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The average annual amount American households spend for daily transportation is \(\$ 6312\) (Money, August 2001). Assume that the amount spent is normally distributed. a. Suppose you learn that \(5 \%\) of American households spend less than \(\$ 1000\) for daily transportation. What is the standard deviation of the amount spent? b. What is the probability that a household spends between \(\$ 4000\) and \(\$ 6000 ?\) c. What is the range of spending for the \(3 \%\) of households with the highest daily transportation cost?

Short Answer

Expert verified
(a) \(\sigma \approx 3221.64\); (b) Probability \(\approx 22.44\%\); (c) Spend over \(12374.1\) dollars.

Step by step solution

01

Understand the Normal Distribution

We know that the amount spent is normally distributed with a mean, \( \mu = 6312 \). We need to find the standard deviation, \( \sigma \), for part (a) and use the normal distribution properties for parts (b) and (c).
02

Calculate the Standard Deviation (Part a)

For part (a), we know that 5% of households spend less than \( 1000 \) dollars. This corresponds to the \( z \)-score of approximately -1.645 (since it is on the left side of the normal distribution curve for 5%). Using the formula for the \( z \)-score: \[ z = \frac{x - \mu}{\sigma} \] where \( x = 1000 \), we have: \[ -1.645 = \frac{1000 - 6312}{\sigma} \] Solving for \( \sigma \), we get: \[ \sigma = \frac{6312 - 1000}{1.645} \approx 3221.64 \]
03

Probability of Spending Between $4000 and $6000 (Part b)

We need to find the probability that a household spends between \( 4000 \) and \( 6000 \). Using the same \( \sigma \) from step 2 and the normal distribution:\[ z_{4000} = \frac{4000 - 6312}{3221.64} \approx -0.72 \] \[ z_{6000} = \frac{6000 - 6312}{3221.64} \approx -0.10 \] Use a standard normal distribution table or calculator to find the probabilities: \( P(z_{4000}) \approx 0.2358 \) and \( P(z_{6000}) \approx 0.4602 \). The probability is \( P(4000 < x < 6000) = 0.4602 - 0.2358 = 0.2244 \).
04

Range for Top 3% (Part c)

For the top 3% of spenders, we look at the 97th percentile of the normal distribution. The \( z \)-score corresponding to the 97th percentile is about 1.881. Thus, we calculate the spending using: \[ x = \mu + z \sigma = 6312 + (1.881 \times 3221.64) \approx 12374.1 \] So, the top 3% of spenders spend more than \( 12374.1 \) dollars.

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Key Concepts

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

Understanding Standard Deviation
Standard deviation is a critical concept in statistics, particularly when dealing with a normal distribution. It is a measure that quantifies the amount of variation or dispersion of a set of values. In simpler terms, it indicates how spread out the numbers in a dataset are around the mean (average).
In a normal distribution, most data points fall close to the mean, and the standard deviation helps to determine how spread out these points are. A smaller standard deviation means that the values tend to be close to the mean, while a larger standard deviation indicates that the numbers are more spread out.
For example, if we have a mean spending of \(6312 as in our problem and we find a standard deviation of around \)3221.64, it tells us that the spending amounts typically vary around this mean by that amount. To calculate the standard deviation when you know the z-score and a data point as in the exercise:
  • Identify your z-score, which indicates the number of standard deviations a data point is from the mean.
  • Use the formula \( z = \frac{x - \mu}{\sigma} \) where \( x \) is your data point, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.
  • Rearrange to solve for \( \sigma \).
This calculation allows us to understand how variable the transportation spending data is.
Deciphering Z-scores
A z-score, also known as a standard score, is essentially a numerical measurement that describes a value's relationship to the mean of a group of values. It is measured in terms of standard deviations from the mean.
Calculating a z-score is an integral part of the normal distribution, as it helps standardize different datasets making analysis comparative and comprehensible. It can tell you exactly how many standard deviations a value is from the mean. In our transportation spending example, the z-score helps determine how unusual a specific spending amount is compared to the average.Here's how to calculate a z-score:
  • Subtract the mean from your data point.
  • Divide the result by the standard deviation: \( z = \frac{x - \mu}{\sigma} \).
The resulting z-score indicates its positioning on the normal distribution curve. For example, a z-score of 0 indicates the data point is exactly at the mean, while a positive or negative z-score shows it is above or below the mean, respectively.When it comes to probabilities, the z-score can also be used to find the percentage of values below, above, or between data points in a normal distribution by referring to standard normal distribution tables or using statistical software.
Percentiles Explained
Percentiles are a statistical concept used to understand and interpret data distributions. They rank data points to indicate the percentage of values within a dataset that lie below a specific point. Understanding percentiles is crucial for grasping how a particular data point compares within the entire dataset.
In simple terms, the nth percentile is the value below which n% of the data falls. For instance, being in the 97th percentile means a data point is greater than or equal to the value of 97% of the remaining data. To find the percentile in a normal distribution:
  • Identify the percentile position you are interested in (e.g., the top 3%, which corresponds to the 97th percentile).
  • Use a standard normal distribution table or calculator to find the corresponding z-score for the desired percentile.
  • Using the z-score, calculate the value (data point) using \( x = \mu + z \sigma \).
For instance, in our problem, the 97th percentile calculation for spending helped us find that the top 3% of households spend more than $12374.1. Understanding percentiles can be vital in making informed decisions based on how extreme or typical a data point is within its distribution.

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