91Ó°ÊÓ

The Medical College Admission Test. Almost all medical schools in the United States require students to take the Medical College Admission Test (MCAT). To estimate the mean score \(\mu\) of those who took the MCAT on your campus, you will obtain the scores of an SRS of students. The scores follow a Normal distribution, and from published information you know that the standard deviation is 10.4. Suppose that (unknown to you) the mean score of those taking the MCAT on your campus is \(500.0\). (a) If you choose one student at random, what is the probability that the student's score is between 495 and 505 ? (b) You sample 25 students. What is the sampling distribution of their average score \(\mathrm{x}^{-} \frac{2}{2}\) ? (c) What is the probability that the mean score of your sample is between 495 and 505 ?

Step by step solution

01

Understanding the Problem

We need to understand what is required. We have a normal distribution of MCAT scores with a mean \(\mu = 500\) and standard deviation \(\sigma = 10.4\). We need to find the probability that a randomly chosen student's score is between 495 and 505, and then analyze the sampling distribution for a sample of 25 students and find the probability that the sample mean is between 495 and 505.

02

Calculate Probability for Single Student

For part (a), we need to calculate the probability that a single student's score is between 495 and 505. Since the distribution is normal, we use the standard normal distribution (Z-distribution). We calculate the Z-scores for 495 and 505:\[ Z_{495} = \frac{495 - 500}{10.4} = -0.4808 \]\[ Z_{505} = \frac{505 - 500}{10.4} = 0.4808 \]Using a standard normal distribution table or calculator, find the probabilities for \(Z_{495}\) and \(Z_{505}\) and the probability that the score is between these Z-scores.

03

Sampling Distribution for Sample Mean

For part (b), when we sample 25 students, the sample mean \(\bar{x}\) follows a normal distribution with mean equal to the population mean and standard deviation \(\frac{\sigma}{\sqrt{n}}\), where \(n = 25\). So, \(\bar{x} \sim N(500, \frac{10.4}{\sqrt{25}}) = N(500, 2.08)\).

04

Calculate Probability for Sample Mean

For part (c), calculate the probability that the sample mean is between 495 and 505. We calculate the Z-scores:\[ Z_{495} = \frac{495 - 500}{2.08} = -2.4038 \]\[ Z_{505} = \frac{505 - 500}{2.08} = 2.4038 \]Using the standard normal distribution table or calculator, find the probabilities for these Z-scores and the overall probability that the sample mean is between 495 and 505.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Normal Distribution

A normal distribution is a common way things are spread out. Think of it as a bell-shaped curve on a graph.
In this context, the MCAT scores of students follow this distribution with a mean, or average score, of 500.
The standard deviation, which is a measure of how spread out the scores are, is 10.4.
The bell curve shows us that most students will score near the mean, and fewer students score very high or very low.
This helps us predict or find probabilities about scores even before we see the data.

• The mean (\( \mu \)) is 500, which is where most scores cluster around.
• The standard deviation (\( \sigma \)) tells us about the spread — here it is 10.4 points.
• We compute the probability of a student's score being between 495 and 505 by finding the Z-scores.

Using Z-scores helps us translate these scores into a standard normal distribution, which simplifies probability calculations.

Sampling Distribution

When you collect data from a few students, you're working with a sample instead of the whole population.
This makes life easier when the population is large or hard to access in full.
So what does the distribution of these sample means look like? That's called the Sampling Distribution.Understanding the behavior of sample means is crucial. Here are some key points:

• The sampling distribution of the sample mean \( \bar{x} \), from 25 students, still forms a normal distribution.
• The mean of the sample means will be the same as the population mean, \( \mu = 500 \).
• The standard deviation of the sample mean, known as the standard error, is \( \frac{10.4}{\sqrt{25}} = 2.08 \).

This tells us the average of these sample means should be centered right about where the population mean is, but with less variability.

Probability Calculation

Calculating probabilities using normal distributions is like predicting outcomes. For example, calculating the likelihood that one score, or a sample mean, falls within a certain range helps make decisions.
How do we calculate this? Enter the Z-score.The Z-score tells how far a data point is from the mean in terms of standard deviations. Here's how:

• For a single score between 495 and 505, the Z-scores are \( Z_{495} = \frac{495 - 500}{10.4} = -0.4808 \) and \( Z_{505} = \frac{505 - 500}{10.4} = 0.4808 \).
• For a sample mean, the Z-scores between 495 and 505 are \( Z_{495} = \frac{495 - 500}{2.08} = -2.4038 \) and \( Z_{505} = \frac{505 - 500}{2.08} = 2.4038 \).

Once you have the Z-scores, you can find the probability of these scores happening using a Z-table or calculator.
This gives you a way to express how extreme or typical a score or a set of scores is.

Central Limit Theorem

The Central Limit Theorem (CLT) is a powerful principle in statistics that can really demystify sampling distributions.
It's why we can assume the sampling distribution of the sample mean is normal, even if the population distribution is not. This theorem holds up no matter what the original population distribution looks like, provided the sample size is sufficiently large.

• It tells us that, regardless of the original shape of the distribution of individual scores, the distribution of the sample mean will approach a normal distribution as sample size increases.
• In our exercise, with a sample size of 25 students, the theorem assures that the distribution will be normal with mean 500 and standard error 2.08.

The CLT is a foundation of many statistical methods, allowing for predictions and decisions based on sample data.
It means that averages of samples behave predictably (like a normal distribution), making things much simpler when dealing with large amounts of data.