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Chapter 8: Problem 18

Suppose that a sample of size 100 is to be drawn from a population with standard deviation \(10 .\) a. What is the probability that the sample mean will be within 2 of the value of \(\mu\) ? b. For this example \((n=100, \sigma=10)\), complete each of the following statements by computing the appropriate value: i. Approximately 95% of the time, \(\bar{x}\) will be within _____ of \(\mu .\) ii. Approximately 0.3% of the time, \(\bar{x}\) will be farther than _____ from\(\mu .\)

Short Answer

Expert verified
a. The probability that the sample mean will be within 2 of the population mean is approximately 95.44%. b.i. Approximately 95% of the time, the sample mean \(\bar{x}\) will be within 1.96 of \(\mu\). b.ii. Approximately 0.3% of the time, the sample mean \(\bar{x}\) will be farther than 2.75 from \(\mu\).

Step by step solution

01

Understanding the Problem

We know the sample size n=100 and the population standard deviation \(\sigma=10\). We need to find the probability that the sample mean \(\bar{x}\) is within 2 of the population mean \(\mu\). Additionally, we need to determine the values for which \(\bar{x}\) will lie within \(\mu\) 95% of the time, and the value from which \(\bar{x}\) will be farther than \(\mu\) 0.3% of the time.
02

Calculate Standard Error

The standard error (SE) can be calculated using the formula SE = \(\sigma /\sqrt{n}\), where \(\sigma\) is the population standard deviation and \(n\) is the sample size. In this case, \(\sigma=10\) and \(n=100\). Thus, SE = \(10/\sqrt{100}=1\). This standard error means that on average, the sample mean should typically deviate from the population mean by 1.
03

Calculate Z-Scores and Apply Central Limit Theorem

The z-score represents the number of standard errors a data point is away from the mean. To find the probability that the sample mean will be within 2 of the population mean, calculate the z-score corresponding to 2 using the formula z = (X - \(\mu\))/SE. Since we want the sample mean to be within 2 of the population mean, X will be \(\mu\)+2 or \(\mu\)-2. The z-scores associated with these are (+2-0)/1 = 2 and (-2-0)/1 = -2. We want the probability that the sample mean is between these z-scores. From the standard normal table, the value corresponding to a z-score of 2 is 0.9772 and for -2 is 0.0228. So, P(-2 < Z < 2) = 0.9772 - 0.0228 = 0.9544. This means there is approximately a 95.44% chance that the sample mean will be within 2 of the population mean.
04

Determining the Range of \(\bar{x}\) for Given Probabilities

To complete the statements, we first rewind to the z-scores. For 95% of the time (or 0.9500 probability), the z-scores from the standard normal table are approximately -1.96 and 1.96. Therefore, \(\bar{x}\) will be within +- \(1.96*SE =1.96*1 = 1.96\) of \(\mu\). Next, for 0.3% of the time (or 0.0030 probability), the z-scores are approximately -2.75 and 2.75 (since probability for z-scores -2.75 and 2.75 is 0.003). Hence, \(\bar{x}\) will be farther than \(2.75*1 = 2.75\) from \(\mu\).

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Key Concepts

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

Central Limit Theorem
The Central Limit Theorem (CLT) is a fundamental principle in statistics. It states that the distribution of the sample mean will approach a normal distribution as the sample size becomes larger, regardless of the shape of the population distribution. This is particularly useful because the normal distribution is well understood and allows us to make inferences about population parameters.

In the given exercise, a sample of size 100 was drawn. According to the CLT, since the sample size is relatively large, the distribution of the sample mean will be approximately normal. This allows us to apply concepts like the standard normal distribution to calculate probabilities about the sample mean.
Standard Error
The standard error (SE) is a measure of how much the sample mean is expected to fluctuate from the true population mean. It is calculated by dividing the population standard deviation by the square root of the sample size: \( SE = \frac{\sigma}{\sqrt{n}} \).

In our specific case, with a standard deviation \( \sigma = 10 \) and a sample size \( n = 100 \), the standard error is \( 1 \). This small value indicates that, generally, the sample mean will be close to the population mean.

This measurement is crucial when using sample data to estimate population parameters, as it helps quantify the precision of the sample mean.
Z-scores
A z-score indicates how many standard deviations an element is from the mean. It is calculated as \( z = \frac{(X - \mu)}{SE} \), where \( X \) is the sample mean, \( \mu \) is the population mean, and \( SE \) is the standard error.

In the exercise, we computed z-scores to find the probability of the sample mean deviating from the population mean by a certain amount. A z-score of 2 means the sample mean is 2 standard deviations above the population mean, and a z-score of -2 means it is 2 standard deviations below.

Using standard normal distribution tables, these z-scores help determine probabilities, such as the chance that the sample mean falls within a specific range of the population mean.
Probability
Probability is the measure of likelihood that an event will occur. In statistics, we often calculate probabilities to make inferences about population parameters from sample statistics.

In this exercise, we considered the probability that the sample mean is within a range of \( \mu \). This was done by determining the z-scores for the specific deviations (e.g., \( -2 \) and \( +2 \)) and finding the corresponding probability from standard normal tables.

The result, a 95.44% probability, suggests that the sample mean will be close to the population mean a high percentage of the time. Probabilities guide our confidence in statistical estimations and predictions.

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Most popular questions from this chapter

Let \(x_{1}, x_{2}, \ldots, x_{100}\) denote the actual net weights (in pounds) of 100 randomly selected bags of fertilizer. Suppose that the weight of a randomly selected bag has a distribution with mean \(50 \mathrm{lb}\) and variance \(1 \mathrm{lb}^{2}\). Let \(\bar{x}\) be the sample mean weight \((n=100)\). a. Describe the sampling distribution of \(\bar{x}\). b. What is the probability that the sample mean is between \(49.75 \mathrm{lb}\) and \(50.25 \mathrm{lb}\) ? c. What is the probability that the sample mean is less than \(50 \mathrm{lb}\) ?

The amount of money spent by a customer at a discount store has a mean of $$\$ 100$$ and a standard deviation of $$\$ 30.$$ What is the probability that a randomly selected group of 50 shoppers will spend a total of more than $$\$ 5300 ?$$ (Hint: The total will be more than \(\$ 5300\) when the sample average exceeds what value?)

A manufacturer of computer printers purchases plastic ink cartridges from a vendor. When a large shipment is received, a random sample of 200 cartridges is selected, and each cartridge is inspected. If the sample proportion of defective cartridges is more than \(.02\), the entire shipment is returned to the vendor. a. What is the approximate probability that a shipment will be returned if the true proportion of defective cartridges in the shipment is .05? b. What is the approximate probability that a shipment will not be returned if the true proportion of defective cartridges in the shipment is .10?

Newsweek (November 23, 1992) reported that 40\% of all U.S. employees participate in "self-insurance" health plans \((\pi=.40)\). a. In a random sample of 100 employees, what is the approximate probability that at least half of those in the sample participate in such a plan? b. Suppose you were told that at least 60 of the \(100 \mathrm{em}\) ployees in a sample from your state participated in such a plan. Would you think \(\pi=.40\) for your state? Explain.

Let \(x\) denote the time (in minutes) that it takes a fifth-grade student to read a certain passage. Suppose that the mean value and standard deviation of \(x\) are \(\mu=2 \mathrm{~min}\) and \(\sigma=0.8\) min, respectively. a. If \(\bar{x}\) is the sample average time for a random sample of \(n=9\) students, where is the \(\bar{x}\) distribution centered, and how much does it spread out about the center (as described by its standard deviation)? b. Repeat Part (a) for a sample of size of \(n=20\) and again for a sample of size \(n=100\). How do the centers and spreads of the three \(\bar{x}\) distributions compare to one another? Which sample size would be most likely to result in an \(\bar{x}\) value close to \(\mu\), and why?
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