the standard error, which is the standard deviation of the sampling distribution, you can determine how much the sample means are expected to vary from the actual population mean, which is critical in finding the probability of certain GPA outcomes.