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Chapter 15: Problem 19

The test grades for a standardized test are normally distributed with a mean of \(50 .\) A grade of 60 represents a z-score of \(1.25 .\) What is the standard deviation of the data?

Short Answer

Expert verified
The standard deviation is 8.

Step by step solution

01

Understand the Z-score Formula

A z-score is a measure of how many standard deviations a data point is from the mean. The formula for a z-score is given by:\[ z = \frac{X - \mu}{\sigma} \]where \( X \) is the data point, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.
02

Substitute Known Values into the Formula

We are given that the mean \( \mu = 50 \), the z-score \( z = 1.25 \), and the grade \( X = 60 \). Substitute these values into the z-score formula:\[ 1.25 = \frac{60 - 50}{\sigma} \]
03

Solve for the Standard Deviation \( \sigma \)

We need to solve the equation for \( \sigma \). First, simplify inside the fraction:\[ 1.25 = \frac{10}{\sigma} \]To isolate \( \sigma \), multiply both sides by \( \sigma \), and then divide both sides by \( 1.25 \):\[ \sigma = \frac{10}{1.25} \]
04

Calculate \( \sigma \)

Divide 10 by 1.25 to find \( \sigma \):\[ \sigma = 8 \]Therefore, the standard deviation of the data is 8.

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Key Concepts

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

Understanding the Z-score
The Z-score is a statistical measure that illustrates how many standard deviations a specific data point is from the mean. It is incredibly useful in understanding the position of a value within a data set.
  • When a Z-score is calculated, it essentially tells us whether the data point is typical for that data set or unusual.
  • A Z-score of 0 means the data point is exactly at the mean.
  • A positive Z-score indicates the data point is above the mean, while a negative Z-score shows it is below.
To compute a Z-score, the formula is:\[ z = \frac{X - \mu}{\sigma} \]Here, \( X \) represents the individual data point, \( \mu \) is the mean of the data set, and \( \sigma \) is the standard deviation. This calculation provides a view of where the specific data point lies in relation to the overall distribution.
Exploring the Normal Distribution
The normal distribution, often referred to as the bell curve, is a common and important concept in statistics. Many different types of measurements conform to this type of distribution, which is symmetrical and centered around the mean.
  • The mean divides the normal distribution into two equal parts, with half of the data points lying on either side.
  • Standard deviation plays a crucial role, determining how spread out the values are from the mean.
  • Approximately 68% of data falls within one standard deviation of the mean, 95% within two, and 99.7% within three standard deviations.
This distribution is not only typical in the natural and social sciences but is also the foundation for many statistical procedures. Recognizing this pattern allows statisticians to make inferences about the probability of events and understand the variability within the data.
Defining the Mean
The mean is one of the most important measures in statistics, serving as a fundamental indicator of the central tendency of a set of numbers. It is calculated by summing all the data points in a set and then dividing by the number of points.
  • The formula for calculating the mean is \( \mu = \frac{\sum X}{N} \), where \( \sum X \) is the sum of all data values and \( N \) is the number of data points.
  • The mean provides a single value that summarizes the entire distribution, making it easier to compare different sets of data.
  • It is a sensitive measure, affected by extreme values known as outliers, which can skew the mean towards larger or smaller than expected values.
In the context of normally distributed data, not only does it act as a point of balance but it also helps in determining how each data point (through calculation of Z-scores) compares with the rest of the dataset.

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Most popular questions from this chapter

In \(3-8,\) find the mean, the median, and the mode for each set of data. $$ \begin{array}{|c|c|}\hline x_{i} & {f_{i}} \\ \hline 95 & {2} \\ {90} & {8} \\\ {85} & {12} \\ {80} & {10} \\ {75} & {9} \\ {70} & {3} \\ {65} & {0} \\\ {60} & {1} \\ \hline\end{array} $$

The given values represent data for a sample. Find the variance and the standard deviation based on this sample. \(\begin{array}{|c|c|}\hline x_{i} & {f_{i}} \\ \hline 1 & {3} \\ {2} & {3} \\\ {3} & {3} \\ {4} & {3} \\ {5} & {3} \\ {6} & {3} \\ {7} & {3} \\\ \hline\end{array}\)

Steve kept a record of the height of a tree that he planted. The heights are shown in the table. $$ \begin{array}{|l|l|l|l|l|l|l|}\hline \text { Age of Tree in Years } & {1} & {3} & {5} & {7} & {9} & {11} & {13} \\ \hline \text { Height in lnches } & {7} & {12} & {15} & {16.5} & {17.8} & {19} & {20} \\ \hline\end{array} $$ a. Write an equation that best fits the data. b. What was the height of the tree after 2 years? c. If the height of the tree continues in this same pattern, how tall will the tree be after 20 years?

To commute to the high school in which Mr. Fedora teaches, he can take either the Line A or the Line B train. Both train stations are the same distance from his house and both stations report that, on average, they run 10 minutes late from the scheduled arrival time. However, the standard deviation for Line A is 1 minute and the standard deviation for Line B is 5 minutes. To arrive at approximately the same time on a regular basis, which train line should Mr. Fedora use? Explain.

The length of time that it takes Ken to drive to work represents a normal distribution with a mean of 25 minutes and a standard deviation of 4.5 minutes. If Ken allows 35 minutes to get to work, what percent of the time can he expect to be late?
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