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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% of the sample observations are between 2500 and 3500. \n b. Approximately 5% of sample observations are outside the interval from 2000 to 4000. \n c. Approximately 34.1% of observations lie between 2000 and 2500. \n d. Chebyshev's rule is not used because the normal distribution provides more accurate percentages as per the empirical rule (68-95-99.7 rule).

Step by step solution

01

Calculation for part a

This involves finding the percentage of sample observations between 2500 and 3500. As the mean is 3000 and standard deviation is 500, this range is exactly one standard deviation below and above the mean. Therefore, using the empirical rule (68-95-99.7%), we conclude that approximately 68% of observations fall within this range.
02

Calculation for part b

Here, we need to find the percentage of sample observations outside the interval 2000 to 4000. Again, this range is two standard deviations below and above the mean. Hence, about 95% of observations fall within this range. So, approximately 100 - 95 = 5% observations are outside this range.
03

Calculation for part c

In this part, finding the approximate percentage of observations between 2000 and 2500 is required. This range falls within the first standard deviation below the mean. Using the characteristics of a normal distribution, we know that about 34.1% of the data falls within the first standard deviation below the mean (half of the 68%).
04

Answer to part d

Chebyshev's theorem is used to estimate the minimum proportion of data within k standard deviations of the mean for any distribution. But for a normal distribution where we know the exact proportions (68-95-99.7 rule), it is better and more accurate to use this than Chebyshev’s rule. Chebyshev’s theorem tends to be a conservative estimate especially for datasets closely following a normal distribution.

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Key Concepts

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

Empirical Rule
The Empirical Rule, often called the 68-95-99.7 rule, is a handy shortcut for understanding the spread of data in a normal distribution. In essence, it tells us how data clusters around the mean. For a data set that follows a normal distribution:
  • Approximately 68% of data points will fall within one standard deviation (\( \pm s \)) from the mean.
  • About 95% will be within two standard deviations (\( \pm 2s \)).
  • Almost 99.7% are within three standard deviations (\( \pm 3s \)).
So, in the concrete specimen exercise, with a mean of 3000 and a standard deviation of 500, the empirical rule helps us quickly determine that about 68% of observations fall between 2500 and 3500.
Standard Deviation
Standard deviation is a measure of how spread out the numbers in a data set are. It is denoted by \( s \) and is a key component when working with the empirical rule. A smaller standard deviation means that the data points are close to the mean, while a larger one indicates they are spread out over a wider range.
In our example of concrete specimens, the standard deviation stands at 500, meaning the compressive strength of most concrete samples deviates 500 units around the mean of 3000. When applying this to the empirical rule:
  • - One standard deviation includes values between 2500 and 3500.
  • - Two standard deviations span from 2000 to 4000.
Understanding standard deviation helps us interpret how typical or atypical certain observations might be in the context of the data.
Histogram Analysis
A histogram is a visual representation of the distribution of numerical data. It uses bars to show the frequency of data intervals or "bins." Analyzing a histogram helps us see patterns, such as skewness, modality, and whether the data follows a particular distribution like the normal distribution.
For the concrete specimens, the histogram closely fits a normal distribution curve. This alignment allows us to confidently apply the empirical rule for calculations regarding the spread of the data.
When analyzing a histogram:
  • Check for symmetry: A normal distribution will have a symmetric histogram.
  • Look for any unusual data points or "outliers" that don't fit the overall pattern.
  • Ensure the intervals are evenly spaced as this affects the accuracy of interpreting patterns.
A well-designed histogram is crucial for accurate data analysis and applying statistical rules.
Chebyshev's Theorem
Chebyshev's Theorem provides a way to estimate the minimum percentage of observations that fall within a certain number of standard deviations from the mean, regardless of the distribution shape. Unlike the empirical rule, Chebyshev's theorem is not restricted to normal distributions, making it more versatile, yet typically more conservative in estimates.
Mathematically, Chebyshev states that at least \(1 - \frac{1}{k^2} \) of the data lies within \( k \) standard deviations of the mean, where \( k > 1 \). For instance, with \( k = 2 \), at least 75% of observations fall within this range in any distribution.
In the original exercise concerning the concrete samples, the dataset follows a normal distribution. This makes the detailed allocations from the empirical rule more precise compared to Chebyshev's broad estimates. Hence, using Chebyshev’s theorem is unnecessary when the data resembles a normal pattern, as it's less efficient and offers wider minimum bounds.

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Most popular questions from this chapter

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An instructor has graded 19 exam papers submitted by students in a class of 20 students, and the average so far is 70 . (The maximum possible score is \(100 .)\) How high would the score on the last paper have to be to raise the class average by 1 point? By 2 points?

An experiment to study the lifetime (in hours) for a certain brand of light bulb involved putting 10 light bulbs into operation and observing them for 1000 hours. Eight of the light bulbs failed during that period, and those lifetimes were recorded. The lifetimes of the two light bulbs still functioning after 1000 hours are recorded as \(1000+\). The resulting sample observations were \(\begin{array}{llllllll}480 & 790 & 1000+ & 350 & 920 & 860 & 570 & 1000+\end{array}\) \(170 \quad 290\) Which of the measures of center discussed in this section can be calculated, and what are the values of those measures?

Research by the Food and Drug Administration (FDA) shows that acrylamide (a possible cancer-causing substance) forms in high-carbohydrate foods cooked at high temperatures and that acrylamide levels can vary widely even within the same brand of food (Associated Press, December 6, 2002). FDA scientists analyzed McDonald's French fries purchased at seven different locations and found the following acrylamide levels: \(\begin{array}{lllllll}497 & 193 & 328 & 155 & 326 & 245 & 270\end{array}\) a. Compute the mean acrylamide level and the seven deviations from the mean. b. Verify that, except for the effect of rounding, the sum of the deviations from mean is equal to 0 for this data set. (If you rounded the sample mean or the deviations, your sum may not be exactly zero, but it should be close to zero if you have computed the deviations correctly.) c. Calculate the variance and standard deviation for this data set.

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