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Suppose that the mean value of interpupillary distance (the distance between the pupils of the left and right eyes) for adult males is \(65 \mathrm{~mm}\) and that the population standard deviation is \(5 \mathrm{~mm}\). a. If the distribution of interpupillary distance is normal and a random sample of \(n=25\) adult males is to be selected, what is the probability that the sample mean distance \(\bar{x}\) for these 25 will be between 64 and \(67 \mathrm{~mm}\) ? At least \(68 \mathrm{~mm}\) ? b. Suppose that a random sample of 100 adult males is to be selected. Without assuming that interpupillary distance is normally distributed, what is the approximate probability that the sample mean distance will be between 64 and 67 \(\mathrm{mm} ?\) At least \(68 \mathrm{~mm} ?\)

Step by step solution

01

Establish the given data

In this problem, the population mean \(\mu\) is 65mm, the population standard deviation \(\sigma\) is 5mm, and the sample size \(n\) is either 25 or 100.

02

Calculate the standard error and Z-score for part a

First, we calculate the standard error (SE) which is \(\sigma / \sqrt{n}\). Next, Z-score is calculated as \((mean - \mu) / SE\), for each boundary point - in this case, for 64mm and 67mm, and for \(at \ least \ 68mm\). The Z-score represents how many standard deviations our data are from the mean.

03

Find Probability for Part a - Sample Size of 25

The probabilities that the sample mean will be between 64mm and 67mm or at least 68mm are found by checking these z-scores on Z-table which gives us the probability under the normal curve up to that calculated z-score point.

04

Calculate the standard error and Z-score for Part b - Sample Size of 100

Repeat the procedure of Step 2 for the sample size of 100.

05

Find Probability for Part b - Sample Size of 100

Repeat the procedure of Step 3 for the new standard errors and calculated Z scores for the sample size of 100. Here we will use the Central Limit theorem to approximate the normal distribution because the distribution is not stated to be normal.

06

Interpret Results

The calculated probabilities represent the likelihood of the sample means respectively being between 64mm and 67mm or being at least 68mm. The larger the sample size, the more the distribution approaches a normal distribution (Central Limit Theorem).

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Key Concepts

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

Sample Mean

The sample mean, denoted \(\bar{x}\), is the average value of a set of observations from a sample. In statistics, it is a critical measure because it provides an estimate of the population mean, \(\mu\).

When working with a sample, you calculate the sample mean by summing up all the observed values and dividing by the number of observations \(n\). Mathematically, this is expressed as:\[ \bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i \]where \(x_i\) represents each observed value.

• **Representative Estimate:** The sample mean serves as an estimate of the actual population mean.
• **Sampling Variability:** Different samples will produce different sample means, showing sampling variability.

Computing the sample mean is a foundational step in inferential statistics because it helps predict the behavior of the population based on the sample data.

Central Limit Theorem

The Central Limit Theorem (CLT) is a key principle in statistics. It states that if you have a large enough sample size, the distribution of the sample mean will be approximately normally distributed, regardless of the population's original distribution.

This principle is powerful because it allows statisticians to make inferences about populations using sample data. The approximation improves as the sample size increases.

• **Sample Size Requirement:** Usually, a sample size \(n\) of 30 or more is considered sufficient for the CLT to apply.
• **Assumptions:** The theorem holds for independent and identically distributed (i.i.d.) samples.

This theorem is particularly helpful when the population distribution is unknown, as it was used in the exercise to approximate the probability of the sample mean for a sample size of 100, allowing us to assume it follows a normal distribution.

Normal Distribution

The normal distribution, also known as the Gaussian distribution, is a bell-shaped curve that is symmetrical around its mean. It is defined by its mean \(\mu\) and standard deviation \(\sigma\).

This distribution is crucial in statistics because many natural phenomena tend to approximate a normal distribution when you sample enough data.

• **Symmetrical Shape:** The curve is symmetric around the mean, meaning the probability of observing a value is equally likely on both sides of the mean.
• **68-95-99.7 Rule:** Approximately 68% of values lie within 1 standard deviation of the mean, 95% within 2, and 99.7% within 3.

In context, knowing that interpupillary distances are normally distributed allowed us to find probabilities directly using Z-scores and tables in the step-by-step solution.

Standard Deviation

The standard deviation, denoted as \(\sigma\), measures the amount of variation or dispersion in a set of values. It is the square root of the variance and provides insight into the spread of the data.

A smaller standard deviation indicates that the data points tend to be close to the mean, whereas a larger standard deviation indicates that the data points are spread out over a wider range of values.

• **Essential Measure:** Standard deviation is vital for understanding data consistency.
• **Empirical Rule:** For a normal distribution, it determines how much data lies within specific intervals around the mean.

In the exercise, knowing the standard deviation of interpupillary distance was crucial for calculating the standard error, which was then used to find the probability of sample means within specific ranges.