/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 134 The test scores for a very large... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The test scores for a very large statistics class have a bell-shaped distribution with a mean of 70 points. a. If \(16 \%\) of all students in the class scored above 85, what is the standard deviation of the scores? b. If \(95 \%\) of the scores are between 60 and 80, what is the standard deviation?

Short Answer

Expert verified
The standard deviation of the class scores for parts a and b are 15 and 5 respectively.

Step by step solution

01

Part a: Identify Z-score from the percentage

First, translate the upper 16% score into a Z-score. You do this by considering the rule of thumb for a normal distribution: approximately 68% of data falls within ±1 standard deviation from the mean, about 95% falls within ±2 standard deviations from the mean, and about 99.7% falls within ±3 standard deviations from the mean. Therefore, 100% - 16% = 84% of data falls at or below this Z-score point. Referring to a Z-table or using a calculator, this Z-score equates roughly to Z = +1.
02

Part a: Calculate Standard Deviation

Next, recalibrate this Z-score in terms of the actual test scores. The Z-score is calculated by \(Z = (X - μ) / σ\), where X is the score, μ is the mean, and σ is the standard deviation. Here, the Z-score is 1, X is 85, and μ is 70. Solving for σ, we have \(σ = (X - μ) / Z = (85 - 70) / 1 = 15\) points. So the standard deviation of the scores for part a is approximately 15.
03

Part b: Identify Z-scores from the percentage

For part b, similarly refer to the Z-score, but realize that the area under 95% is within two points, 60 and 80, so to calculate the Z-score, subtract the mean from the points and divide by the standard deviation. For 60, this works out to \((60 - 70) / σ\), and for 80 it is \((80 - 70) / σ\). But note that 95% of the curve is within ±2 standard deviations, so, in effect, 60 indicates -2 standard deviations and 80 indicates +2 standard deviations.
04

Part b: Calculate Standard Deviation

Knowing that the standard deviation for 60 points equates to -2 and for 80 points equates to +2, take either equivalent and solve for σ again. The result is \(σ = (X - μ) / Z\), which for X = 60, μ = 70, and Z = -2 results in \(σ = (60 - 70) / -2 = 5\). For X = 80, μ = 70, and Z = 2 it results in \(σ = (80 - 70) / 2 = 5\). Thus, the standard deviation for part b is 5 points.

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 91Ó°ÊÓ!

Key Concepts

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

Normal Distribution
The normal distribution is a fundamental concept in statistics, characterized by its bell-shaped curve. It serves as a model for understanding how data tends to be distributed in a wide range of natural phenomena. In a normal distribution, most data points are clustered around the mean, with fewer occurrences appearing as one moves further away from the mean. This symmetry about the mean helps in various analyses and predictions. The curve of a normal distribution is determined by two main parameters: the mean (average) and the standard deviation (a measure of spread). The mean signifies the center of the distribution, while the standard deviation indicates how much the data spreads out from the mean. These parameters make it easier to calculate probabilities and analyze data distributions within a given dataset. In many real-world applications, test scores and other measurements tend to follow a normal distribution, making this an essential concept for students and professionals alike.
Z-score Calculation
A Z-score is a statistical measure that describes a value's position relative to the mean of a group of values. It is expressed in terms of standard deviations from the mean. The formula for calculating a Z-score is: \[ Z = \frac{X - μ}{σ} \] where:
  • \(X\) is the value,
  • \(μ\) is the mean,
  • \(σ\) is the standard deviation.
The Z-score helps in understanding how far a particular point is from the average. A Z-score of 0 indicates that the value is exactly at the mean. A positive Z-score suggests that the value is above the mean, while a negative Z-score indicates it is below the mean. This calculation is widely used in statistics to compare results from a data point to a "normal" point where the mean and standard deviation are known. It's also essential for identifying outliers and performing various hypothesis tests.
Bell-Shaped Distribution
A bell-shaped distribution, often related to the normal distribution, describes a symmetrical probability distribution with a peak at the mean. Its shape resembles a bell, indicating that data near the mean are more frequent in occurrence than data far from the mean. This shape is crucial for statistical analysis because it implies that data follows a known pattern where most values cluster around the center. Such distributions occur widely across various fields, including psychology, biology, and economics. The concept of a bell-shaped curve is vital because it aligns with the empirical rule, which states that for a normal distribution:
  • Approximately 68% of values lie within one standard deviation of the mean,
  • About 95% lie within two standard deviations,
  • And around 99.7% lie within three standard deviations.
Understanding a bell-shaped distribution can help predict probabilities and make inferences about the population from which the sample data is drawn.
Mean and Standard Deviation Relationship
The relationship between the mean and standard deviation is central to understanding data distribution in statistics. The mean provides a measure of the central tendency, or the "average," of a dataset. It is the point at which the dataset balances, providing a simple representation of data. On the other hand, the standard deviation represents the spread or variability within the dataset. It quantifies how much the data points typically deviate from the mean. A smaller standard deviation indicates that the data points tend to be close to the mean, resulting in a steeper curve in a normal distribution. Conversely, a larger standard deviation implies that the data points are more spread out, creating a flatter curve. The relationship between these two concepts is crucial in various statistical applications. For example, when interpreting test scores, knowing both the mean and standard deviation helps determine if a score is extraordinary or typical within the given context. This relationship guides many analyses, such as determining cut-off points for percentiles in standardized testing.

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

Consider the following two data sets. \(\begin{array}{llllrl}\text { Data Set I: } & 12 & 25 & 37 & 8 & 41 \\ \text { Data Set II: } & 19 & 32 & 44 & 15 & 48\end{array}\) Note that each value of the second data set is obtained by adding 7 to the corresponding value of the first data set. Calculate the standard deviation for each of these two data sets using the formula for sample data. Comment on the relationship between the two standard deviations.

The following data set lists the number of women from each of 10 different countries who were on the Rolex Women's World Golf Rankings Top 25 list as of March 31,2009 . The data, entered in that order, are for the following countries: Australia, Brazil, England, Japan, Korea, Mexico, Norway, Sweden, Taiwan, and United States. \(\begin{array}{lllllllll}2 & 1 & 1 & 2 & 9 & 1 & 1 & 2 & 2 & 4\end{array}\) a. Calculate the mean and median for these data. b. Identify the outlier in this data set. Drop the outlier and recalculate the mean and median. Which of these two summary measures changes by a larger amount when you drop the outlier? c. Which is the better summary measure for these data, the mean or the median? Explain.

Consider the following two data sets. \(\begin{array}{llrlrl}\text { Data Set I: } & 4 & 8 & 15 & 9 & 11 \\ \text { Data Set II: } & 8 & 16 & 30 & 18 & 22\end{array}\) Notice that each value of the second data set is obtained by multiplying the corresponding value of the first data set by 2. Calculate the mean for each of these two data sets. Comment on the relationship between the two means.

The following data give the numbers of minor penalties accrued by each of the 30 National Hockey League franchises during the \(2007-08\) regular season. \(\begin{array}{llllllllll}318 & 336 & 337 & 339 & 362 & 363 & 366 & 369 & 372 & 375 \\ 378 & 381 & 384 & 385 & 386 & 387 & 390 & 393 & 395 & 403 \\ 405 & 409 & 417 & 431 & 433 & 434 & 438 & 444 & 461 & 480\end{array}\) a. Calculate the values of the three quartiles and the interquartile range. b. Find the approximate value of the 57 th percentile. c. Calculate the percentile rank of 417 .

Refer to Exercise \(3.115\). Suppose the times taken to learn the basics of this word processor by all students have a bell-shaped distribution with a mean of 200 minutes and a standard deviation of 20 minutes. a. Using the empirical rule, find the percentage of students who will learn the basics of this word processor in i. 180 to 220 minutes ii. 160 to 240 minutes "b. Using the empirical rule, find the interval that contains the time taken by \(99.7 \%\) of all students to learn this word processor.

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.