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The College Board American College Testing Program reported a population mean SAT score of \(\mu=1020\) (The World Almanac 2003). Assume that the population standard deviation is \(\sigma=100\) a. What is the probability that a random sample of 75 students will provide a sample mean SAT score within 10 of the population mean? b. What is the probability a random sample of 75 students will provide a sample mean SAT score within 20 of the population mean?

Short Answer

Expert verified
(a) 0.6156, (b) 0.9164

Step by step solution

01

Understand the Problem

We are given that the population mean SAT score \( \mu = 1020 \) and the population standard deviation \( \sigma = 100 \). We need to find the probability that the sample mean is within a certain range from the population mean using a sample of 75 students. First, calculate the standard error of the mean.
02

Calculate the Standard Error of the Mean

The standard error (SE) of the mean can be calculated using the formula: \( \text{SE} = \frac{\sigma}{\sqrt{n}} \), where \( \sigma = 100 \) and \( n = 75 \). Calculate:\[ \text{SE} = \frac{100}{\sqrt{75}} \approx 11.55 \]
03

Determine Z-scores for Part (a)

For part (a), we want the sample mean to be within 10 of the population mean, so the sample mean should be in the interval \[1010, 1030\]. Calculate the Z-scores:For \( x = 1010 \) :\[ z = \frac{1010 - 1020}{11.55} \approx -0.87 \]For \( x = 1030 \) :\[ z = \frac{1030 - 1020}{11.55} \approx 0.87 \]
04

Find Probability for Part (a)

Using the standard normal distribution table, find the probabilities corresponding to \( z = -0.87 \) and \( z = 0.87 \). \( P(Z < -0.87) \approx 0.1922 \)\( P(Z < 0.87) \approx 0.8078 \)Thus, \( P(1010 < \bar{x} < 1030) = P(Z < 0.87) - P(Z < -0.87) \approx 0.8078 - 0.1922 = 0.6156 \)
05

Determine Z-scores for Part (b)

For part (b), we want the sample mean to be within 20 of the population mean, so the sample mean should be in the interval \[1000, 1040\]. Calculate the Z-scores:For \( x = 1000 \) :\[ z = \frac{1000 - 1020}{11.55} \approx -1.73 \]For \( x = 1040 \) :\[ z = \frac{1040 - 1020}{11.55} \approx 1.73 \]
06

Find Probability for Part (b)

Using the standard normal distribution table, find the probabilities corresponding to \( z = -1.73 \) and \( z = 1.73 \). \( P(Z < -1.73) \approx 0.0418 \)\( P(Z < 1.73) \approx 0.9582 \)Thus, \( P(1000 < \bar{x} < 1040) = P(Z < 1.73) - P(Z < -1.73) \approx 0.9582 - 0.0418 = 0.9164 \)

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Key Concepts

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

Probability
Probability is a way to express the likelihood of an event occurring. In scientific studies, it helps to understand and predict outcomes. When you say the probability of rolling a die and getting a four is 1/6, it means that with one roll, there is one way to get a four and six possible outcomes.
In the context of our exercise, probability is used to determine how likely it is for the sample mean of SAT scores to fall within a certain range around the population mean. Here, we are calculating the probability for two scenarios: the sample mean being within 10 and 20 points of the population mean of 1020 points. To find these probabilities, we use statistical tools like Z-scores and the normal distribution curve. Understanding probabilities helps gauge the reliability and validity of the sample mean to represent the population mean.
Standard Error
The Standard Error (SE) is a measure of the variability or dispersion of a sample mean from the population mean. In simpler terms, it shows how far the sample mean might deviate from the actual population mean.
The formula to calculate SE is \( \frac{\sigma}{\sqrt{n}} \), where \( \sigma \) is the standard deviation of the population, and \( n \) represents the sample size. For our example, the population standard deviation is 100, and the sample size is 75. Therefore, the SE is approximately 11.55.
This calculation tells us about the precision of our sample mean as an estimate of the population mean. A smaller SE indicates a more accurate estimate. Understanding SE is crucial for statistical inference as it assists in making predictions and decisions based on data analysis.
Normal Distribution
The normal distribution, often represented as a bell curve, is a common probability distribution in statistics. It is symmetric around the mean, indicating that data near the mean are more frequent in occurrence than data far from the mean.
One key feature of normal distribution is that about 68% of data points fall within one standard deviation of the mean, about 95% fall within two standard deviations, and about 99.7% fall within three standard deviations. This property makes it incredibly useful for statistical analysis, like predicting outcomes and understanding data behavior.
In our exercise, we employ the normal distribution to calculate the probability of the sample mean falling within a specific range. Using the curve, we interpret Z-scores to find these probabilities. The normal distribution aids us in understanding how likely a sample mean is to deviate from the true population mean.
Z-score
The Z-score is a metric that quantifies the number of standard deviations a data point is from the population mean. It converts individual scores onto a standard scale, taking into account standard deviation and mean.
To calculate a Z-score, the formula used is: \( z = \frac{x - \mu}{\text{SE}} \). Here, \( x \) is the data point, \( \mu \) is the mean, and SE is the standard error. For example, a score of 1030 with a population mean of 1020 and an SE of 11.55 gives a Z-score of about 0.87.
Z-scores enable statisticians to compare data points from different distributions or within the same distribution. In this exercise, Z-scores are pivotal for determining the probability of the sample mean being close to the population mean. With them, we convert scores onto the normal distribution table to find probabilities, hence making informed statistical predictions.

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Most popular questions from this chapter

Suppose a simple random sample of size 50 is selected from a population with \(\sigma=10\) Find the value of the standard error of the mean in each of the following cases (use the finite population correction factor if appropriate). a. The population size is infinite. b. The population size is \(N=50,000\). c. The population size is \(N=5000\). d. The population size is \(N=500\).

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