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A sample of concrete specimens of a certain type is selected, and the compressive strength of each specimen is determined. The mean and standard deviation are calculated as \(\bar{x}=3000\) and \(s=500\), and the sample histogram is found to be well approximated by a normal curve. a. Approximately what percentage of the sample observations are between 2500 and \(3500 ?\) b. Approximately what percentage of sample observations are outside the interval from 2000 to 4000 ? c. What can be said about the approximate percentage of observations between 2000 and \(2500 ?\) d. Why would you not use Chebyshev's Rule to answer the questions posed in Parts (a)-(c)?

Short Answer

Expert verified
a. Approximately 68.27% of sample observations fall between 2500 and 3500.b. Approximately 4.55% of sample observations are outside the interval from 2000 to 4000.c. Approximately 27.18% of observations fall between 2000 and 2500.d. The reason not to use Chebyshev's Rule in this scenario is because the distribution is known to be normal, so the Empirical Rule (68-95-99.7 rule) provides a more accurate estimation.

Step by step solution

01

Understand Z-score theory

In a normal distribution, a Z-score is a measure of how many standard deviations an element is from the mean. It's calculated as \(Z = \frac{x - \bar{x}}{s}\), where \(x\) is an individual data point, \(\bar{x}\) is the mean, and \(s\) is the standard deviation.
02

Calculate the Z-scores for part a

Here, \(\bar{x}=3000\) and \(s=500\). The Z-scores for \(x=2500\) and \(x=3500\) are calculated as \(Z = \frac{2500 - 3000}{500} = -1\) and \(Z = \frac{3500 - 3000}{500} = 1\).
03

Translate Z-scores to percentage values for part a

Looking up these Z-scores in a standard normal table or using a Z-score calculator, the proportion of specimens between \(Z = -1\) and \(Z = 1\) corresponds to approximately 68.27% of the sample observations.
04

Repeat the same calculation for part b

The Z-scores for \(x=2000\) and \(x=4000\) are \(Z = \frac{2000 - 3000}{500} = -2\) and \(Z = \frac{4000 - 3000}{500} = 2\). The percentage of observations outside the interval from 2000 to 4000 is 100% - percentage of observations within \(Z = -2\) and \(Z = 2\), which according to the normal table is 100% - 95.45% = 4.55%.
05

Deduction for part c

The percentage of observations between 2000 and 2500 equals the percentage of observations within \(Z = -2\) and \(Z = -1\), which is 95.45% - 68.27% = 27.18%.
06

Explanation for part d

Chebyshev's Rule can be applied to any distribution (normal or not). However, if the distribution is known to be normal (as it is in this case), the Empirical Rule (68-95-99.7 rule) provides a more accurate estimation for data within certain numbers of standard deviations from the mean. Therefore, the empirical rule is the preferred method in this context.

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Key Concepts

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

Z-score
A Z-score, also known as a standard score, is a numerical measurement that describes a value's relationship to the mean of a group of values. When you're dealing with a normal distribution, the Z-score indicates how many standard deviations an element is from the mean. This is crucial because it standardizes your data on a uniform scale, making comparisons very convenient. To calculate the Z-score for any data point, you use the formula:\[ Z = \frac{x - \bar{x}}{s} \]where:
  • \(x\) is the raw score or data point,
  • \(\bar{x}\) is the mean of your dataset,
  • \(s\) represents the standard deviation.
For example, in our exercise involving concrete specimens with a mean compressive strength of 3000 and a standard deviation of 500, the Z-score helps determine how much a particular strength deviates from the average. By understanding and calculating Z-scores, we relate percentages of data points in a specific interval, such as determining how many specimens lie between 2500 and 3500 in compressive strength.
Empirical Rule
The Empirical Rule, also known as the 68-95-99.7 rule, is an important guideline in statistics relating to the normal distribution. This rule is incredibly useful due to its simplicity and applicability to data that form a bell-shaped curve. According to the Empirical Rule:
  • Approximately 68% of data lies within one standard deviation of the mean.
  • About 95% of data falls within two standard deviations.
  • Almost 99.7% is within three standard deviations.
This rule is applicable for data following a normal distribution shape, like the exercise with the concrete specimens. Using this rule, we could find that around 68% of observations lie between strengths of 2500 and 3500, corresponding to one standard deviation on either side of the mean. This clear rule of thumb helps in estimating the spread of data and understanding the distribution of measurements in practical, simple terms.
Chebyshev's Rule
Chebyshev's Rule is a statistical theorem that provides bounds for any distribution, not just the normal. This gives it wide applicability, especially when you do not know or cannot assume a normal distribution for your data.Chebyshev’s theorem states that:For any dataset, regardless of distribution, at least \((1 - \frac{1}{k^2}) \times 100\%\) of observations fall within \(k\) standard deviations of the mean. This holds true for any \(k > 1\).For example:
  • With \(k = 2\), at least 75% of data falls within two standard deviations of the mean.
  • With \(k = 3\), at least 88.9% of data lies within three standard deviations.
While this rule gives a conservative estimate and is applicable universally, it lacks precision compared to the Empirical Rule for normal distributions. Hence, if data is normally distributed, like our concrete specimens, the Empirical Rule is preferred due to its more accurate predictions.
Standard Deviation
Standard deviation is a fundamental measure in statistics that reflects the amount of variation or dispersion in a set of data values. It provides insight into how spread out the values in your dataset are in relation to the mean. By definition: Standard deviation is calculated as the square root of the variance. The variance itself is the average of the squared differences from the mean. A lower standard deviation means the data points are close to the mean, whereas a higher standard deviation indicates the data points are spread out over a larger range of values. In our example, with a standard deviation of 500, it explains the average deviation of the concrete specimens' compressive strengths from their mean value of 3000. The ability to compute this is crucial for statistical analysis, offering a better understanding of your data's reliability and variability. Understanding the role of standard deviation aids in interpreting normal distribution curves and applying rules like the Empirical and Z-scores effectively.

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Most popular questions from this chapter

An advertisement for the "30-in. Wonder" that appeared in the September 1983 issue of the journal Packaging claimed that the 30 -in. Wonder weighs cases and bags up to \(110 \mathrm{lb}\) and provides accuracy to within \(0.25 \mathrm{oz}\). Suppose that a 50 -oz weight was repeatedly weighed on this scale and the weight readings recorded. The mean value was \(49.5\) oz, and the standard deviation was \(0.1\). What can be said about the proportion of the time that the scale actually showed a weight that was within \(0.25 \mathrm{oz}\) of the true value of 50 oz? (Hint: Use Chebyshev's Rule.)

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Suppose that the distribution of scores on an exam is closely described by a normal curve with mean 100 . The 16 th percentile of this distribution is 80 . a. What is the 84 th percentile? b. What is the approximate value of the standard deviation of exam scores? c. What \(z\) score is associated with an exam score of 90 ? d. What percentile corresponds to an exam score of \(140 ?\) e. Do you think there were many scores below 40 ? Explain.

Mobile homes are tightly constructed for energy conservation. This can lead to a buildup of indoor pollutants. The paper "A Survey of Nitrogen Dioxide Levels Inside Mobile Homes" (Journal of the Air Pollution Control Association \([1988]: 647-651\) ) discussed various aspects of NO, concentration in these structures. a. In one sample of mobile homes in the Los Angeles area, the mean \(\mathrm{NO}_{2}\) concentration in kitchens during the summer was \(36.92 \mathrm{ppb}\), and the standard deviation was 11.34. Making no assumptions about the shape of the \(\mathrm{NO}_{2}\) distribution, what can be said about the percentage of observations between \(14.24\) and \(59.60 ?\) b. Inside what interval is it guaranteed that at least \(89 \%\) of the concentration observations will lie? c. In a sample of non-Los Angeles mobile homes, the average kitchen \(\mathrm{NO}_{2}\) concentration during the winter was \(24.76 \mathrm{ppb}\), and the standard deviation was \(17.20 .\) Do these values suggest that the histogram of sample observations did not closely resemble a normal curve? (Hint: What is \(\bar{x}-2 s ?\)

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