91Ó°ÊÓ

Suppose that when the \(\mathrm{pH}\) of a certain chemical compound is \(5.00\), the \(\mathrm{pH}\) measured by a randomly selected beginning chemistry student is a random variable with mean \(5.00\) and standard deviation 2. A large batch of the compound is subdivided and a sample given to each student in a morning lab and each student in an afternoon lab. Let \(\bar{X}=\) the average \(\mathrm{pH}\) as determined by the morning students and \(\bar{Y}=\) the average \(\mathrm{pH}\) as determined by the afternoon students. a. If \(\mathrm{pH}\) is a normal variable and there are 25 students in each lab, compute \(P(-.1 \leq \bar{X}-\bar{Y} \leq .1)\). [Hinr: \(\bar{X}-\bar{Y}\) is a linear combination of normal variables, so is normally distributed. Compute \(\mu_{\bar{X}-\bar{Y}}\) and \(\sigma_{\bar{X}-\bar{Y}}\).] b. If there are 36 students in each lab, but pH determinations are not assumed normal, calculate (approximately) \(P(-.1 \leq \bar{X}-\bar{Y} \leq .1)\).

Step by step solution

01

Understand the variables and distributions

Given that the pH measured by students is a normal random variable with mean \(5.00\) and standard deviation \(2\), we're considering \(\bar{X}\) and \(\bar{Y}\) as sample means from the morning and afternoon labs, respectively. Each lab has 25 students. The distribution of \(\bar{X}\) and \(\bar{Y}\) will also be normal by the Central Limit Theorem, with parameters based on the number of students.

02

Calculate mean and standard deviation of \(\bar{X}\) and \(\bar{Y}\)

The mean of \(\bar{X}\) and \(\bar{Y}\) is the same as the population mean: \(\mu_{\bar{X}} = \mu_{\bar{Y}} = 5.00\). The standard deviation is given by \(\sigma_{\bar{X}} = \sigma_{\bar{Y}} = \frac{2}{\sqrt{25}} = 0.40 \).

03

Calculate mean and standard deviation of \(\bar{X} - \bar{Y}\)

Since \(\bar{X}\) and \(\bar{Y}\) have the same mean, \(\mu_{\bar{X} - \bar{Y}} = \mu_{\bar{X}} - \mu_{\bar{Y}} = 0\). The standard deviation is \(\sigma_{\bar{X} - \bar{Y}} = \sqrt{\sigma_{\bar{X}}^2 + \sigma_{\bar{Y}}^2} = \sqrt{0.40^2 + 0.40^2} = 0.566\).

04

Calculate the probability for normal distribution (Part a)

Using the standard normal distribution and the found standard deviation \(0.566\), convert the interval \(-0.1 \leq \bar{X} - \bar{Y} \leq 0.1\) to z-scores. \[ z = \frac{x - \mu}{\sigma} \Rightarrow z_1 = \frac{-0.1 - 0}{0.566}, z_2 = \frac{0.1 - 0}{0.566} \] results in \(z_1 = -0.177\) and \(z_2 = 0.177\). Find \(P(-0.177 \leq Z \leq 0.177)\) using z-tables or normal distributions tools, \(P \approx 0.141\).

05

Re-evaluate with Central Limit Theorem (Part b)

For 36 students, use the CLT to approximate these distributions as normal even without assuming a normal original population. The standard deviation updates to \(\sigma_{\bar{X} - \bar{Y}} = \sqrt{\left(\frac{2}{\sqrt{36}}\right)^2 + \left(\frac{2}{\sqrt{36}}\right)^2} = 0.471\), resulting in new z-scores \(z_1 = \frac{-0.1}{0.471} = -0.212\) and \(z_2 = \frac{0.1}{0.471} = 0.212\). Using normal distribution tools: \(P(-0.212 \leq Z \leq 0.212) \approx 0.168\).

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Key Concepts

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

Normal Distribution

The concept of a normal distribution is crucial for understanding many statistical analyses, including this exercise. Normal distribution, also known as Gaussian distribution, is a bell-shaped curve symmetrical around its mean. This means that data tends to cluster around a central point and taper off symmetrically on both sides.

Key characteristics include:

• The mean, median, and mode of a normal distribution are all equal.
• The curve is defined mathematically by its mean (average) and standard deviation.
• Approximately 68% of the data within a normal distribution lies within one standard deviation from the mean, 95% within two, and 99.7% within three, a rule commonly referred to as the Empirical Rule or the 68-95-99.7 Rule.

Understanding these properties helps in analyzing data that fits this model, as it allows for predictions and estimations about probabilities within that dataset.

Standard Deviation

Standard deviation is a measure of the amount of variation or dispersion in a set of values. It tells us how much the individual data points deviate from the mean value of the dataset.

In a normal distribution, the standard deviation is incredibly informative:

• It helps determine the spread and clustering of the data around the mean.
• A small standard deviation indicates that the data points are close to the mean, while a large standard deviation signifies that they are spread out over a wider range.

Mathematically, the standard deviation is the square root of the variance, which is the average of the squared differences from the mean. In this example, the standard deviation assists in finding the spread of the sample means and is pivotal in calculating probabilities using z-scores.

Sample Means

Sample means refer to the average values computed from a sample of observations taken from a larger population. In this exercise, \\(\bar{X}\) and \\(\bar{Y}\) represent the sample means of pH measurements by morning and afternoon lab students respectively.

Calculating the sample mean from a set of data is straightforward and involves summing all observations and dividing by the number of observations. Importantly, the Central Limit Theorem (CLT) indicates that, regardless of the population's distribution, the distribution of sample means approaches a normal distribution as the sample size increases, typically becoming normal for samples sizes of 30 or more.

• This property makes the sample mean a useful estimator for population parameters.
• Sample means allow us to construct confidence intervals and hypothesis tests.

Using the sample mean simplifies complex datasets, making it easier to draw meaningful conclusions about a larger population.

Z-scores

A z-score measures how many standard deviations an element is from the mean. It is a dimensionless quantity, meaning it has no units and expresses distance from the mean in terms of the number of standard deviations.

This is essential for standardizing different data sets.

• A positive z-score indicates the data point is above the mean.
• A negative z-score indicates it is below the mean.
• A z-score of 0 signifies that the data point is exactly on the mean.

In the given exercise, converting the range of differences between the sample means to z-scores helps quantify the probability of those differences occurring, leveraging normal distribution tables to find probable outcomes effectively. This conversion simplifies the task of comparing data against standard normal distributions, which is why z-scores are widely used in statistical analyses.