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The scores on the Psychomotor Development Index (PDI), a scale of infant development, have a normal population distribution with mean 100 and standard deviation 15. An infant is selected at random. a. Find the \(z\) -score for a PDI value of 90 . b. A study uses a random sample of 225 infants. Using the sampling distribution of the sample mean PDI, find the \(z\) -score corresponding to a sample mean of 90 . c. Explain why a PDI value of 90 is not surprising, but a sample mean PDI score of 90 for 225 infants would be surprising.

Short Answer

Expert verified
A PDI value of 90 is common for one infant, but very rare as an average for 225 infants.

Step by step solution

01

Understanding the Z-score Formula

The Z-score of a value tells us how many standard deviations away it is from the mean. It is calculated using the formula:\[ z = \frac{x - \mu}{\sigma} \]where \(x\) is the value, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.
02

Calculating the Z-score for a Single PDI Value

To find the Z-score for a PDI value of 90:\[ z = \frac{90 - 100}{15} = \frac{-10}{15} = -0.67 \]Thus, the Z-score for a PDI value of 90 is \(-0.67\).
03

Setting Up the Sampling Distribution

When using a sample, the standard deviation of the sampling distribution of the sample mean is given by the standard error,\[ SE = \frac{\sigma}{\sqrt{n}} \]where \(n\) is the sample size. Here, \(\sigma = 15\) and \(n = 225\).
04

Calculating the Standard Error

Calculate the standard error using the sample size:\[ SE = \frac{15}{\sqrt{225}} = \frac{15}{15} = 1 \]So, the standard error is 1.
05

Calculating the Z-score for the Sample Mean

Now, find the Z-score for the sample mean of 90:\[ z = \frac{90 - 100}{1} = -10 \]Therefore, the Z-score for the sample mean of 90 is \(-10\).
06

Interpreting the Results

A Z-score of \(-0.67\) for a single PDI value of 90 is within one standard deviation from the mean, which is common, implying that a single score of 90 is not surprising. However, a Z-score of \(-10\) for a sample mean indicates a highly unusual event, as it is extremely far from the mean.

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Key Concepts

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

Normal Distribution
In statistics, the concept of a normal distribution is fundamental. It often mirrors real-world situations. Imagine it as a bell-shaped curve where most occurrences take place near the mean. This mean is an average value around which data is distributed.
The essential properties of a normal distribution include:
  • Symmetry: distribution is symmetric about the mean.
  • Mean, median, and mode are equal.
  • Curve is bell-shaped and approaches but never touches the horizontal axis.
A normal distribution is characterized by its mean and standard deviation. These two components give a complete description of the distribution. For example, in the PDI score case, the distribution has a mean of 100 and a standard deviation of 15. This setup allows us to make inferences about where a particular score falls relative to the average.
Z-score
The Z-score is a statistical measurement that tells us how far a particular value is from the mean. It's expressed in terms of standard deviations.
To compute a Z-score for a value, use the formula:\[ z = \frac{x - \mu}{\sigma} \]where:
  • \(x\) is the particular score.
  • \(\mu\) is the mean of the distribution.
  • \(\sigma\) is the standard deviation.
For a PDI value of 90, the Z-score is \(-0.67\), indicating that this score is below the mean but within the typical range one might expect in a normal distribution. In simple terms, a Z-score like \(-0.67\) indicates the value isn't unusual since it's less than one standard deviation away from the mean.
Sampling Distribution
A sampling distribution refers to the probability distribution of a statistic obtained from a large number of samples drawn from a specific population.
In practice, we create a sampling distribution of the sample mean to understand how the mean of data samples from the population varies. When you take a large enough sample, the sampling distribution of the sample mean will tend to have a normal distribution, even if the original data is not perfectly normal. This is known as the Central Limit Theorem.
The sampling distribution helps us understand the precision of our sample mean in representing the population mean. For instance, with a sample size of 225 from the PDI distribution, we can create a distribution of sample means. This distribution has its own mean (same as the population mean) and a smaller standard deviation, known as the standard error.
Standard Error
The standard error provides a way to measure the variability of the sample mean compared to the population mean. It essentially tells us how precisely the sample mean estimates the population mean.
The formula to calculate standard error is:\[ SE = \frac{\sigma}{\sqrt{n}} \]where:
  • \(\sigma\) is the standard deviation of the population.
  • \(n\) is the sample size.
In the given exercise, with a sample size of 225 and a standard deviation of 15, the standard error becomes 1. This small standard error means the sample mean is a very precise estimate of the population mean. Hence, when a sample mean has a Z-score of \(-10\), it becomes a surprising result, indicating that this sample mean is remarkably far from what we would expect for the average sample mean, making it an unusual occurrence.

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Most popular questions from this chapter

The sampling distribution of a sample mean for a random sample size of 100 describes a. How sample means tend to vary from random sample to random sample of size 100 . b. How observations tend to vary from person to person in a random sample of size 100 . c. How the data distribution looks like the population distribution when the sample size is larger than 30 . d. How the standard deviation varies among samples of size 100 .

A car dealer offers a \( 500\) discount to customers if they agree to buy a car immediately without doing further research. Suppose \(30%\) of all customers who visit him accept this offer. Depending on whether or not a given customer accepts the offer, let \(X\) be either 1 or 0 , respectively. a. If \(n=5\) customers, find the probability distribution of the proportion of customers who will accept the offer. (Hint: List all possible values for the sample proportion and their chances of occurring.) b. Referring to part a, what are the mean and standard deviation of the sample proportion? c. Repeat part b for a group of \(n=10\) customers and \(n=100\) customers. d. What happens to the mean and standard deviation of the sample proportion as \(n\) increases?

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How would you explain to someone who has never studied statistics what a sampling distribution is? Explain by using the example of polls of 1000 Canadians for estimating the proportion who think the prime minister is doing a good job.

Based on data from the 2010 Major League Baseball season, \(X=\) number of home runs the San Francisco Giants hit in a game has a mean of 1.0 and a standard deviation of 1.0 . a. Do you think \(X\) has a normal distribution? Why or why not? b. Suppose that this year \(X\) has the same distribution. Report the shape, mean, and standard deviation of the sampling distribution of the mean number of home runs the team will hit in its 162 games. c. Based on the answer to part b, find the probability that the mean number of home runs per game in this coming season will exceed 1.50 .

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