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The average reading speed of students completing a speed-reading course is 450 words per minute (wpm). If the standard deviation is \(70 \mathrm{wpm}\), find the \(z\) score associated with each of the following reading speeds. a. \(320 \mathrm{wpm}\) c. \(420 \mathrm{wpm}\) b. 475 wpm d. 610 wpm

Short Answer

Expert verified
The z-scores for reading speeds 320 wpm, 420 wpm, 475 wpm, and 610 wpm are approximately -1.86, -0.43, 0.36, and 2.29 respectively.

Step by step solution

01

Identify the Mean and Standard Deviation

Given in the problem, the average (mean) reading speed is \( 450 \) words per minute, and the standard deviation is \( 70 \) words per minute.
02

Calculate the Z-score for 320 wpm

Use the formula \[ z = \frac{x - \mu}{\sigma} \]. Substitute \( x = 320 \), \( \mu = 450 \), and \( \sigma = 70 \) into the formula to get \[ z = \frac{320 - 450}{70} \].
03

Calculate the Z-score for 420 wpm

Substitute \( x = 420 \), \( \mu = 450 \), and \( \sigma = 70 \) into the formula to get \[ z = \frac{420 - 450}{70} \].
04

Calculate the Z-score for 475 wpm

Substitute \( x = 475 \), \( \mu = 450 \), and \( \sigma = 70 \) into the formula to get \[ z = \frac{475 - 450}{70} \].
05

Calculate the Z-score for 610 wpm

Substitute \( x = 610 \), \( \mu = 450 \), and \( \sigma = 70 \) into the formula to get \[ z = \frac{610 - 450}{70} \].

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Key Concepts

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

Standard Deviation
Understanding the concept of standard deviation is crucial when dealing with any form of data that varies. It measures how spread out the numbers in a data set are. In a more technical term, standard deviation is the average distance from the mean of the data set. A small standard deviation indicates that the values in a data set are close to the mean of the data set, while a large standard deviation indicates that the values are spread out over a wider range.

Regarding the z score calculation exercise, knowing the standard deviation allows us to determine how far a particular reading speed is from the average speed, which is essential for finding the z score. If a student reads at 320 wpm, and the standard deviation is 70 wpm, this means the student's speed is more than one standard deviation below the average. Recognizing this helps to contextualize the student's performance.
Normal Distribution
The concept of normal distribution, often referred to as the 'bell curve', is a foundational idea in statistics. It represents a distribution where most of the data points cluster around a central point, known as the mean, with values tapering off symmetrically on both sides as they move away from the mean.

In terms of our exercise, the reading speeds of students are likely to follow a normal distribution, assuming a large and random sample of students. This is important because the z score calculation assumes that the distribution of the variable (reading speed, in this case) is normal. Under a normal distribution, specific z scores correspond to percentages that represent the likelihood of observing a value at or below a given point; an important concept in statistical analysis and interpretation of the z score.
Statistical Analysis
Statistical analysis involves collecting, reviewing, and interpreting data to discover underlying patterns and trends. It is an essential tool for making informed decisions based on quantitative data. In our z score exercise, statistical analysis would involve examining how different reading speeds compare to the average reading speed once the z scores are calculated.

By calculating the z score, we can perform a statistical analysis to identify how individual students' reading speeds differ from the average or expected speed. A z score tells us how many standard deviations an observation is from the mean, which is critical information in understanding whether a reading speed is typical, exceptionally high, or low. This information can be used to assess students' progress and even tailor instruction to improve reading skills.

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Most popular questions from this chapter

The article "Can We Really Walk Straight?" \((\) American Journal of Physical Anthropology \([1992]: 19-\) 27) reported on an experiment in which each of 20 healthy men was asked to walk as straight as possible to a target \(60 \mathrm{~m}\) away at normal speed. Consider the following data on cadence (number of strides per second): \(\begin{array}{llllllll}0.95 & 0.85 & 0.92 & 0.95 & 0.93 & 0.86 & 1.00 & 0.92 \\\ 0.85 & 0.81 & 0.78 & 0.93 & 0.93 & 1.05 & 0.93 & 1.06 \\ 1.06 & 0.96 & 0.81 & 0.96 & & & & \end{array}\) Use the methods developed in this chapter to summarize the data; include an interpretation or discussion wherever appropriate. (Note: The author of the paper used a rather sophisticated statistical analysis to conclude that people cannot walk in a straight line and suggested several explanations for this.)

Suppose that your statistics professor returned your first midterm exam with only a \(z\) score written on it. She also told you that a histogram of the scores was approximately normal. How would you interpret each of the following \(z\) scores? a. \(2.2\) d. \(1.0\) b. \(0.4\) e. 0 c. \(1.8\)

Mobile homes are tightly constructed for energy conservation. This can lead to a buildup of indoor pollutants. The paper "A Survey of Nitrogen Dioxide Levels Inside Mobile Homes" (Journal of the Air Pollution Control Association \([1988]: 647-651\) ) discussed various aspects of NO, concentration in these structures. a. In one sample of mobile homes in the Los Angeles area, the mean \(\mathrm{NO}_{2}\) concentration in kitchens during the summer was \(36.92 \mathrm{ppb}\), and the standard deviation was 11.34. Making no assumptions about the shape of the \(\mathrm{NO}_{2}\) distribution, what can be said about the percentage of observations between \(14.24\) and \(59.60 ?\) b. Inside what interval is it guaranteed that at least \(89 \%\) of the concentration observations will lie? c. In a sample of non-Los Angeles mobile homes, the average kitchen \(\mathrm{NO}_{2}\) concentration during the winter was \(24.76 \mathrm{ppb}\), and the standard deviation was \(17.20 .\) Do these values suggest that the histogram of sample observations did not closely resemble a normal curve? (Hint: What is \(\bar{x}-2 s ?\)

The San Luis Obispo Telegram-Tribune (November 29,1995 ) reported the values of the mean and median salary for major league baseball players for \(1995 .\) The values reported were \(\$ 1,110,766\) and \(\$ 275,000\). a. Which of the two given values do you think is the mean and which is the median? Explain your reasoning. b. The reported mean was computed using the salaries of all major league players in \(1995 .\) For the 1995 salaries, is the reported mean the population mean \(\mu\) or the sample mean \(\bar{x}\) ? Explain.

USA Today (May 9,2006 ) published the accompanying average weekday circulation for the six month period ending March 31,2006 for the top 20 newspapers in the country: \(\begin{array}{rrrrr}2,272,815 & 2,049,786 & 1,142,464 & 851,832 & 724,242 \\\ 708,477 & 673,379 & 579,079 & 513,387 & 438,722 \\ 427,771 & 398,329 & 398,246 & 397,288 & 365,011 \\ 362,964 & 350,457 & 345,861 & 343,163 & 323,031\end{array}\) a. Which of the mean or the median do you think will be larger for this data set? Explain. b. Compute the values of the mean and the median of this data set. c. Of the mean and median, which does the best job of describing a typical value for this data set? d. Explain why it would not be reasonable to generalize from this sample of 20 newspapers to the population of daily newspapers in the United States.

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