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Chapter 3: Problem 21

The scores of adults on an IQ test are approximately Normally distributed, with mean 100 and standard deviation 15. Alysha scores 135 on such a test. Her \(z\)-score is about a. \(1.33 .\) b. \(2.33\) c. \(6.33 .\)

Short Answer

Expert verified
Her z-score is 2.33, so the answer is b.

Step by step solution

01

Understanding the Z-score Formula

The z-score formula is given by \( z = \frac{x - \mu}{\sigma} \) where \( x \) is the individual's score, \( \mu \) is the mean of the distribution, and \( \sigma \) is the standard deviation.
02

Substitute the Given Values

We know Alysha's score \( x = 135 \), the mean \( \mu = 100 \), and the standard deviation \( \sigma = 15 \). Substitute these values into the z-score formula.
03

Calculate the Z-score

Using the formula: \[ z = \frac{135 - 100}{15} \] compute the value of \( z \).
04

Simplify the Expression

First, subtract the mean from Alysha's score: \( 135 - 100 = 35 \). Then divide 35 by the standard deviation 15 to get \( z = \frac{35}{15} = 2.33 \).
05

Compare with Choices

Compare the calculated z-score \( 2.33 \) to the given options: 1.33, 2.33, and 6.33.

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Key Concepts

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

Z-score
The z-score is a statistical measure that tells us how many standard deviations an individual data point is from the mean. It's a way to understand the position of a score within a distribution.
  • For Alysha's case, her IQ test score of 135 is converted into a z-score to see how unusual it is compared to the average score.
  • The formula for calculating a z-score is: \[ z = \frac{x - \mu}{\sigma}\] where:
    • \(x\) is the individual score.
    • \(\mu\) is the mean of the distribution.
    • \(\sigma\) is the standard deviation.
The z-score tells us if Alysha's score is below, above, or at the mean and quantitatively how significant it is. A z-score of 2.33 means Alysha scored well above the average by about 2.33 standard deviations.
Mean and Standard Deviation
The mean and standard deviation are critical descriptive statistics in understanding the properties of a distribution.
  • The mean, often referred to as the "average," is the central point of a data set. In this IQ test context, the mean is 100, meaning that most scores will hover around this value.
  • The standard deviation is a measure of how spread out the numbers are in a set of data. Here it's 15, which indicates how much the IQ scores deviate on average from the mean. A smaller standard deviation would mean scores are more bunched together, while a larger one indicates widespread scores.
Together, these two values provide a complete picture of the distribution: the mean tells us about the central tendency, while the standard deviation informs us about the variability.
IQ Test Scores
IQ test scores are usually designed to follow a normal distribution. This is why understanding the mean, standard deviation, and z-score helps in gauging one's IQ score effectively.
  • A normal distribution is symmetric and bell-shaped, where most of the data points cluster around the mean.
  • For IQ tests, the mean is standardized to 100, and the standard deviation is typically set to 15, which reflects the range of scores most test-takers fall into.
Alysha's score of 135, with a z-score of 2.33, indicates that she performed significantly better than the mean. This suggests that only a small percentage of people score as high as Alysha, making her result quite exceptional in the population context.

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Most popular questions from this chapter

Are the Data Normal? A cidity of Rainfall. Exercise \(3.31\) (page 93 ) concerns the acidity (measured by \(\mathrm{pH}\) ) of rainfall. A sample of 105 rainwater specimens had mean \(\mathrm{pH} 5.43\), standard deviation \(0.54\), and five-number summary \(4.33,5.05,5.44,5.79,6.81 .14\) a. Compare the mean and median and also the distances of the two quartiles from the median. Does it appear that the distribution is quite symmetric? Why? b. If the distribution is really \(N(5.43,0.54)\), what proportion of observations would be less than \(5.05\) ? Less than \(5.79\) ? Do these proportions suggest that the distribution is close to Normal? Why?

Osteoporosis. Osteoporosis is a condition in which the bones become brittle due to loss of minerals. To diagnose osteoporosis, an elaborate apparatus measures bone mineral density (BMD). BMD is usually reported in standardized form. The standardization is based on a population of healthy young adults. The World Health Organization (WHO) criterion for osteoporosis is a BMD lower than \(2.5\) standard deviations below the mean for healthy young adults. BMD measurements in a population of people similar in age and sex roughly follow a Normal distribution. a. What percentage of healthy young adults have osteoporosis by the WHO criterion? b. Women aged \(70-79\) are, of course, not young adults. The mean BMD in this age is about \(-2\) on the standard scale for young adults. Suppose the standard deviation is the same as for young adults. What percentage of this older population have osteoporosis?

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 Medical College Admissions Test. Almost all medical schools in the United States require students to take the Medical College Admission Test (MCAT). The total score of the four sections on the test ranges from 472 to 528 . In spring of 2019 , the mean score was \(500.9\), with a standard deviation of \(10.6\). a. What proportion of students taking the MCAT had a score over 510 ? b. What proportion had scores between 505 and 515 ?

Grading Managers. In Exercise 3.44, we saw that Ford Motor Company once graded its managers in such a way that the top \(10 \%\) received an A grade, the bottom \(10 \%\) a C, and the middle \(80 \%\) a B. Let's suppose that performance scores follow a Normal distribution. How many standard deviations above and below the mean do the \(A / B\) and \(B / C\) cutoffs lie? (Use the standard Normal distribution to answer this question.)
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