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Chapter 15: Problem 20

Scores on the Critical Reading part of the SAT exam in a recent year were roughly Normal with mean 495 and standard deviation 118. You choose an SRS of 100 students and average their SAT Critical Reading scores. If you do this many times, the standard deviation of the average scores you get will be close to (a) 118 . (b) \(118 / 100=1.18\). (c) \(118 / 100=11.8^{118 / \sqrt{100}}=11.8\)

Short Answer

Expert verified
The correct answer is (c) 11.8.

Step by step solution

01

Identify Known Quantities

We know that the SAT Critical Reading scores are normally distributed with a mean \(\mu = 495\) and a standard deviation \(\sigma = 118\). We also have a sample size \(n = 100\).
02

Apply Central Limit Theorem

According to the Central Limit Theorem, the distribution of the sample mean will be approximately normal with mean \(\mu = 495\) and the standard deviation of the sample mean, called the standard error, is given by \(\frac{\sigma}{\sqrt{n}}\).
03

Calculate Standard Deviation of Sample Mean

Use the formula for standard error: \(\text{Standard Error} = \frac{\sigma}{\sqrt{n}} = \frac{118}{\sqrt{100}}\). Simplify to get \(\frac{118}{10} = 11.8\).
04

Select the Correct Option

The calculated standard deviation of the average scores is 11.8, which corresponds to option (c).

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Key Concepts

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

Normal Distribution
Normal Distribution is a common pattern in the world of statistics, often used to represent real-world data. It is characterized by its bell-shaped curve.

In the case of the SAT exam scores, the normal distribution implies that most students score around the average mark, with fewer students obtaining extremely high or low scores.
  • In the example of the SAT Critical Reading scores, these are normally distributed with a mean (\( \mu \)) of 495 and a standard deviation (\( \sigma \)) of 118.
This representation helps in predicting the probability of a certain score occurring within the distribution.
The Central Limit Theorem further suggests that even if the original scores are not perfectly normal, the averages of samples will tend towards a normal distribution as the sample size (\( n \)) becomes large.
sample mean
The sample mean is a critical term in statistics, particularly when dealing with large data sets like the SAT exam scores. It is the average score calculated from a sample of students.
  • The formula for the sample mean is:\[ \bar{x} = \frac{\sum_{i=1}^{n} x_i}{n}\]where \(n\) is the number of data points, and\( x_i \)is each individual score in the sample.
The importance of the sample mean is that it provides an estimate of the population mean.

According to the Central Limit Theorem, when you take many samples and calculate their averages, the distribution of these means will approach a normal distribution, regardless of the original distribution.
This plays a crucial role when determining how SAT scores could vary among different groups of students.
standard error
Standard error gives an idea of the variability of the sample mean compared to the population mean.
  • It is calculated using the formula:\[ \text{Standard Error (SE)} = \frac{\sigma}{\sqrt{n}}\]where:
    \(\sigma\) is the standard deviation of the population,
    \(n\) is the sample size.
For the SAT scores example, the standard error helps us understand how much the average score of the group of students (sample) is expected to differ from the overall average (population mean).

In our exercise, using a standard deviation of 118 and a sample size of 100, the standard error calculates to 11.8. This indicates precision in our estimate of the sample mean, illustrating lesser variability.
SAT exam scores
SAT exam scores are used as a measure of a student's readiness for college. In this example, we focus on the Critical Reading part of the SAT exam scores.
  • The given problem states that these scores are normally distributed with a mean score of 495 and a standard deviation of 118.
  • These scores allow us to apply statistical concepts such as Normal Distribution, sample mean, and standard error to determine the reliability and differences in students' performances.
The SAT scores are collected into a larger data set where educators can analyze patterns among students nationwide.

Analyzing these scores using the concept of normal distribution lets us understand the standard performance and predict outcomes based on sample averages.
Interpreting SAT results using statistical methods is beneficial, particularly for grouping statistics over many years to draw conclusions about trends and changes in educational standards.

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Most popular questions from this chapter

Statistics anxiety. What can teachers do to alleviate statistics anxiety in their students? To explore this question, statistics anxiety for students in two classes was compared. In one class, the instructor lectured in a formal manner, including dressing formally. In the other, the instructor was less formal, dressed informally, was more personal, used humor, and called on students by their first names. Anxiety was measured using a questionnaire. Higher scores indicate a greater level of anxiety. The mean anxiety score for students in the formal lecture class was \(25.40\); in the informal class the mean was 20.41. For each of the boldface numbers, indicate whether it is a parameter or a statistic. Explain your answer.

Glucose testing, continued. Shelia's measured glucose level one hour after having a sugary drink varies according to the Normal distribution with \(\mu=122\) \(\mathrm{mg} / \mathrm{dL}\) and \(\sigma=12 \mathrm{mg} / \mathrm{dL}\). What is the level \(L\) such that there is probability only \(0.05\) that the mean glucose level of four test results falls above L? (Hint: This requires a backward Normal calculation. See page 91 in Chapter 3 if you need to review.)

Testing glass. How well materials conduct heat matters when designing houses. As a test of a new measurement process, 10 measurements are made on pieces of glass known to have conductivity 1 . The average of the 10 measurements is 1.07. For each of the boldface numbers, indicate whether it is a parameter or a statistic. Explain your answer.

Playing the numbers: A gambler gets chance outcomes. The law of large numbers tells us what happens in the long run. Like many games of chance, the numbers racket has outcomes that vary considerably-one three-digit number wins \(\$ 600\) and all others win nothing - that gamblers never reach "the long run." Even after many bets, their average winnings may not be close to the mean. For the numbers racket, the mean payout for single bets is \(\$ 0.60\) (60 cents) and the standard deviation of payouts is about \(\$ 18.96\). If Joe plays 350 days a year for 40 years, he makes 14,000 bets. (a) What are the mean and standard deviation of the average payout \(\mathrm{x}^{-} \bar{x}\) that Joe receives from his 14,000 bets? (b) The central limit theorem says that his average payout is approximately Nomal with the mean and standard deviation you found in part (a). What is the approximate probability that Joe's average payout per bet is between \(\$ 0.50\) and \(\$ 0.70\) ? You see that Joe's average may not be very close to the mean \(\$ 0.60\) even after 14,000 bets.

Larger Sample, More Accurate Estimate. Suppose that, in fact, the total cholesterol level of all men aged 20-34 follows the Normal distribution with mean \(\mu=182\) milligrams per deciliter (mg/dL) and standard deviation \(\sigma=37\) \(\mathrm{mg} / \mathrm{dL}\). (a) Choose an SRS of 100 men from this population. What is the sampling distribution of \(\mathrm{x}^{-} \vec{x}\) ? What is the probability that \(\mathrm{x}^{-} \vec{x}\) takes a value between 180 and \(184 \mathrm{mg} / \mathrm{dL}\) ? This is the probability that \(\mathrm{x}^{-} \bar{x}\) estimates \(\mu\) within \(\pm 2\) \(\mathrm{mg} / \mathrm{dL}\). (b) Choose an SRS of 1000 men from this population. Now what is the probability that \(\mathrm{x}^{-}=\)falls within \(\pm 2 \mathrm{mg} / \mathrm{dL}\) of \(\mu\) ? The larger sample is much more likely to give an accurate estimate of \(\mu\).
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