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According to a government study, among adults in the 25 - to 34 -year age group, the mean amount spent per year on reading and entertainment is \(\$ 1,994 .\) Assume that the distribution of the amounts spent follows the normal distribution with a standard deviation of \(\$ 450 .\) a. What percent of the adults spend more than \(\$ 2,500\) per year on reading and entertainment? b. What percent spend between \(\$ 2,500\) and \(\$ 3,000\) per year on reading and entertainment? c. What percent spend less than \(\$ 1,000\) per year on reading and entertainment?

Short Answer

Expert verified
a) 13.14%, b) 11.89%, c) 1.36%

Step by step solution

01

Understanding the Problem

The problem involves a normal distribution with a mean of $1,994 and a standard deviation of $450. We need to find the percentages of adults based on various spending intervals.
02

Standardizing the First Boundary

For part (a), standardize the value \(2,500 using the formula \( z = \frac{x - \mu}{\sigma} \), where \( x \) is \)2,500, \( \mu \) is \(1,994, and \( \sigma \) is \)450. Calculate: \( z = \frac{2500 - 1994}{450} \approx 1.12 \).
03

Finding the First Percent

Use a standard normal distribution table or calculator to find the area to the right of \( z = 1.12 \). The table gives an area (to the left) of approximately 0.8686. Therefore, the percent more than $2,500 is \( 1 - 0.8686 = 0.1314 \) or 13.14%.
04

Standardizing the Second Boundary

For part (b), calculate the \( z \)-scores for \(2,500 and \)3,000. Previously calculated \( z = 1.12 \) for \(2,500. Now calculate for \)3,000: \( z = \frac{3000 - 1994}{450} \approx 2.24 \).
05

Finding the Second Percent

Find the area for \( z = 2.24 \), which is approximately 0.9875. The area between the two \( z \)-scores 1.12 and 2.24 is \( 0.9875 - 0.8686 \approx 0.1189 \) or 11.89%.
06

Standardizing the Third Boundary

For part (c), calculate the \( z \)-score for $1,000: \( z = \frac{1000 - 1994}{450} \approx -2.21 \).
07

Finding the Third Percent

Find the area to the left of \( z = -2.21 \), which is approximately 0.0136 or 1.36%.

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Key Concepts

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

Standard Deviation
Standard deviation is a key measure in statistics that helps us understand the amount of variation or dispersion in a set of values. When we apply it to a normal distribution, it tells us how spread out the data points are around the mean (average) value.
In the context of the exercise, with a mean annual spending of $1,994 on reading and entertainment, and a standard deviation of $450, the standard deviation indicates that most adults spend within $450 more or less than the mean. This gives us a range of spending for the majority of people in the group.
If the standard deviation is small, the data points are close to the mean, indicating less variability. A larger standard deviation means the data is more spread out. This tool is crucial for understanding how typical or atypical a particular data point is, relative to the rest of the data.
Z-Score
A Z-score is a statistical measurement that describes a value's position relative to the mean of a group of values. It is expressed in terms of standard deviations.
The formula to calculate a Z-score is: \[ z = \frac{x - \mu}{\sigma} \] where :
  • \( x \) is the value you're examining,
  • \( \mu \) is the mean,
  • \( \sigma \) is the standard deviation.
In the exercise, when calculating the Z-score for \(2,500, we used the mean \)1,994 and standard deviation \(450. The Z-score tells us how far, in units of standard deviations, \)2,500 is from the mean. By knowing the Z-score, we can use a standard normal distribution table to find the probability or percentage of data points falling above or below that Z-score. In this way, Z-scores can effectively standardize different data sets on a uniform scale, which makes comparisons and interpretations more intuitive.
Calculating Percentages
To calculate the percentages of the data within certain ranges of a normal distribution, we often use Z-scores as an intermediary step. Once we have standardized our data with Z-scores, we use a standard normal distribution table (or calculator) to find the percentage of data points to the left or right of a given Z-score.
  • In cases like the exercise, where we need to find the percentage of adults spending more than a certain amount, we first determine the area to the left of the Z-score and subtract it from total one to find the desired percentage to the right.
  • For spending ranges, like between $2,500 and $3,000, we find and subtract the percentages for each boundary using their respective Z-scores.
  • For amounts less than a specific figure, such as $1,000, we directly use the area to the left of the relevant Z-score.
Calculating these percentages requires understanding the normal distribution's structure and knowing how to leverage the cumulative probability that Z-scores provide. This approach helps in quantifying and interpreting real-world scenarios with precision and ease.

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