Problem 38
Question
A set of data items is normally distributed with a mean of 60 and a standard deviation of 8 . In Exercises 33-48, convert each data item to a z-score. 72
Step-by-Step Solution
Verified Answer
The z-score of 72 is 1.5.
1Step 1: Identifying the given parameters
Identify the mean, standard deviation and the data item. The mean (\( \mu \)) is 60, the standard deviation (\( \sigma \)) is 8 and the data item (x) is 72.
2Step 2: Apply the z-score formula
Apply the formula for the z-score: \( z = \frac{x - \mu}{\sigma} \). Substituting the given values into the formula will give us \( z = \frac{72 - 60}{8} \).
3Step 3: Calculate the z-score
Solving the expression \( z = \frac{72 - 60}{8} \) gives us \( z = \frac{12}{8} = 1.5 \). The z-score of 72 is 1.5.
Key Concepts
Understanding the Normal DistributionThe Role of Standard DeviationStatistical Analysis and the Z-Score
Understanding the Normal Distribution
When discussing statistical analysis, the concept of a normal distribution is fundamental. It's a probability distribution that is perfectly symmetrical, resembling the shape of a bell, hence it's often referred to as the bell curve. The highest point on the curve, or the peak, represents the mean, median, and mode of the data set, which are all equal in a perfectly normal distribution. When data is normally distributed, it indicates that values are more likely to occur closer to the mean, becoming less likely as they move farther away.
The normal distribution is exceptionally useful because of its predictability. For instance, in a standard normal distribution approximately 68% of the data falls within one standard deviation from the mean, 95% within two standard deviations, and 99.7% within three standard deviations. This characteristic is key to various fields such as psychology, finance, and other sciences where data analysis is crucial.
The normal distribution is exceptionally useful because of its predictability. For instance, in a standard normal distribution approximately 68% of the data falls within one standard deviation from the mean, 95% within two standard deviations, and 99.7% within three standard deviations. This characteristic is key to various fields such as psychology, finance, and other sciences where data analysis is crucial.
The Role of Standard Deviation
Standard deviation is a critical component in understanding how spread out your data is around the mean. Imagine you're in a room with 10 people and you want to know how varied their heights are. If most of them are around the same height, with very little difference, you could say that the variation, or standard deviation, is low. Conversely, if the heights are all over the place, with a significant difference between the shortest and the tallest, the standard deviation is high.
In the context of the provided exercise, a standard deviation of 8 means that the values in this normally distributed dataset typically vary by 8 units from the mean (60). Understanding standard deviation is essential for interpreting the spread of data, making it a cornerstone for statistical analysis and helping to identify outliers or unusual occurrences in a data set.
In the context of the provided exercise, a standard deviation of 8 means that the values in this normally distributed dataset typically vary by 8 units from the mean (60). Understanding standard deviation is essential for interpreting the spread of data, making it a cornerstone for statistical analysis and helping to identify outliers or unusual occurrences in a data set.
Statistical Analysis and the Z-Score
Statistical analysis is the process by which we derive meaning from data. It involves collecting, summarizing, and interpreting data to make informed decisions or hypotheses about a certain population. A z-score is a statistical tool that helps you understand how a single data point relates to the rest of the dataset. It's expressed in terms of standard deviations from the mean.
In the exercise, converting a data item into a z-score allows us to identify its position within the normal distribution. A z-score of 1.5, as in the solution, indicates that the data item (72) is 1.5 standard deviations above the mean since it's a positive value. If it were negative, it would mean below the mean. By using z-scores, you can compare data points from different normal distributions or assess the rarity or commonality of a particular observation.
In the exercise, converting a data item into a z-score allows us to identify its position within the normal distribution. A z-score of 1.5, as in the solution, indicates that the data item (72) is 1.5 standard deviations above the mean since it's a positive value. If it were negative, it would mean below the mean. By using z-scores, you can compare data points from different normal distributions or assess the rarity or commonality of a particular observation.
Other exercises in this chapter
Problem 37
In Exercises 37-44, find the midrange for each group of data items. \(7,4,3,2,8,5,1,3\)
View solution Problem 38
Explain how to use a table showing \(z\)-scores and percentiles to determine the percentage of data items between two \(z\)-scores.
View solution Problem 38
The data sets give the ages of the first six U.S. presidents and the last six U.S. presidents (through Donald Trump). AGE OF FIRST SIX U.S. AGE OF LAST SIX U.S.
View solution Problem 38
In Exercises 37-44, find the midrange for each group of data items. \(11,6,4,0,2,1,12,0,0\)
View solution