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Chapter 3: Problem 24

The scores of adults on an IQ test are approximately Normal with mean 100 and standard deviation 15. Alysha scores 135 on such a test. She scores higher than what percent of all adults? (a) About \(5 \%\) (b) About \(95 \%\) (c) About \(99 \%\)

Short Answer

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(c) About 99%

Step by step solution

01

Define the Problem

We need to determine the percentage of adults that Alysha scores higher than, given her IQ score of 135, and the IQ distribution has a mean (\(\mu\)) of 100 and a standard deviation (\(\sigma\)) of 15.
02

Calculate the Z-Score

Using the formula for a Z-score, we calculate Alysha's Z-score: \[ Z = \frac{X - \mu}{\sigma} \] where \(X = 135\), \(\mu = 100\), and \(\sigma = 15\). Thus, \[ Z = \frac{135 - 100}{15} = \frac{35}{15} \approx 2.33 \]
03

Use the Z-Table

Look up the Z-score of 2.33 in the standard normal distribution (Z) table. The score corresponds to the percentile of the population Alysha scores higher than. A Z-score of 2.33 corresponds to roughly 0.9901, meaning she scores higher than about 99.01% of all adults.
04

Interpret the Result

The percentile score from the Z-table is interpreted as the probability that Alysha's score is higher than that percentage of the population. Since the Z-table gives the area to the left, the result implies Alysha scored better than approximately 99% of the population.

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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 key concept in statistics, especially when dealing with the normal distribution. It helps us to determine how far, and in what direction, a data point is from the mean, measured in standard deviations. To calculate a Z-score, we use the formula:\[ Z = \frac{X - \mu}{\sigma} \]Where:
  • \(X\) is the value of the data point.
  • \(\mu\) is the mean of the data set.
  • \(\sigma\) is the standard deviation of the data set.
In this exercise, Alysha's IQ score is 135, with a known population mean of 100 and a standard deviation of 15. Plugging these numbers into our formula, we get:\[ Z = \frac{135 - 100}{15} = \frac{35}{15} \approx 2.33 \]Thus, Alysha's Z-score is approximately 2.33, indicating her score is 2.33 standard deviations above the mean.
IQ Test Scores
IQ test scores are designed to assess human intelligence. They are typically modeled using a normal distribution, where most people score around the average.
  • The mean IQ score is typically set at 100.
  • The standard deviation is often 15, which implies that most people's scores will fall within 15 points of the mean, either above or below.
A higher IQ score suggests better than average cognitive abilities, while a lower score indicates below-average abilities. In this context, Alysha's score of 135 is significantly above the average, placing her as a unique statistical individual within this distribution model.
Percentile Interpretation
In the context of a normal distribution, the percentile represents the position of a data point relative to the rest of the data set. It tells us the percentage of scores that fall below a particular value. When we calculate Alysha's score's Z-score and look it up in a Z-table, we find the percentile, which in Alysha's case is 0.9901. This means 99.01% of the population scores at or below Alysha's score. The interpretation is straightforward:
  • Alysha scores higher than 99% of the population.
  • This is a representation of how rare her score is within the general population.
The percentile rank is a valuable metric in assessing one's score compared to a broader population, especially in educational and psychological assessments.

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Most popular questions from this chapter

The distribution of hours of sleep per week night, 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 week night by college students? (a) \(5.5\) and \(7.5\) hours per week night (b) \(4.5\) and \(7.5\) hours per week night (c) \(4.5\) and \(8.5\) hours per week night

Body mass index. Your body mass index (BMI) is your weight in kilograms divided by the square of your height in meters. Many online BMI calculators allow you to enter weight in pounds and height in inches. High BMI is a common but controversial indicator of overweight or obesity. A study by the National Center for Health Statistics found that the BMI of American young men (ages 20-29) is approximately Normal with mean \(26.8\) and standard deviation 5.2. 12 (a) People with BMI less than \(18.5\) are often classified as "underweight." What percent of men aged 20-29 are underweight by this criterion? (b) People with BMI over 30 are often classified as "obese." What percent of men aged \(20-29\) are obese by this criterion?

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.

Use the Normal Table. Use Table A to find the proportion of observations from a standard Normal distribution that satisfies each of the following statements. In each case, sketch a standard Normal curve and shade the area under the curve that is the answer to the question. (a) \(z<-0.42\) (b) \(z>-1.58\) (c) \(z<2.12\) (d) \(-0.42

What's your percentile? Reports on a student's test score such as the SAT or a child's height or weight usually give the percentile as well as the actual value of the variable. The percentile is just the cumulative proportion stated as a percent: the percent of all values of the variable that were lower than this one. The upper arm lengths of females in the United States are approximately Normal with mean \(35.8 \mathrm{~cm}\) and standard deviation \(2.1 \mathrm{~cm}\), and those for males are approximately Normal with mean \(39.1 \mathrm{~cm}\) and standard deviation \(2.3 \mathrm{~cm}\). (a) Cecile, a 73-year-old female in the United States, has an upper arm length of \(33.9 \mathrm{~cm}\). What is her percentile? (b) Measure your upper arm length to the nearest tenth of a centimeter, referring to Exercise \(3.5\) (page 84 ) for the measurement instructions. What is your arm length in centimeters? What is your percentile?
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