Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Consider the IQ model \(N(100,16)\) one last time. a) What IQ represents the 15 th percentile? b) What IQ represents the 98 th percentile? c) What's the IQR of the IQs?

Short Answer

Expert verified
15th percentile: 95.84, 98th percentile: 108.2, IQR: 5.4.

Step by step solution

01

Understand the Problem

We have a normal distribution of IQ scores with a mean of 100 and a variance of 16. This means the standard deviation is \(\sigma = \sqrt{16} = 4\). We are tasked with finding IQ scores that correspond to certain percentiles.
02

Find the 15th Percentile IQ

To find the 15th percentile, we will use the Z-score formula for a normal distribution: \(Z = \frac{X - \mu}{\sigma}\). We need to determine the Z-score for the 15th percentile using a standard normal distribution table or calculator. The Z-score corresponding to the 15th percentile is approximately \(-1.04\). Now apply it to the formula: \(X = \mu + Z \cdot \sigma = 100 + (-1.04) \cdot 4 = 100 - 4.16 = 95.84\). The 15th percentile IQ is approximately 95.84.
03

Find the 98th Percentile IQ

To find the 98th percentile, we use the Z-score for the 98th percentile which is approximately \(2.05\). Plug this Z-score into our formula: \(X = \mu + Z \cdot \sigma = 100 + 2.05 \cdot 4 = 100 + 8.2 = 108.2\). The 98th percentile IQ is approximately 108.2.
04

Calculate the Interquartile Range (IQR)

IQR is the difference between the 75th percentile (Q3) and the 25th percentile (Q1). First, find the Z-score for the 25th percentile, which is \(-0.675\), and the Z-score for the 75th percentile, which is \(0.675\). Calculate these IQ values: \(Q1 = 100 + (-0.675) \cdot 4 = 97.3\) and \(Q3 = 100 + 0.675 \cdot 4 = 102.7\). Thus, the IQR is \(Q3 - Q1 = 102.7 - 97.3 = 5.4\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Percentiles
Percentiles are a way to understand and interpret where a particular value stands in a distribution of data. In the context of a normal distribution like the IQ scores we've been working with, a percentile tells you what percentage of the population scores below a certain value. For example, the 15th percentile indicates that 15% of people have an IQ score below this value.
To find the IQ at the 15th percentile, you'd use the Z-score table to find which Z-score corresponds to 15%. In our example, it is \(-1.04\).
Plug it into the formula, \(X = \mu + Z \cdot \sigma\), where \(\mu\) is the mean and \(\sigma\) is the standard deviation. You calculate: \(100 + (-1.04) \cdot 4 = 95.84\) IQ points for the 15th percentile.
Similarly, for the 98th percentile, meaning 98% of people have an IQ below that score, you find a Z-score of \(2.05\). This calculation yields \(108.2\) IQ points for the 98th percentile.
Standard Deviation
Standard deviation (SD) is a key concept in stats that tells us how much variation there is in a set of data. In a normal distribution, most values cluster around the mean, and the standard deviation quantifies how spread out the numbers are.
A smaller SD means the numbers are closer to the mean. A larger SD indicates more spread. For the IQ model \(N(100,16)\), the standard deviation is \(\sigma = \sqrt{16} = 4\). This means most IQ scores are within 4 points above or below the mean of 100.
In practice, the standard deviation helps us understand how many data points lie within certain distances from the mean. About 68% of the data lies within one SD, \(68\%\) between \(96\) to \(104\) IQ points. \(95\%\) are within two SDs (\(92\) to \(108\) IQ points).
This makes SD a critical component in calculating Z-scores and percentiles.
Interquartile Range (IQR)
The Interquartile Range (IQR) is a measure of statistical dispersion and a better indicator of variability than the range because it is less affected by outliers. IQR is the difference between the first quartile \((Q1)\) and the third quartile \((Q3)\), essentially covering the middle 50% of the data.
For our IQ example, we calculated \(Q1\) at the 25th percentile and \(Q3\) at the 75th percentile. First, find their Z-scores, \(-0.675\) for \(Q1\) and \(0.675\) for \(Q3\). Use those to calculate the IQ values:
  • \(Q1 = 100 + (-0.675) \cdot 4 = 97.3\)
  • \(Q3 = 100 + 0.675 \cdot 4 = 102.7\)
The IQR is \(Q3 - Q1 = 102.7 - 97.3 = 5.4\), meaning the middle half of IQ scores fall within this 5.4-point range.
Utilizing IQR helps analysts understand the core spread of data and is particularly useful when identifying outliers.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An incoming freshman took her college's placement exams in French and mathematics. In French, she scored 82 and in math 86 . The overall results on the French exam had a mean of 72 and a standard deviation of 8 , while the mean math score was 68 , with a standard deviation of \(12 .\) On which exam did she do better compared with the other freshmen?

A forester measured 27 of the trees in a large woods that is up for sale. He found a mean diameter of \(10.4\) inches and a standard deviation of \(4.7\) inches. Suppose that these trees provide an accurate description of the whole forest and that a Normal model applies. a) Draw the Normal model for tree diameters. b) What size would you expect the central \(95 \%\) of all trees to be? c) About what percent of the trees should be less than an inch in diameter? d) About what percent of the trees should be between \(5.7\) and \(10.4\) inches in diameter? e) About what percent of the trees should be over 15 inches in diameter?

Each year thousands of high school students take either the SAT or the ACT, standardized tests used in the college admissions process. Combined SAT Math and Verbal scores go as high as 1600, while the maximum ACT composite score is 36 . Since the two exams use very different scales, comparisons of performance are difficult. A convenient rule of thumb is \(S A T=40 \times A C T+150\); that is, multiply an ACT score by 40 and add 150 points to estimate the equivalent SAT score. An admissions officer reported the following statistics about the ACT scores of 2355 students who applied to her college one year. Find the summaries of equivalent SAT scores. Lowest score \(=19 \quad\) Mean \(=27\) Standard deviation \(=3\) \(\mathrm{Q} 3=30\) Median \(=28 \quad \mathrm{IQR}=6\)

The first Stat exam had a mean of 80 and a standard deviation of 4 points; the second had a mean of 70 and a standard deviation of 15 points. Reginald scored an 80 on the first test and an 85 on the second. Sara scored an 88 on the first but only a 65 on the second. Although Reginald's total score is higher, Sara feels she should get the higher grade. Explain her point of view.

Assume the cholesterol levels of adult American women can be described by a Normal model with a mean of \(188 \mathrm{mg} / \mathrm{dL}\) and a standard deviation of \(24 .\) a) Draw and label the Normal model. b) What percent of adult women do you expect to have cholesterol levels over \(200 \mathrm{mg} / \mathrm{dL}\) ? c) What percent of adult women do you expect to have cholesterol levels between 150 and \(170 \mathrm{mg} / \mathrm{dL}\) ? d) Estimate the IQR of the cholesterol levels. e) Above what value are the highest \(15 \%\) of women's cholesterol levels?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free