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.A Sur prising Calculation. Changing the mean and standard deviation of a Normal distribution by a moderate amount can greatly change the percentage of observations in the tails. Suppose a college is looking for applicants with either SAT Math or Evidence-Based Reading and Writing (ERW) scores 780 and above. a. In 2018, the scores on the math SAT followed the \(N(528,117)\) distribution. What percentage scored 780 or better? b. The ERW scores that year had the \(N(531,104)\) distribution. What percentage scored 780 or better? You see that the percentage of students with math SAT scores above 780 is almost two times the percentage of students with such high ERW scores.

Short Answer

Expert verified
1.54% for Math and 0.83% for ERW.

Step by step solution

01

Understand the Distribution

For part (a), the SAT Math scores follow a normal distribution with a mean of 528 and a standard deviation of 117. In part (b), the ERW scores have a mean of 531 and a standard deviation of 104. We need to calculate the percentage of scores 780 or better in each case.
02

Calculate the Z-score for Math

For the math SAT scores, the Z-score can be calculated using the formula: \[ Z = \frac{X - \mu}{\sigma} \]where \( X = 780 \), \( \mu = 528 \), and \( \sigma = 117 \).\[ Z_{math} = \frac{780 - 528}{117} = \frac{252}{117} = 2.1547 \]
03

Find the Percentage for Math

Using the standard normal distribution table, find the probability of a Z-score greater than 2.1547. This value corresponds to the area to the right of the Z-score in the standard normal distribution. The table gives us approximately 0.9846, corresponding to the left tail, so \[ P(Z > 2.1547) = 1 - 0.9846 = 0.0154 \]Thus, 1.54% of students scored 780 or better in math.
04

Calculate the Z-score for ERW

For the ERW scores, use the same Z-score formula:\[ Z_{ERW} = \frac{780 - 531}{104} = \frac{249}{104} = 2.3942 \]
05

Find the Percentage for ERW

Using the standard normal distribution table again, find the probability of a Z-score greater than 2.3942. This value corresponds to the area to the right of the Z-score in the standard normal distribution. The table gives us approximately 0.9917 for the left tail,\[ P(Z > 2.3942) = 1 - 0.9917 = 0.0083 \]Thus, 0.83% of students scored 780 or better in ERW.

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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, especially when dealing with normal distributions. It helps us understand how far a specific data point is from the mean of a distribution, in terms of standard deviations.

The formula for calculating a Z-score is:
  • \( Z = \frac{X - \mu}{\sigma} \)
where:
  • \( X \) is the value you'd like to compare,
  • \( \mu \) is the mean of the distribution,
  • \( \sigma \) is the standard deviation.
Once you have a Z-score, you can use it to find probabilities and percentages using a standard normal distribution table. This helps determine how unusual or common a score is within a given distribution.
SAT scores
The SAT, or Scholastic Assessment Test, is a standardized test commonly used for college admissions in the United States. It comprises several sections, typically including Math and Evidence-Based Reading and Writing (ERW). Each section is scored separately, and these scores are often analyzed using statistical methods.

In our example, we are interested in how many students scored 780 or higher in the SAT Math and ERW sections. The scores are normally distributed, which enables us to use the properties of the normal distribution to calculate probabilities of obtaining certain scores. By understanding the distributions of these scores, colleges can set benchmarks and compare applicants effectively.
standard deviation
Standard deviation is a measure that describes the amount of variation or dispersion in a set of values. In simpler terms, it tells us how spread out numbers are in a dataset.

A smaller standard deviation means the values are clustered closely around the mean, while a larger one indicates that they are spread out over a wider range of values.

In a normal distribution, about 68% of values lie within one standard deviation of the mean, about 95% lie within two standard deviations, and about 99.7% fall within three standard deviations. Understanding standard deviation helps in knowing the extent of normal deviations you might expect from the mean.
probability
Probability is a measure of the likelihood that a particular event will happen. In the context of normal distributions, probability helps us find out how likely it is that a score falls within a certain range.

We use the Z-score and the properties of the normal distribution to calculate these probabilities. By looking up a Z-score in a standard normal distribution table, you can determine the probability associated with a specific score.

For example, in our SAT scenario, we are interested in finding the probability that scores are 780 or above. The normal distribution table tells us the area to the left of the Z-score, and we subtract this from 1 to find the area (probability) to the right, which corresponds to scores of 780 and above.

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Most popular questions from this chapter

To completely specify the shape of a Normal distribution, you must give a. the mean and the standard deviation. b. the five-number summary. c. the median and the quartiles.

Cholesterol. Low density lipoprotein, or LDL, is the main source of cholesterol buildup and blockage in the arteries. This is why LDL is known as "bad cholesterol." LDL is measured in milligrams per deciliter of blood, or mg/dL. In a population of adults at risk for cardiovascular problems, the distribution of LDL levels is Normal, with a mean of 123 \(\mathrm{mg} / \mathrm{dL}\) and a standard deviation of \(41 \mathrm{mg} / \mathrm{dL}\). If an individual's LDL is at least 1 standard deviation or more above the mean, he or she will be monitored carefully by a doctor. What percentage of individuals from this population will have LDL levels 1 or more standard deviations above the mean? Use the 68 -95-99.7 rule.

Upper Arm Lengths. The upper arm length of males over 20 years old in the United States is approximately Normal with mean \(39.1\) centimeters \((\mathrm{cm})\) and standard deviation \(5.0 \mathrm{~cm}\). Use the 68-95-99.7 rule to answer the following questions. (Start by making a sketch like Figure 3.10.) a. What range of lengths covers the middle \(99.7 \%\) of this distribution? b. What percentage of men over 20 have upper arm lengths greater than \(44.1 \mathrm{~cm}\) ?

The proportion of observations from a standard Normal distribution that take values greater than \(1.78\) is about a. \(0.9554 .\) b. \(0.0446 .\) c. \(0.0375 .\)

The distribution of hours of sleep per weeknight among college students is found to be Normally distributed, with a mean of \(6.5\) hours and a standard deviation of 1 hour. What range contains the middle \(95 \%\) of hours slept per weeknight by college students? a. \(5.5\) and \(7.5\) hours per weeknight b. \(4.5\) and \(7.5\) hours per weeknight c. \(4.5\) and \(8.5\) hours per weeknight

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