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The average U.S. yearly per capita consumption of citrus fruit is 26.8 pounds. Suppose that the distribution of fruit amounts consumed is bell-shaped with a standard deviation equal to 4.2 pounds. What percentage of Americans would you expect to consume more than 31 pounds of citrus fruit per year?

Short Answer

Expert verified
15.87% of Americans consume more than 31 pounds of citrus fruit annually.

Step by step solution

01

Understanding the Distribution and Question

The problem describes a bell-shaped distribution of citrus fruit consumption with a mean of 26.8 pounds and a standard deviation of 4.2 pounds. We need to find the percentage of people consuming more than 31 pounds per year. This requires understanding that our question involves a normal distribution.
02

Calculate the Z-Score

The Z-score determines how many standard deviations away 31 pounds is from the mean. Calculate it using: \[Z = \frac{{X - ext{mean}}}{{ ext{standard deviation}}} = \frac{{31 - 26.8}}{4.2}.\]The Z-score represents a standardized value that can be used to find probabilities.
03

Solve for the Z-Score

Substitute the values into the formula to solve for the Z-score:\[Z = \frac{{31 - 26.8}}{4.2} = \frac{4.2}{4.2} = 1.\]
04

Find the Probability Using the Z-Score

The Z-score of 1.0 indicates that 31 pounds is one standard deviation above the mean. We use a standard normal distribution table or calculator to find the probability of Z being less than 1. According to the table, the probability for Z < 1 is approximately 0.8413, indicating 84.13% of people consume less than 31 pounds.
05

Calculate the Percentage Consuming More Than 31 Pounds

To find the percentage consuming more than 31 pounds, subtract the less-than probability from 1: \[1 - 0.8413 = 0.1587.\] So, 15.87% of Americans are expected to consume more than 31 pounds of citrus fruit per year.

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Key Concepts

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

Understanding Z-Score
Have you ever wondered how we can determine whether a specific data point is common or unusual in a dataset? This is done using a concept called the "Z-score." A Z-score tells us how many standard deviations a data point is from the mean of a dataset. It helps us understand whether the data point lies close to the mean or is more extreme. To calculate the Z-score, we use the formula:
  • \[ Z = \frac{{X - \text{mean}}}{{\text{standard deviation}}} \]
In our citrus fruit consumption example, the Z-score was calculated to see how far consuming 31 pounds deviates from the average. By knowing the Z-score, you can assess how likely or unlikely it is for someone to consume that amount within the given population.
What is Standard Deviation?
Standard deviation is a key concept in statistics that measures the amount of variation or dispersion in a set of values. In simple terms, it helps us understand how spread out the data points are in a dataset. A small standard deviation means the data points are close to the mean, while a large standard deviation indicates that the data points are spread out over a wide range of values.
In the citrus fruit consumption example, the standard deviation is 4.2 pounds. This tells us how much the individual consumption values deviate from the average of 26.8 pounds. Knowing the standard deviation is crucial when we want to determine probabilities and compare different situations or datasets.
Understanding Mean in Context
The mean, often called the average, is a fundamental concept in statistics. It gives us a central value for a set of numbers, allowing people to understand what is typical within a dataset. To calculate the mean, sum up all the data points and divide by the number of data points.
  • For example, the mean consumption of citrus fruit among Americans is 26.8 pounds.
This means that if you were to distribute all the consumed fruit equally among the population, each person would get 26.8 pounds. The mean provides a basis from which we can compare individual data points using other statistical measures like standard deviation and Z-scores.
Different Approaches to Probability
Probability is the study of how likely an event is to occur. It ranges from 0 to 1, where 0 means an event will not happen, and 1 means it will certainly happen. Understanding probability helps us make predictions and informed decisions based on data. In our context, after finding the Z-score, we used it to look up a probability in a standard normal distribution table. This tells us the likelihood of a data-point falling under certain criteria. For instance, the probability of consuming less than 31 pounds of citrus fruit was found to be 0.8413. Therefore, the probability of consuming more than 31 pounds was 1 minus 0.8413, which equals 0.1587. This allows us to understand that 15.87% of Americans might consume more than 31 pounds.

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Most popular questions from this chapter

For this data set, find the mean and standard deviation of the variable. The data represent the serum cholesterol levels of 30 individuals. Count the number of data values that fall within 2 standard deviations of the mean. Compare this with the number obtained from Chebyshev’s theorem. Comment on the answer. \(\begin{array}{lllll}{211} & {240} & {255} & {219} & {204} \\ {200} & {212} & {193} & {187} & {205} \\ {256} & {203} & {210} & {221} & {249} \\ {231} & {212} & {236} & {204} & {187} \\ {201} & {247} & {206} & {187} & {200} \\\ {237} & {227} & {221} & {192} & {196}\end{array}\)

Define percentile rank.

The data show the population (in thousands) for a recent year of a sample of cities in South Carolina. \(\begin{array}{llllll}{29} & {26} & {15} & {13} & {17} & {58} \\ {14} & {25} & {37} & {19} & {40} & {67} \\ {23} & {10} & {97} & {12} & {129} & {} \\\ {27} & {20} & {18} & {120} & {35} \\ {66} & {21} & {11} & {43} & {22}\end{array}\) Find the data value that corresponds to each percentile. a. 40th percentile b. 75th percentile c. 90th percentile d. 30th percentile Using the same data, find the percentile corresponding to the given data value. e. 27 f. 40 g. 58 h. 67

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