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The heights of the kindergarten children mentioned in Example 7.6 (p. 328 ) are approximately normally distributed with \(\mu=39\) and \(\sigma=2\). a. If an individual kindergarten child is selected at random, what is the probability that he or she has a height between 38 and 40 inches? b. \(\quad\) A classroom of 30 of these children is used as a sample. What is the probability that the class mean \(\bar{x}\) is between 38 and 40 inches? c. If an individual kindergarten child is selected at random, what is the probability that he or she is taller than 40 inches? d. A classroom of 30 of these kindergarten children is used as a sample. What is the probability that the class mean \(\bar{x}\) is greater than 40 inches?

Short Answer

Expert verified
a. The probability that a random individual child has a height between 38 and 40 inches is approx 0.383. b. The probability that the mean height of a classroom of 30 children lies between 38 and 40 inches is nearly 1. c. The probability that a random individual child is taller than 40 inches is approx 0.3085. d. The probability that the mean height of a classroom of 30 children exceeds 40 inches is almost 0.

Step by step solution

01

Compute Z-scores for Individual Child

Z-score standardizes a data point from a normal distribution. Z-score for a value x is computed using formula \(Z = (x - \mu) / \sigma\). Here, for a value 38: \(Z = (38 - 39) / 2 = -0.5\) and for a value 40: \(Z = (40 - 39) / 2 = 0.5\).
02

Compute Probability for Individual Child

The probability between two Z-Scores can be calculated from the standard normal distribution table. For the range Z=-0.5 to Z=0.5, the probability is 0.1915 + 0.1915 = 0.383 using the Z-table.
03

Compute Z-scores for Sample Mean

When dealing with sample mean, the formula to compute Z-Score changes slightly due to Central Limit Theorem (CLT). It becomes \(Z = (x - \mu) / (\sigma/\sqrt{n})\), where n represents sample size. Here, for a value 38: \(Z = (38 - 39) / (2/\sqrt{30}) = -4.08\) and for a value 40: \(Z = (40 - 39) / (2/\sqrt{30}) = 4.08\).
04

Compute Probability for Sample Mean

Again, using the Z-table, but this time for the range Z=-4.08 to Z=4.08, the probability is nearly 1 (as it's very certain the mean height is between 38 and 40 inches).
05

Compute Z-score for Child Taller than 40 inches

The taller than 40 inches condition is translated into Z > 0.5.
06

Compute Probability Child Taller than 40 inches

The probability Z > 0.5 is the complement of Z < 0.5. Hence, the probability is 0.5 - 0.1915 = 0.3085.
07

Compute Z-score for Sample Mean > 40 inches

This condition is translated into Z > 4.08.
08

Compute Probability Sample Mean > 40 inches

This probability is almost 0. It's very unlikely for the mean height to be over 40 inches, considering the provided mean and standard deviation.

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Key Concepts

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

Z-Score Calculation
The z-score is a crucial concept in statistics that allows us to understand how far away a particular data point is from the mean of a dataset. In the context of a normal distribution, which is symmetrically shaped and characterized by a bell curve, the z-score can tell us exactly how many standard deviations our data point is from the mean.

Take the example of kindergarten children’s heights. To determine how unusual a child's height is within the population, we calculate the z-score using the formula:
column_startcolumn_endZ = (x - \text{\mu}) / \text{\sigma}

where \(x\) is the data point (child's height), \(\mu\) is the mean average height, and \(\sigma\) is the standard deviation. By calculating the z-score, we transform the individual heights into standardized values that can be compared across different scales. This process is essential when we attempt to determine the probability of a child falling within a certain height range or being taller than a specific height. Knowing how to calculate and interpret z-scores is a key skill in various fields from education to professional research.

When solving exercises that involve z-scores, remember that a positive z-score indicates a value above the mean, and a negative z-score points to a value below the mean. Additionally, larger absolute values of the z-score mean that the data point is further from the mean, which could indicate a more unusual observation within the given distribution.
Central Limit Theorem
The Central Limit Theorem (CLT) is a fundamental principle in the field of statistics that describes the shape of the distribution of sample means. It tells us that, regardless of the population's distribution, as long as the sample size is sufficiently large, the distribution of the sample means will tend to be normally distributed.

Understanding the CLT is critical when we work with samples, as is the case when calculating the probability that the class mean height is between two values. According to CLT, the distribution of the mean of the 30 children's heights can be approximated by a normal distribution with the same mean as the population (\(\mu\)) and a standard error equal to the population's standard deviation divided by the square root of the sample size (\(\sigma / \sqrt{n}\)).

This results in a new z-score formula for sample means:
\[Z = (\bar{x} - \mu) / (\sigma/\sqrt{n})\]

With the CLT providing this foundation, we're able to use the same standard normal distribution tables that apply to individual data points to now explore the behavior of sample means. Moreover, the theorem provides assurance of the stability of results when sampling repeatedly, which is why it's so widely relied upon in various disciplines such as economics, psychology, and biology to infer population parameters.
Standard Normal Distribution Table
The standard normal distribution table, also known as the z-table, is a reference for statisticians and students alike to find the probability of a data point falling within certain areas of a standard normal distribution. This table is predicated on the standard normal distribution, characterized by a mean (\(\text{\mu}\)) of 0 and a standard deviation (\(\text{\sigma}\)) of 1.

When we have calculated our z-scores, we turn to this table to determine probabilities. For example, if we wish to know the likelihood of a kindergarten child being between 38 and 40 inches tall, we first compute the z-scores for these heights. Then, by looking up these z-scores in the standard normal distribution table, we find the probabilities associated with each z-score. Finally, we can add or subtract these probabilities to find the overall chance of a height falling within a range or being beyond a specific value.

The standard normal distribution table is incredibly useful because it transforms complex integration problems into a simple lookup task. However, it's essential to comprehend the key aspects of z-scores and how they relate to the areas under the curve of a normal distribution before utilizing the table. This way, the process is not just a mechanical procedure but a meaningful analysis of data within a statistical framework.

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Most popular questions from this chapter

a. Use a computer to draw 500 random samples, each of size \(20,\) from the normal probability distribution with mean 80 and standard deviation 15. b. Find the mean for each sample. c. Construct a frequency histogram of the 500 sample means. d. Describe the sampling distribution shown in the histogram in part \(c,\) including the mean and standard deviation.

A researcher wants to take a simple random sample of about \(5 \%\) of the student body at each of two schools. The university has approximately 20,000 students, and the college has about 5000 students. Identify each of the following as true or false and justify your answer. a. The sampling variability is the same for both schools. b. The sampling variability for the university is higher than that for the college. c. The sampling variability for the university is lower than that for the college. d. No conclusion about the sampling variability can be stated without knowing the results of the study.

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